5
INTEGRALS
5.1 Areas and Distances
1. (a) Since f is increasing, we can obtain a lower estimate by using
left endpoints. We are instructed to use five rectangles, so n = 5.
5
L5 =
f (xi−1 ) ∆x
[∆x =
i=1
b−a
n
=
10 − 0
5
= 2]
= f (x0 ) · 2 + f (x1 ) · 2 + f (x2 ) · 2 + f (x3 ) · 2 + f(x4 ) · 2
= 2 [f (0) + f (2) + f (4) + f (6) + f (8)]
≈ 2(1 + 3 + 4.3 + 5.4 + 6.3) = 2(20) = 40
Since f is increasing, we can obtain an upper estimate by using
right endpoints.
5
R5 =
f(xi ) ∆x
i=1
= 2 [f(x1 ) + f (x2 ) + f (x3 ) + f (x4 ) + f(x5 )]
= 2 [f(2) + f (4) + f (6) + f (8) + f (10)]
≈ 2(3 + 4.3 + 5.4 + 6.3 + 7) = 2(26) = 52
Comparing R5 to L5 , we see that we have added the area of the rightmost upper rectangle, f (10) · 2, to the sum and
subtracted the area of the leftmost lower rectangle, f (0) · 2, from the sum.
(b) L10 =
10
f (xi−1 ) ∆x
[∆x =
i=1
10 − 0
10
= 1]
= 1 [f (x0 ) + f (x1 ) + · · · + f (x9 )]
= f (0) + f(1) + · · · + f (9)
≈ 1 + 2.1 + 3 + 3.7 + 4.3 + 4.9 + 5.4 + 5.8 + 6.3 + 6.7
= 43.2
10
R10 =
i=1
f (xi ) ∆x = f (1) + f (2) + · · · + f (10)
= L10 + 1 · f (10) − 1 · f (0)
add rightmost upper rectangle,
subtract leftmost lower rectangle
= 43.2 + 7 − 1 = 49.2
185
186
¤
CHAPTER 5
3. (a) R4 =
INTEGRALS
4
f(xi ) ∆x
∆x =
i=1
π/2 − 0
π
=
4
8
4
f (xi ) ∆x
=
i=1
= [f (x1 ) + f (x2 ) + f(x3 ) + f (x4 )] ∆x
= cos π8 + cos
2π
8
+ cos 3π
+ cos 4π
8
8
π
8
≈ (0.9239 + 0.7071 + 0.3827 + 0) π8 ≈ 0.7908
Since f is decreasing on [0, π/2], an underestimate is obtained by using the right endpoint approximation, R4 .
(b) L4 =
4
4
f (xi−1 ) ∆x =
i=1
f (xi−1 ) ∆x
i=1
= [f(x0 ) + f (x1 ) + f (x2 ) + f (x3 )] ∆x
= cos 0 + cos π8 + cos
2π
8
+ cos 3π
8
π
8
≈ (1 + 0.9239 + 0.7071 + 0.3827) π8 ≈ 1.1835
L4 is an overestimate. Alternatively, we could just add the area of the leftmost upper rectangle and subtract the area of the
rightmost lower rectangle; that is, L4 = R4 + f (0) · π8 − f π2 · π8 .
2 − (−1)
=1 ⇒
3
R3 = 1 · f (0) + 1 · f(1) + 1 · f (2) = 1 · 1 + 1 · 2 + 1 · 5 = 8.
5. (a) f (x) = 1 + x2 and ∆x =
2 − (−1)
= 0.5 ⇒
6
R6 = 0.5[f (−0.5) + f (0) + f (0.5) + f (1) + f (1.5) + f (2)]
∆x =
= 0.5(1.25 + 1 + 1.25 + 2 + 3.25 + 5)
= 0.5(13.75) = 6.875
(b) L3 = 1 · f(−1) + 1 · f(0) + 1 · f (1) = 1 · 2 + 1 · 1 + 1 · 2 = 5
L6 = 0.5[f (−1) + f (−0.5) + f (0) + f (0.5) + f (1) + f(1.5)]
= 0.5(2 + 1.25 + 1 + 1.25 + 2 + 3.25)
= 0.5(10.75) = 5.375
(c) M3 = 1 · f (−0.5) + 1 · f (0.5) + 1 · f (1.5)
= 1 · 1.25 + 1 · 1.25 + 1 · 3.25 = 5.75
M6 = 0.5[f(−0.75) + f (−0.25) + f(0.25)
+ f (0.75) + f (1.25) + f (1.75)]
= 0.5(1.5625 + 1.0625 + 1.0625 + 1.5625 + 2.5625 + 4.0625)
= 0.5(11.875) = 5.9375
(d) M6 appears to be the best estimate.
SECTION 5.1
AREAS AND DISTANCES
¤
187
7. Here is one possible algorithm (ordered sequence of operations) for calculating the sums:
1 Let SUM = 0, X_MIN = 0, X_MAX = 1, N = 10 (depending on which sum we are calculating),
DELTA_X = (X_MAX - X_MIN)/N, and RIGHT_ENDPOINT = X_MIN + DELTA_X.
2 Repeat steps 2a, 2b in sequence until RIGHT_ENDPOINT > X_MAX.
2a Add (RIGHT_ENDPOINT)^4 to SUM.
2b Add DELTA_X to RIGHT_ENDPOINT.
At the end of this procedure, (DELTA_X)·(SUM) is equal to the answer we are looking for. We find that
R10 =
R100 =
1 10
10 i=1
1 100
100 i=1
i
10
4
i
100
≈ 0.2533, R30 =
1 30
30 i=1
i
30
4
≈ 0.2170, R50 =
1
50
50
i=1
4
i
50
≈ 0.2101, and
4
≈ 0.2050. It appears that the exact area is 0.2.
The following display shows the program SUMRIGHT and its output from a TI-83 Plus calculator. To generalize the
program, we have input (rather than assign) values for Xmin, Xmax, and N. Also, the function, x4 , is assigned to Y1 , enabling
us to evaluate any right sum merely by changing Y1 and running the program.
9. In Maple, we have to perform a number of steps before getting a numerical answer. After loading the student package
[command: with(student);] we use the command
left_sum:=leftsum(1/(xˆ2+1),x=0..1,10 [or 30, or 50]); which gives us the expression in summation
notation. To get a numerical approximation to the sum, we use evalf(left_sum);. Mathematica does not have a special
command for these sums, so we must type them in manually. For example, the first left sum is given by
(1/10)*Sum[1/(((i-1)/10)ˆ2+1)],{i,1,10}], and we use the N command on the resulting output to get a
numerical approximation.
In Derive, we use the LEFT_RIEMANN command to get the left sums, but must define the right sums ourselves.
(We can define a new function using LEFT_RIEMANN with k ranging from 1 to n instead of from 0 to n − 1.)
(a) With f (x) =
1 n
1
, 0 ≤ x ≤ 1, the left sums are of the form Ln =
x2 + 1
n i=1
L30 ≈ 0.7937, and L50 ≈ 0.7904. The right sums are of the form Rn =
R30 ≈ 0.7770, and R50 ≈ 0.7804.
1
i−1 2
n
1 n
n i=1
+1
. Specifically, L10 ≈ 0.8100,
1
i 2
n
+1
. Specifically, R10 ≈ 0.7600,
188
¤
CHAPTER 5
INTEGRALS
(b) In Maple, we use the leftbox (with the same arguments as left_sum) and rightbox commands to generate the
graphs.
left endpoints, n = 10
left endpoints, n = 30
left endpoints, n = 50
right endpoints, n = 10
right endpoints, n = 30
right endpoints, n = 50
(c) We know that since y = 1/(x2 + 1) is a decreasing function on (0, 1), all of the left sums are larger than the actual area,
and all of the right sums are smaller than the actual area. Since the left sum with n = 50 is about 0.7904 < 0.791 and the
right sum with n = 50 is about 0.7804 > 0.780, we conclude that 0.780 < R50 < exact area < L50 < 0.791, so the
exact area is between 0.780 and 0.791.
11. Since v is an increasing function, L6 will give us a lower estimate and R6 will give us an upper estimate.
L6 = (0 ft/s)(0.5 s) + (6.2)(0.5) + (10.8)(0.5) + (14.9)(0.5) + (18.1)(0.5) + (19.4)(0.5) = 0.5(69.4) = 34.7 ft
R6 = 0.5(6.2 + 10.8 + 14.9 + 18.1 + 19.4 + 20.2) = 0.5(89.6) = 44.8 ft
13. Lower estimate for oil leakage: R5 = (7.6 + 6.8 + 6.2 + 5.7 + 5.3)(2) = (31.6)(2) = 63.2 L.
Upper estimate for oil leakage: L5 = (8.7 + 7.6 + 6.8 + 6.2 + 5.7)(2) = (35)(2) = 70 L.
15. For a decreasing function, using left endpoints gives us an overestimate and using right endpoints results in an underestimate.
We will use M6 to get an estimate. ∆t = 1, so
M6 = 1[v(0.5) + v(1.5) + v(2.5) + v(3.5) + v(4.5) + v(5.5)] ≈ 55 + 40 + 28 + 18 + 10 + 4 = 155 ft
For a very rough check on the above calculation, we can draw a line from (0, 70) to (6, 0) and calculate the area of the
triangle: 12 (70)(6) = 210. This is clearly an overestimate, so our midpoint estimate of 155 is reasonable.
17. f (x) =
√
4
x, 1 ≤ x ≤ 16.
∆x = (16 − 1)/n = 15/n and xi = 1 + i ∆x = 1 + 15i/n.
n
A = lim Rn = lim
n→∞
n→∞ i=1
19. f (x) = x cos x, 0 ≤ x ≤
n
f(xi ) ∆x = lim
n→∞ i=1
π
.
2
n→∞
n→∞ i=1
1+
∆x = ( π2 − 0)/n =
n
A = lim Rn = lim
4
n
f(xi ) ∆x = lim
n→∞ i=1
15i 15
·
.
n
n
π
/n
2
and xi = 0 + i ∆x =
iπ
iπ
cos
2n
2n
·
π
.
2n
π
i/n.
2
SECTION 5.2
n
21. lim
n→∞ i=1
n
A = lim
n→∞ i=1
with ∆x =
π
4
f (x∗i ) ∆x = lim
n
n→∞ i=1
23. (a) y = f (x) = x5 . ∆x =
tan
π
. Note that this answer is not unique, since the expression for the area is
4n
iπ
4n
n
n→∞
CAS
i5 =
(c) lim
n→∞
n
n→∞ i=1
f (xi ) ∆x = lim
n→∞ i=1
2i
n
5
·
n 32i5
64
2
2
= lim
· = lim
n n→∞ i=1 n5 n n→∞ n6
n
i5 .
i=1
2
2
2
n2 + 2n + 1 2n2 + 2n − 1
64 n (n + 1) 2n + 2n − 1
64
=
lim
·
6
n
12
12 n→∞
n2 · n2
25. y = f (x) = cos x. ∆x =
1
16
2
lim 1 + + 2
3 n→∞
n n
b
b−0
bi
= and xi = 0 + i ∆x = .
n
n
n
n
A = lim Rn = lim
n→∞
π
,
2
, where k is any integer.
n2 (n + 1)2 2n2 + 2n − 1
12
=
If b =
π
4
2
2−0
2i
=
and xi = 0 + i ∆x = .
n
n
n
A = lim Rn = lim
i=1
n→∞ i=1
n
f (xi ) ∆x = lim
n→∞ i=1
cos
bi
n
then A = sin π2 = 1.
2+
1
2
− 2
n
n
=
16
3
·1·2=
⎡
1
⎢ b sin b 2n + 1
b CAS
= lim ⎢
·
n→∞ ⎣
n
b
2n sin
2n
−
32
3
⎤
b ⎥
⎥ CAS
= sin b
2n ⎦
5.2 The Definite Integral
1. f (x) = 3 − 12 x, 2 ≤ x ≤ 14. ∆x =
14 − 2
b−a
=
= 2.
n
6
Since we are using left endpoints, x∗i = xi−1 .
6
L6 =
189
π/4 − 0
π
iπ
=
, xi = 0 + i ∆x =
, and x∗i = xi , the expression for the area is
n
4n
4n
the same for the function y = tan(x − kπ) on the interval kπ, kπ +
n
¤
π
iπ
tan
can be interpreted as the area of the region lying under the graph of y = tan x on the interval 0, π4 ,
4n
4n
since for y = tan x on 0,
(b)
THE DEFINITE INTEGRAL
f (xi−1 ) ∆x
i=1
= (∆x) [f (x0 ) + f (x1 ) + f (x2 ) + f (x3 ) + f (x4 ) + f(x5 )]
= 2[f (2) + f (4) + f (6) + f (8) + f (10) + f (12)]
= 2[2 + 1 + 0 + (−1) + (−2) + (−3)] = 2(−3) = −6
The Riemann sum represents the sum of the areas of the two rectangles above the x-axis minus the sum of the areas of the
three rectangles below the x-axis; that is, the net area of the rectangles with respect to the x-axis.
190
¤
CHAPTER 5
INTEGRALS
5
3. M5 =
f(xi ) ∆x
i=1
[x∗i = xi = 12 (xi−1 + xi ) is a midpoint and ∆x = 1]
= 1 [f(1.5) + f (2.5) + f (3.5)
[f (x) =
+ f(4.5) + f (5.5)]
√
x − 2]
≈ −0.856759
The Riemann sum represents the sum of the areas of the two rectangles
above the x-axis minus the sum of the areas of the three rectangles below
the x-axis.
5. ∆x = (b − a)/n = (8 − 0)/4 = 8/4 = 2.
(a) Using the right endpoints to approximate
8
0
f (x) dx, we have
4
i=1
f (xi ) ∆x = 2[f (2) + f (4) + f (6) + f(8)] ≈ 2[1 + 2 + (−2) + 1] = 4.
(b) Using the left endpoints to approximate
8
0
f (x) dx, we have
4
i=1
f (xi−1 ) ∆x = 2[f (0) + f (2) + f(4) + f (6)] ≈ 2[2 + 1 + 2 + (−2)] = 6.
(c) Using the midpoint of each subinterval to approximate
8
0
f (x) dx, we have
4
i=1
f (xi ) ∆x = 2[f (1) + f (3) + f (5) + f(7)] ≈ 2[3 + 2 + 1 + (−1)] = 10.
7. Since f is increasing, L5 ≤
25
0
f (x) dx ≤ R5 .
Lower estimate = L5 =
5
f (xi−1 ) ∆x = 5[f (0) + f(5) + f (10) + f (15) + f (20)]
i=1
= 5(−42 − 37 − 25 − 6 + 15) = 5(−95) = −475
Upper estimate = R5 =
5
f (xi ) ∆x = 5[f (5) + f(10) + f (15) + f (20) + f (25)]
i=1
= 5(−37 − 25 − 6 + 15 + 36) = 5(−17) = −85
9. ∆x = (10 − 2)/4 = 2, so the endpoints are 2, 4, 6, 8, and 10, and the midpoints are 3, 5, 7, and 9. The Midpoint Rule
gives
10
2
√
x3 + 1 dx ≈
4
f (xi ) ∆x = 2
i=1
√
√
√
√
33 + 1 + 53 + 1 + 73 + 1 + 93 + 1 ≈ 124.1644.
11. ∆x = (1 − 0)/5 = 0.2, so the endpoints are 0, 0.2, 0.4, 0.6, 0.8, and 1, and the midpoints are 0.1, 0.3, 0.5, 0.7, and 0.9.
The Midpoint Rule gives
1
0
sin(x2 ) dx ≈
5
i=1
f(xi ) ∆x = 0.2 sin(0.1)2 + sin(0.3)2 + sin(0.5)2 + sin(0.7)2 + sin(0.9)2 ≈ 0.3084.
13. In Maple, we use the command with(student); to load the sum and box commands, then
m:=middlesum(sin(xˆ2),x=0..1,5); which gives us the sum in summation notation, then M:=evalf(m); which
SECTION 5.2
THE DEFINITE INTEGRAL
¤
gives M5 ≈ 0.30843908, confirming the result of Exercise 11. The command middlebox(sin(xˆ2),x=0..1,5)
generates the graph. Repeating for n = 10 and n = 20 gives M10 ≈ 0.30981629 and M20 ≈ 0.31015563.
π
0
15. We’ll create the table of values to approximate
sin x dx by using the
n
program in the solution to Exercise 5.1.7 with Y1 = sin x, Xmin = 0,
5
1.933766
10
1.983524
50
1.999342
100
1.999836
Xmax = π, and n = 5, 10, 50, and 100.
The values of Rn appear to be approaching 2.
n
17. On [2, 6], lim
n→∞ i=1
n
19. On [1, 8], lim
n→∞ i=1
21. Note that ∆x =
1−
4+
x2i
x2i
6
∆x =
2
8
1
√
2x + x2 dx.
n
n→∞ i=1
−1
6
n→∞ n
= lim
= lim
n→∞
n
f (xi ) ∆x = lim
1 + 3 −1 +
n→∞ i=1
n
n
(−2) +
i=1
i=1
6i
n
18i
6
18
= lim
−2n +
n→∞ n
n
n
6 n
18i
6
= lim
−2 +
n→∞
n
n i=1
n
n
i
i=1
6
108 n(n + 1)
18 n(n + 1)
−2n +
·
= lim −12 + 2 ·
n→∞
n
n
2
n
2
= lim
n→∞
0
1−x
dx.
4 + x2
6
5 − (−1)
6i
= and xi = −1 + i ∆x = −1 + .
n
n
n
(1 + 3x) dx = lim
2
2
2x∗i + (x∗i )2 ∆x =
5
23. Note that ∆x =
Rn
−12 + 54
n+1
1
= lim −12 + 54 1 +
n→∞
n
n
= −12 + 54 · 1 = 42
2
2−0
2i
= and xi = 0 + i ∆x = .
n
n
n
2 − x2 dx = lim
n
n→∞ i=1
n
f (xi ) ∆x = lim
n→∞ i=1
2
4
2n − 2
n→∞ n
n
n
= lim
= lim
n→∞
4−
i=1
i2
2−
= lim
n→∞
4 n + 1 2n + 1
·
·
3
n
n
4i2
n2
4−
= lim
n→∞
2
n
2
n→∞ n
n
= lim
i=1
2−
4
n2
n
i2
i=1
8 n(n + 1)(2n + 1)
·
n3
6
4−
4
1
1+
3
n
2+
1
n
= 4−
4
3
·1·2=
4
3
191
192
¤
CHAPTER 5
1
2−1
= and xi = 1 + i ∆x = 1 + i(1/n) = 1 + i/n.
n
n
25. Note that ∆x =
2
INTEGRALS
x3 dx = lim
n
n→∞ i=1
1
1
n→∞ n4
n
f (xi ) ∆x = lim
n→∞ i=1
n
n→∞
1 n n+i
n→∞ n i=1
n
1
n→∞ n4
i=1
n
n
n3 +
i=1
i=1
1+
1 n + 1 2n + 1
1 (n + 1)2
3 n+1
·
+ ·
·
+ ·
2
n
2
n
n
4
n2
= lim
1+
3
1
1+
2
n
+
1
1
1+
2
n
2+
1
n
+
a(b − a) n
(b − a)2
b−a n
b−a
i = lim
1+
a+
n→∞
n→∞
n i=1
n
n
n2
i=1
6
so
2
i3
i=1
=1+
3
+
2
1
2
·2+
1
4
= 3.75
n
i
i=1
a(b − a)
(b − a)2 n(n + 1)
(b − a)2
n+
·
=
a
(b
−
a)
+
lim
n→∞
n
n2
2
2
= a(b − a) + 12 (b − a)2 = (b − a) a + 12 b − 12 a = (b − a) 12 (b + a) =
29. f (x) =
2
1
1
1+
4
n
x dx = lim
n→∞
n
n
n
n
1
i + 3n
i2 +
i3
n · n3 + 3n2
n4
i=1
i=1
i=1
= lim
= lim
3ni2 +
i=1
3 n(n + 1)(2n + 1)
1 n2 (n + 1)2
3 n(n + 1)
+ 3 ·
+ 4 ·
·
2
n
2
n
6
n
4
n→∞
a
n
3n2 i +
1+
n→∞
27.
3
= lim
= lim
n→∞
b
1
n
n3 + 3n2 i + 3ni2 + i3 = lim
= lim
= lim
3
i
n
1+
1
2
1+
1
n
b2 − a2
4
x
6−2
4i
= . Using Theorem 4, we get x∗i = xi = 2 + i ∆x = 2 + ,
, a = 2, b = 6, and ∆x =
1 + x5
n
n
n
n
x
dx = lim Rn = lim
n→∞
n→∞ i=1
1 + x5
4i
n
4i
1+ 2+
n
2+
5
·
4
.
n
31. ∆x = (π − 0)/n = π/n and x∗i = xi = πi/n.
π
n
sin 5x dx = lim
n→∞ i=1
0
33. (a) Think of
so
(b)
5
0
2
0
2
0
π
n
(sin 5xi )
n
= lim
n→∞ i=1
sin
5πi
n
π CAS
5π
1
cot
= π lim
n→∞ n
n
2n
CAS
= π
=
2
5
f(x) dx as the area of a trapezoid with bases 1 and 3 and height 2. The area of a trapezoid is A = 12 (b + B)h,
f(x) dx = 12 (1 + 3)2 = 4.
f (x) dx =
2
0
f (x) dx +
trapezoid
=
1
2 (1
3
2
f (x) dx +
5
3
rectangle
+ 3)2 +
3·1
f (x) dx
triangle
1
2
+
·2·3
= 4 + 3 + 3 = 10
(c)
7
5
f (x) dx is the negative of the area of the triangle with base 2 and height 3.
(d)
9
7
f (x) dx is the negative of the area of a trapezoid with bases 3 and 2 and height 2, so it equals
7
5
f(x) dx = − 12 · 2 · 3 = −3.
− 12 (B + b)h = − 12 (3 + 2)2 = −5. Thus,
9
0
2
5π
f (x) dx =
5
0
f (x) dx +
7
5
f (x) dx +
9
7
f (x) dx = 10 + (−3) + (−5) = 2.
SECTION 5.2
35.
3
0
1
2x
THE DEFINITE INTEGRAL
¤
− 1 dx can be interpreted as the area of the triangle above the x-axis
minus the area of the triangle below the x-axis; that is,
1
1
2 (1) 2
37.
− 12 (2)(1) =
1
4
0
− 1 = − 34 .
√
9 − x2 dx can be interpreted as the area under the graph of
√
f (x) = 1 + 9 − x2 between x = −3 and x = 0. This is equal to one-quarter
0
−3
1+
the area of the circle with radius 3, plus the area of the rectangle, so
√
0
1 + 9 − x2 dx = 14 π · 32 + 1 · 3 = 3 + 94 π.
−3
39.
2
−1
|x| dx can be interpreted as the sum of the areas of the two shaded
triangles; that is, 12 (1)(1) + 12 (2)(2) =
1
2
+
= 52 .
4
2
0
41.
π
π
43.
1
(5
0
45.
4
(2x2
1
sin2 x cos4 x dx = 0 since the limits of intergration are equal.
− 6x2 ) dx =
1
0
5 dx − 6
− 3x + 1) dx = 2
4
1
1
0
x2 dx = 5(1 − 0) − 6
x2 dx − 3
4
1
x dx +
4
1
1
3
=5−2 =3
1 dx
= 2 · 13 (43 − 13 ) − 3 · 12 (42 − 12 ) + 1(4 − 1) =
47.
49.
2
−2
f (x) dx +
9
[2f (x)
0
5
2
−1
−2
f (x) dx −
+ 3g(x)] dx = 2
9
0
f (x) dx =
5
−2
f (x) dx +
=
5
−1
f (x) dx
f (x) dx + 3
9
0
2
0
= 22.5
[by Property 5 and reversing limits]
f (x) dx
[Property 5]
g(x) dx = 2(37) + 3(16) = 122
51. Using Integral Comparison Property 8, m ≤ f(x) ≤ M
2m ≤
−2
−1
45
2
2
0
⇒ m(2 − 0) ≤
f(x) dx ≤ M(2 − 0) ⇒
f (x) dx ≤ 2M .
√
√
1 + x2 ≤ 2 and
√
√
1 √
1 √
1[1 − (−1)] ≤ −1 1 + x2 dx ≤ 2 [1 − (−1)] [Property 8]; that is, 2 ≤ −1 1 + x2 dx ≤ 2 2.
53. If −1 ≤ x ≤ 1, then 0 ≤ x2 ≤ 1 and 1 ≤ 1 + x2 ≤ 2, so 1 ≤
55. If 1 ≤ x ≤ 4, then 1 ≤
57. If
π
4
≤x≤
π
,
3
√
x ≤ 2, so 1(4 − 1) ≤
then 1 ≤ tan x ≤
√
3, so 1
59. For −1 ≤ x ≤ 1, 0 ≤ x4 ≤ 1 and 1 ≤
or 2 ≤
61.
1
−1
√
√
1 + x4 dx ≤ 2 2.
√
√
x4 + 1 ≥ x4 = x2 , so
3
1
π
3
−
4
1
π
4
√
x dx ≤ 2(4 − 1); that is, 3 ≤
≤
π/3
π/4
tan x dx ≤
√
3
√
√
1 + x4 ≤ 2, so 1[1 − (−1)] ≤
√
x4 + 1 dx ≥
3
1
x2 dx =
1
3
33 − 13 =
π
3
1
−1
26
.
3
−
4
1
π
4
√
x dx ≤ 6.
or
π
12
≤
π/3
π/4
tan x dx ≤
√
√
1 + x4 dx ≤ 2 [1 − (−1)]
π
12
√
3.
193
¤
194
CHAPTER 5
INTEGRALS
63. Using right endpoints as in the proof of Property 2, we calculate
b
a
n
cf (x) dx = lim
n→∞ i=1
n
n
cf (xi ) ∆x = lim c
n→∞
f (xi ) ∆x = c lim
f (xi ) ∆x = c
n→∞ i=1
i=1
b
a
f (x) dx.
65. Since − |f (x)| ≤ f (x) ≤ |f (x)|, it follows from Property 7 that
−
b
a
|f (x)| dx ≤
b
a
b
a
f (x) dx ≤
b
a
|f(x)| dx ⇒
f (x) dx ≤
b
a
|f (x)| dx
Note that the definite integral is a real number, and so the following property applies: −a ≤ b ≤ a ⇒ |b| ≤ a for all real
numbers b and nonnegative numbers a.
67. To show that f is integrable on [0, 1] , we must show that lim
n
n→∞ i=1
the interval [0, 1] into n equal subintervals 0,
subinterval, then we obtain the Riemann sum
1
1 2
n−1
,
, 1 . If we choose x∗i to be a rational number in the ith
,
, ... ,
n
n n
n
n
i=1
f (x∗i ) ·
choose x∗i to be an irrational number. Then we get
n
i=1
n
lim
n→∞ i=1
f (x∗i ) ·
f (x∗i ) ∆x exists. Let n denote a positive integer and divide
n
1
1
= 0, so lim
f (x∗i ) · = lim 0 = 0. Now suppose we
n→∞
n
n n→∞
i=1
f (x∗i ) ·
n
1
=
n
i=1
1·
1
1
= n · = 1 for each n, so
n
n
n
1
= lim 1 = 1. Since the value of lim
f (x∗i ) ∆x depends on the choice of the sample points x∗i , the
n→∞ i=1
n n→∞
limit does not exist, and f is not integrable on [0, 1].
69. lim
n
n→∞ i=1
n
lim
n→∞ i=1
is
1
0
n i4
n
i4
i
1
=
lim
· = lim
n→∞ i=1 n4
n5
n n→∞ i=1 n
4
1
. At this point, we need to recognize the limit as being of the form
n
f (xi ) ∆x, where ∆x = (1 − 0)/n = 1/n, xi = 0 + i ∆x = i/n, and f (x) = x4 . Thus, the definite integral
x4 dx.
71. Choose xi = 1 +
2
1
i
√
and x∗i = xi−1 xi =
n
x−2 dx = lim
n→∞
1 n
n i=1 1 +
n
= lim n
n→∞
= lim n
n→∞
= lim n
n→∞
i=1
1+
i−1
n
i
. Then
n
n
1
i−1
n
1+
1+
i
n
= lim n
n→∞
1
1
−
n+i−1
n+i
i=1
1
(n + i − 1)(n + i)
[by the hint] = lim n
n→∞
n−1
i=0
n
1
1
−
n + i i=1 n + i
1
1
1
1
1
1
+
+··· +
−
+ ··· +
+
n
n+1
2n − 1
n+1
2n − 1
2n
1
1
−
n
2n
= lim 1 −
n→∞
1
2
=
1
2
5.3 The Fundamental Theorem of Calculus
1. One process undoes what the other one does. The precise version of this statement is given by the Fundamental Theorem of
Calculus. See the statement of this theorem and the paragraph that follows it on page 320.
SECTION 5.3
3. (a) g(x) =
x
0
f (t) dt.
g(0) =
0
0
f (t) dt = 0
g(1) =
1
0
f (t) dt = 1 · 2 = 2
g(2) =
2
0
g(3) =
f (t) dt =
1
2
1
0
g(6) = g(3) +
=7+ −
[rectangle],
2
1
f (t) dt +
3
2
f (t) dt = g(1) +
2
1
f (t) dt
[rectangle plus triangle],
·1·2=5
f (t) dt = g(2) +
6
3
1
2
195
(d)
=2+1·2+
3
0
¤
THE FUNDAMENTAL THEOREM OF CALCULUS
f (t) dt = 5 +
1
2
· 1 · 4 = 7,
f (t) dt [the integral is negative since f lies under the x-axis]
·2·2+1·2
=7−4 =3
(b) g is increasing on (0, 3) because as x increases from 0 to 3, we keep adding more area.
(c) g has a maximum value when we start subtracting area; that is, at x = 3.
(a) By FTC1 with f (t) = t2 and a = 1, g(x) =
5.
x 2
t
1
dt ⇒
g 0 (x) = f (x) = x2 .
(b) Using FTC2, g(x) =
7. f (t) =
t3
1
and g(x) =
+1
x
1
x 2
t
1
1 3 x
t 1
3
dt =
= 13 x3 −
1
3
⇒ g 0 (x) = x2 .
1
1
dt, so by FTC1, g0 (x) = f (x) = 3
. Note that the lower limit, 1, could be any
t3 + 1
x +1
real number greater than −1 and not affect this answer.
9. f (t) = t2 sin t and g(y) =
π
11. F (x) =
x
13. Let u =
h0 (x) =
d
dx
2
19.
−1
d
dx
x
√
d
1 + sec t dt ⇒ F 0 (x) = −
dx
π
x
π
√
√
1 + sec t dt = − 1 + sec x
du
1
dh du
1
dh
. Then
= − 2 . Also,
=
, so
x
dx
x
dx
du dx
d
dx
1/x
sin4 t dt =
2
tan x
t+
0
1
1−3x
d
du
2
√
d
t dt =
du
sin4 t dt ·
− sin4 (1/x)
du
du
= sin4 u
=
.
dx
dx
x2
u
t+
0
√
du
=
t dt ·
dx
u+
√ du
=
u
dx
tan x +
√
tan x sec2 x.
dy
dy dw
dw
= −3. Also,
=
, so
dx
dx
dw dx
u3
d
du =
1 + u2
dw
x3 − 2x dx =
u
dy du
du
dy
= sec2 x. Also,
=
, so
dx
dx
du dx
17. Let w = 1 − 3x. Then
y0 =
sin t dt, so by FTC1, g0 (y) = f (y) = y 2 sin y.
√
1 + sec t dt = −
15. Let u = tan x. Then
y0 =
y 2
t
2
x4
− x2
4
1
w
2
=
−1
u3
d
dw
=−
du ·
1 + u2
dx
dw
24
− 22
4
−
w
1
u3
w3
dw
3(1 − 3x)3
=−
du ·
(−3) =
2
2
1+u
dx
1+w
1 + (1 − 3x)2
(−1)4
− (−1)2
4
= (4 − 4) −
1
4
− 1 = 0 − − 34 =
3
4
196
¤
CHAPTER 5
21.
4
(5
1
23.
1
0
2
1
27.
2
0
3
dt = 3
t4
5 9/5
x
9
2
9
1
1
2
(2x
0
9
1
=
2
3
sec2 t dt = tan t
33.
2
(1
1
+ 2y)2 dy =
cos x
f(x) dx =
1
9
dx =
1
2
3
= −1
2
0
1
−1
8
128
7
= 4+
=
7
8
− (0 + 0) =
(x1/2 − x−1/2 ) dx =
− 2 = 12 − − 43 =
2 3/2
x
3
156
7
− 2x1/2
9
1
40
3
= tan π4 − tan 0 = 1 − 0 = 1
+ 4y + 4y 2 ) dy = y + 2y 2 + 43 y 3
if 0 ≤ x < π/2
if π/2 ≤ x ≤ π
π/2
0
1
+ x6 ) dx = x2 + 17 x7
π/4
0
2
(1
1
2
1
3
−3 t3
=
· 27 − 2 · 3 −
π/4
0
sin x
2
x
1
√ −√
x
x
31.
π
0
t−3
−3
= (20 − 16 + 64) − (5 − 1 + 1) = 68 − 5 = 63
5
9
−0=
1
x−1
√ dx =
x
35. If f (x) =
5
9
=
0
t−4 dt = 3
x(2 + x5 ) dx =
29.
4
1
− 2t + 3t2 ) dt = 5t − t2 + t3
x4/5 dx =
25.
INTEGRALS
sin x dx +
2
1
= 2+8+
32
3
− 1+2+
4
3
π
π/2
= − cos π2 + cos 0 + sin π − sin π2
=
62
3
−
13
3
=
49
3
then
π
π/2
cos x dx = − cos x
π/2
0
+ sin x
= −0 + 1 + 0 − 1 = 0
Note that f is integrable by Theorem 3 in Section 5.2.
37. f (x) = x−4 is not continuous on the interval [−2, 1], so FTC2 cannot be applied. In fact, f has an infinite discontinuity at
x = 0, so
1
−2
x−4 dx does not exist.
39. f (θ) = sec θ tan θ is not continuous on the interval [π/3, π], so FTC2 cannot be applied. In fact, f has an infinite
discontinuity at x = π/2, so
π
π/3
sec θ tan θ dθ does not exist.
41. From the graph, it appears that the area is about 60. The actual area is
27
0
x1/3 dx =
3 4/3
4x
27
=
0
3
4
· 81 − 0 =
243
4
= 60.75. This is
3
4
of the
area of the viewing rectangle.
43. It appears that the area under the graph is about
2
3
of the area of the viewing
rectangle, or about 23 π ≈ 2.1. The actual area is
π
0
sin x dx = [− cos x]π0 = (− cos π) − (− cos 0) = − (−1) + 1 = 2.
SECTION 5.3
45.
2
−1
x3 dx =
3x
47. g(x) =
2x
1 4 2
4 x −1
=4−
u2 − 1
du =
u2 + 1
1
4
=
0
u2 − 1
du +
u2 + 1
2x
15
4
2
g 0 (x) = −
49. y =
x3
√
x
¤
197
= 3.75
3x
0
u2 − 1
du = −
u2 + 1
2
2x
0
u2 − 1
du +
u2 + 1
2
3x
0
u2 − 1
du ⇒
u2 + 1
2
(3x) − 1 d
4x − 1
9x − 1
(2x) − 1 d
·
(2x) +
·
(3x) = −2 · 2
+3· 2
(2x)2 + 1 dx
(3x)2 + 1 dx
4x + 1
9x + 1
√
t sin t dt =
1
√
x
√
t sin t dt +
x3
1
√
t sin t dt = −
√
√
d √
d
( x ) + x3/2 sin(x3 ) ·
y 0 = − 4 x (sin x ) ·
x3
dx
dx
√
sin x
= 3x7/2 sin(x3 ) − √
2 4x
x2
x
f (t) dt ⇒ F 0 (x) = f (x) =
51. F (x) =
1
00
THE FUNDAMENTAL THEOREM OF CALCULUS
0
F (x) = f (x) =
1 + (x2 )4 d 2
x =
·
x2
dx
53. (a) The Fresnel function S(x) =
x
0
sin
1
√
x3 √
t sin t dt + 1
t sin t dt ⇒
√
√
4
x sin x
√
=−
+ x3/2 sin(x3 )(3x2 )
2 x
√
x
1
√
1 + u4
du
u
t2
since f (t) =
1
√
1 + u4
du
u
⇒
√
√
√
√
2 1 + x8
1 + x8
. So F 00 (2) = 1 + 28 = 257.
· 2x =
x2
x
π 2
t
2
dt has local maximum values where 0 = S 0 (x) = sin
π 2
t
2
and
S 0 changes from positive to negative. For x > 0, this happens when π2 x2 = (2n − 1)π [odd multiples of π] ⇔
√
x2 = 2(2n − 1) ⇔ x = 4n − 2, n any positive integer. For x < 0, S 0 changes from positive to negative where
√
π 2
x = 2nπ [even multiples of π] ⇔ x2 = 4n ⇔ x = −2 n. S 0 does not change sign at x = 0.
2
(b) S is concave upward on those intervals where S 00 (x) > 0. Differentiating our expression for S 0 (x), we get
S 00 (x) = cos
2n −
1
2
π<
π 2
x
2
π 2
x
2
. For x > 0, S 00 (x) > 0 where cos( π2 x2 ) > 0 ⇔ 0 < π2 x2 < π2 or
√
√
π, n any integer ⇔ 0 < x < 1 or 4n − 1 < x < 4n + 1, n any positive integer.
2 π2 x = πx cos
< 2n +
1
2
π 2
x
2
For x < 0, S 00 (x) > 0 where cos( π2 x2 ) < 0 ⇔ 2n − 32 π < π2 x2 < 2n − 12 π, n any integer ⇔
√
√
√
√
4n − 3 < |x| < 4n − 1 ⇒
4n − 3 < −x < 4n − 1 ⇒
4n − 3 < x2 < 4n − 1 ⇔
√
√
√
√
− 4n − 3 > x > − 4n − 1, so the intervals of upward concavity for x < 0 are − 4n − 1, − 4n − 3 , n any
√ √
√
√
√
3, 5 , − 7, − 5 ,
positive integer. To summarize: S is concave upward on the intervals (0, 1), − 3, −1 ,
√
7, 3 , . . . .
(c) In Maple, we use plot({int(sin(Pi*tˆ2/2),t=0..x),0.2},x=0..2);. Note that
Maple recognizes the Fresnel function, calling it FresnelS(x). In Mathematica, we use
Plot[{Integrate[Sin[Pi*tˆ2/2],{t,0,x}],0.2},{x,0,2}]. In Derive, we load the utility file
198
¤
CHAPTER 5
INTEGRALS
FRESNEL and plot FRESNEL_SIN(x). From the graphs, we see that
x
0
sin
π 2
2t
dt = 0.2 at x ≈ 0.74.
55. (a) By FTC1, g0 (x) = f (x). So g0 (x) = f (x) = 0 at x = 1, 3, 5, 7, and 9. g has local maxima at x = 1 and 5 (since f = g0
changes from positive to negative there) and local minima at x = 3 and 7. There is no local maximum or minimum at
x = 9, since f is not defined for x > 9.
1
0
(b) We can see from the graph that
g(5) =
5
0
3
1
f dt = g(1) −
3
1
f dt <
f dt +
5
3
f dt <
5
3
f dt <
f dt , and g(9) =
9
0
7
5
9
7
f dt <
f dt = g(5) −
7
5
f dt . So g(1) =
f dt +
9
7
1
0
f dt ,
f dt . Thus,
g(1) < g(5) < g(9), and so the absolute maximum of g(x) occurs at x = 9.
(c) g is concave downward on those intervals where g 00 < 0. But
(d)
g (x) = f (x), so g (x) = f (x), which is negative on
0
00
(approximately)
0
, (4, 6) and (8, 9). So g is concave downward
1
,2
2
on these intervals.
57. lim
n
n→∞ i=1
i3
1−0
= lim
n→∞
n4
n
n
i=1
i
n
3
1
=
x3 dx =
0
x4
4
1
1
4
=
0
59. Suppose h < 0. Since f is continuous on [x + h, x], the Extreme Value Theorem says that there are numbers u and v in
[x + h, x] such that f (u) = m and f (v) = M , where m and M are the absolute minimum and maximum values of f on
[x + h, x]. By Property 8 of integrals, m(−h) ≤
x
x+h
f(t) dt ≤ M(−h); that is, f(u)(−h) ≤ −
Since −h > 0, we can divide this inequality by −h: f (u) ≤
g(x + h) − g(x)
1
=
h
h
1
h
f (t) dt ≤ f (v)(−h).
x+h
x
x+h
x
x+h
x
f (t) dt for h 6= 0, and hence f (u) ≤
f (t) dt ≤ f (v). By Equation 2,
g(x + h) − g(x)
≤ f (v), which is Equation 3 in the
h
case where h < 0.
√
√
x ⇒ f 0 (x) = 1/(2 x ) > 0 for x > 0 ⇒ f is increasing on (0, ∞). If x ≥ 0, then x3 ≥ 0, so
√
1 + x3 ≥ 1 for x ≥ 0. Next let
1 + x3 ≥ 1 and since f is increasing, this means that f 1 + x3 ≥ f (1) ⇒
61. (a) Let f (x) =
g(t) = t2 − t ⇒ g 0 (t) = 2t − 1 ⇒ g 0 (t) > 0 when t ≥ 1. Thus, g is increasing on (1, ∞). And since g(1) = 0,
√
√
g(t) ≥ 0 when t ≥ 1. Now let t = 1 + x3 , where x ≥ 0. 1 + x3 ≥ 1 (from above) ⇒ t ≥ 1 ⇒ g(t) ≥ 0 ⇒
√
√
1 + x3 − 1 + x3 ≥ 0 for x ≥ 0. Therefore, 1 ≤ 1 + x3 ≤ 1 + x3 for x ≥ 0.
1
1√
1
(b) From part (a) and Property 7: 0 1 dx ≤ 0 1 + x3 dx ≤ 0 (1 + x3 ) dx ⇔
1
1
1√
1√
x 0 ≤ 0 1 + x3 dx ≤ x + 14 x4 0 ⇔ 1 ≤ 0 1 + x3 dx ≤ 1 + 14 = 1.25.
SECTION 5.4
63. 0 <
x4
5
199
x2
x2
1
< 4 = 2 on [5, 10], so
2
+x +1
x
x
10
0≤
¤
INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM
x4
10
x2
dx <
+ x2 + 1
5
10
1
1
dx = −
x2
x
5
x
65. Using FTC1, we differentiate both sides of 6 +
a
=−
1
1
− −
10
5
=
1
= 0.1.
10
√
f (t)
f (x)
1
dt = 2 x to get 2 = 2 √
2
t
x
2 x
a
To find a, we substitute x = a in the original equation to obtain 6 +
a
⇒ f(x) = x3/2 .
√
√
f(t)
dt = 2 a ⇒ 6 + 0 = 2 a ⇒
2
t
√
3 = a ⇒ a = 9.
t
0
67. (a) Let F (t) =
f (s) ds. Then, by FTC1, F 0 (t) = f(t) = rate of depreciation, so F (t) represents the loss in value over the
interval [0, t].
(b) C(t) =
1
A+
t
t
f (s) ds =
0
A + F (t)
represents the average expenditure per unit of t during the interval [0, t],
t
assuming that there has been only one overhaul during that time period. The company wants to minimize average
expenditure.
(c) C(t) =
1
A+
t
t
0
f (s) ds . Using FTC1, we have C 0 (t) = −
t
C 0 (t) = 0 ⇒ t f (t) = A +
9
69.
1
1
1
dx =
2x
2
√
3/2
71.
1/2
73.
1
−1
9
1
1
dx =
x
6
√
dt = 6
1 − t2
eu+1 du = eu+1
√
9
1
2
ln |x|
3/2
1/2
1
−1
0
f(s) ds ⇒ f(t) =
1
= 12 (ln 9 − ln 1) =
1
√
dt = 6 sin−1 t
1 − t2
√
1
A+
t
1
2
3/2
1/2
t
1
A+
t2
0
1
f (s) ds + f (t).
t
t
f (s) ds = C(t).
0
ln 9 − 0 = ln 91/2 = ln 3
= 6 sin−1
√
3
2
− sin−1
1
2
=6
= e2 − e0 = e2 − 1 [or start with eu+1 = eu e1 ]
5.4 Indefinite Integrals and the Net Change Theorem
1.
d √ 2
d
x +1+C =
dx
dx
3.
d
sin x −
dx
1
3
x2 + 1
sin3 x + C =
1/2
+C =
1
2
x2 + 1
−1/2
x
· 2x + 0 = √
2
x +1
d
sin x − 13 (sin x)3 + C = cos x −
dx
1
3
· 3(sin x)2 (cos x) + 0
= cos x(1 − sin2 x) = cos x(cos2 x) = cos3 x
5.
7.
(x2 + x−2 ) dx =
x−1
1
x3
1
+
+ C = x3 − + C
3
−1
3
x
x4 − 12 x3 + 14 x − 2 dx =
1 x4
1 x2
x5
−
+
− 2x + C = 15 x5 − 18 x4 + 18 x2 − 2x + C
5
2 4
4 2
π
3
−
π
6
=6
π
6
=π
200
¤
CHAPTER 5
INTEGRALS
(1 − t)(2 + t2 ) dt =
9.
x3 − 2
x
11.
√
x
t3
t4
t2
+
−
+ C = 2t − t2 + 13 t3 − 14 t4 + C
2
3
4
(2 − 2t + t2 − t3 ) dt = 2t − 2
x3
2x1/2
−
x
x
dx =
13.
(θ − csc θ cot θ) dθ = 12 θ2 + csc θ + C
15.
(1 + tan2 α) dα =
√
x1/2
x3
−2
+ C = 13 x3 − 4 x + C
3
1/2
(x2 − 2x−1/2 ) dx =
dx =
sec2 α dα = tan α + C
cos x + 12 x dx = sin x + 14 x2 + C. The members of the family
17.
in the figure correspond to C = −5, 0, 5, and 10.
19.
2
(6x2
0
21.
0
(5y 4
−3
23.
2
(3u
−2
− 4x + 5) dx = 6 · 13 x3 − 4 · 12 x2 + 5x
− 6y 2 + 14) dy = 5
+ 1)2 du =
2
−2
1 5
y
5
−6
1 3
y
3
2
0
= 2x3 − 2x2 + 5x
+ 14y
0
−3
2
0
= (16 − 8 + 10) − 0 = 18
= y 5 − 2y 3 + 14y
9u2 + 6u + 1 du = 9 · 13 u3 + 6 · 12 u2 + u
2
−2
0
−3
= 0 − (−243 + 54 − 42) = 231
= 3u3 + 3u2 + u
2
−2
= (24 + 12 + 2) − (−24 + 12 − 2) = 38 − (−14) = 52
25.
4
1
√
t (1 + t) dt =
−1
27.
4y 3 +
−2
29.
31.
33.
1
0
4
1
5/x dx =
π
(4 sin θ
0
π/4
35.
0
2
y3
√
√
3
4
x+ x
x
4 1/2
(t
1
+ t3/2 ) dt =
1
(x4/3
0
√
5
4
1
64
x−1/2 dx =
1
√
√
5 2 x
−2
=
π/4
0
64
π
0
cos2 θ
1
+
cos2 θ
cos2 θ
1
x1/2
+
π/4
0
= 2x1/2 + 65 x5/6
1
y2
+ 49 x9/4
+
−1
−2
64
5
+
2
5
π/4
dθ =
14
3
=
3
7
+
4
9
−0 =
1
4
+
62
5
=
256
15
= − 63
4
55
63
(sec2 θ + 1) dθ
0
π
4
64
− (0 + 0) = 1 +
π
4
192
5
64
x−1/2 + x(1/3) − (1/2) dx =
1
= 16 +
=
= (1 − 1) − 16 −
1
0
2
3
−
= (4 − 0) − (−4 − 0) = 8
dx =
64
1
16
3
=
1
√
√
5 (2 · 2 − 2 · 1) = 2 5
= tan π4 +
x1/3
x1/2
4
= y4 −
3 7/3
7x
1
1 + cos2 θ
dθ =
cos2 θ
1
−1
4
− 3 cos θ) dθ = − 4 cos θ − 3 sin θ
√
1+ 3x
√
dx =
x
+ 25 t5/2
+ x5/4 ) dx =
= tan θ + θ
37.
1 −2
y
−2
dy = 4 · 14 y 4 + 2 ·
dx =
2 3/2
t
3
1
− 2+
6
5
= 14 +
186
5
=
256
5
(x−1/2 + x−1/6 ) dx
SECTION 5.4
39.
41.
√
√
4
5
x5 + x4 dx =
1
0
2
−1
1
(x5/4
0
0
[x
−1
(x − 2 |x|) dx =
1
2
=3 0−
x9/5
x9/4
+
9/4
9/5
+ x4/5 ) dx =
− 2(−x)] dx +
INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM
2
[x
0
1
=
+ 59 x9/5
0
0
−1
− 2(x)] dx =
4 9/4
9x
1
=
0
2
(−x) dx
0
3x dx +
4
9
=3
+
5
9
¤
201
−0 =1
1 2 0
x −1
2
1 2 2
x 0
2
−
− (2 − 0) = − 72 = −3.5
43. The graph shows that y = x + x2 − x4 has x-intercepts at x = 0 and at
x = a ≈ 1.32. So the area of the region that lies under the curve and above the
x-axis is
a
(x
0
45. A =
a
0
+ x2 − x4 ) dx =
1 2
x
2
+ 13 x3 − 15 x5
=
1 2
a
2
+ 13 a3 − 15 a5 − 0 ≈ 0.84
2
0
2y − y 2 dy = y 2 − 13 y 3
2
0
= 4−
8
3
−0=
4
3
47. If w0 (t) is the rate of change of weight in pounds per year, then w(t) represents the weight in pounds of the child at age t. We
10
5
know from the Net Change Theorem that
w0 (t) dt = w(10) − w(5), so the integral represents the increase in the child’s
weight (in pounds) between the ages of 5 and 10.
49. Since r(t) is the rate at which oil leaks, we can write r(t) = −V 0 (t), where V (t) is the volume of oil at time t. [Note that the
minus sign is needed because V is decreasing, so V 0 (t) is negative, but r(t) is positive.] Thus, by the Net Change Theorem,
120
0
r(t) dt = −
120
0
V 0 (t) dt = − [V (120) − V (0)] = V (0) − V (120), which is the number of gallons of oil that leaked
from the tank in the first two hours (120 minutes).
51. By the Net Change Theorem,
5000
1000
R0 (x) dx = R(5000) − R(1000), so it represents the increase in revenue when
production is increased from 1000 units to 5000 units.
b
a
53. In general, the unit of measurement for
f (x) dx is the product of the unit for f (x) and the unit for x. Since f (x) is
measured in newtons and x is measured in meters, the units for
100
0
f (x) dx are newton-meters. (A newton-meter is
abbreviated N·m and is called a joule.)
55. (a) Displacement =
3
(3t
0
3
0
(b) Distance traveled =
− 5) dt =
|3t − 5| dt =
= 5t −
57. (a) v 0 (t) = a(t) = t + 4
3 2
t
2
3 2 5/3
t 0
2
+
− 5t
5/3
(5
0
3 2
t
2
3
0
=
10
0
=
500
3
59. Since m0 (x) = ρ(x), m =
|v(t)| dt =
10
0
− 5t
1 2
t
2
− 15 = − 32 m
− 3t) dt +
3
5/3
⇒ v(t) = 12 t2 + 4t + C
(b) Distance traveled =
27
2
=
25
3
3
(3t
5/3
−
3
2
·
− 5) dt
25
9
+
27
2
− 15 −
3
2
·
25
9
−
25
3
=
41
6
m
⇒ v(0) = C = 5 ⇒ v(t) = 12 t2 + 4t + 5 m/s
+ 4t + 5 dt =
10 1 2
t
2
0
+ 4t + 5 dt =
1 3
t
6
+ 2t2 + 5t
10
0
+ 200 + 50 = 416 23 m
4
0
ρ(x) dx =
4
0
9+2
√
x dx = 9x + 43 x3/2
4
= 36 +
0
32
3
−0 =
140
3
= 46 23 kg.
202
¤
CHAPTER 5
INTEGRALS
100
0
61. Let s be the position of the car. We know from Equation 2 that s(100) − s(0) =
v(t) dt. We use the Midpoint Rule for
0 ≤ t ≤ 100 with n = 5. Note that the length of each of the five time intervals is 20 seconds =
20
3600
hour =
1
180
hour.
So the distance traveled is
100
0
v(t) dt ≈
1
180 [v(10)
1
180 (38
+ v(30) + v(50) + v(70) + v(90)] =
247
180
+ 58 + 51 + 53 + 47) =
≈ 1.4 miles.
63. From the Net Change Theorem, the increase in cost if the production level is raised
4000
2000
from 2000 yards to 4000 yards is C(4000) − C(2000) =
4000
2000
4000
2000
C 0 (x) dx =
C 0 (x) dx.
4000
2000
3 − 0.01x + 0.000006x2 dx = 3x − 0.005x2 + 0.000002x3
= 60,000 − 2,000 = $58,000
65. (a) We can find the area between the Lorenz curve and the line y = x by subtracting the area under y = L(x) from the area
under y = x. Thus,
coefficient of inequality =
=
(b) L(x) =
5 2
x
12
+
7
x
12
1
0
area between Lorenz curve and line y = x
=
area under line y = x
1
0
[x − L(x)] dx
[x2/2]10
⇒ L(50%) = L
1
2
=
5
48
+
1
0
=
7
24
=
[x − L(x)] dx
1
0
[x − L(x)] dx
=2
1/2
1
0
x dx
[x − L(x)] dx
19
48
= 0.39583, so the bottom 50% of the households receive
1
0
x−
at most about 40% of the income. Using the result in part (a),
1
0
coefficient of inequality = 2
1 5
(x
0 12
=2
67.
[x − L(x)] dx = 2
− x2 ) dx =
1 2
2x
5
6
5 2
x
12
− 13 x3
−
1
0
7
x
12
=
5
6
dx = 2
1
2
−
1
3
1
0
=
5
6
5
x
12
−
5 2
x
12
1
6
=
5
36
dx
(sin x + sinh x) dx = − cos x + cosh x + C
x2 + 1 +
69.
√
1/ 3
71.
0
x2
1
+1
1/
t2 − 1
dt =
t4 − 1
=
dx =
√
3
0
π
6
x3
+ x + tan−1 x + C
3
t2 − 1
dt =
(t2 + 1)(t2 − 1)
−0 =
1/
√
0
3
1
dt = arctan t
t2 + 1
1/
0
√
3
√
= arctan 1/ 3 − arctan 0
π
6
5.5 The Substitution Rule
1. Let u = 3x. Then du = 3 dx, so dx =
cos 3x dx =
cos u
1
3
du =
1
3
1
3
du. Thus,
cos u du =
1
3
sin u + C =
1
3
sin 3x + C. Don’t forget that it is often very easy to check
an indefinite integration by differentiating your answer. In this case,
d
dx
1
3
sin 3x + C = 13 (cos 3x) · 3 = cos 3x, the
desired result.
3. Let u = x3 + 1. Then du = 3x2 dx and x2 dx =
x2
x3 + 1 dx =
√
u
1
3
du =
1
3
du, so
1 2
1 u3/2
+ C = · u3/2 + C = 29 (x3 + 1)3/2 + C.
3 3/2
3 3
SECTION 5.5
THE SUBSTITUTION RULE
¤
203
5. Let u = cos θ. Then du = − sin θ dθ and sin θ dθ = −du, so
cos3 θ sin θ dθ =
u3 (−du) = −
u4
+ C = − 14 cos4 θ + C.
4
7. Let u = x2 . Then du = 2x dx and x dx =
1
2
9. Let u = 3x − 2. Then du = 3 dx and dx =
du, so
1
3
x sin(x2 ) dx =
sin u
du, so (3x − 2)20 dx =
u20
11. Let u = 2x + x2 . Then du = (2 + 2x) dx = 2(1 + x) dx and (x + 1) dx =
(x + 1)
√
u
2x + x2 dx =
1
2
du =
1 u3/2
+C =
2 3/2
2x + x2
1
3
3/2
1
2
du = − 12 cos u + C = − 12 cos(x2 ) + C.
1
3
du =
1
2
du, so
1
3
1 21
· 21
u +C =
1
(3x − 2)21
63
+ C.
+ C.
√
Or: Let u = 2x + x2 . Then u2 = 2x + x2 ⇒ 2u du = (2 + 2x) dx ⇒ u du = (1 + x) dx, so
√
(x + 1) 2x + x2 dx = u · u du = u2 du = 13 u3 + C = 13 (2x + x2 )3/2 + C.
13. Let u = πt. Then du = π dt and dt =
1
π
du, so
sin πt dt =
sin u
1
π
du =
1
(− cos u)
π
+ C = − π1 cos πt + C.
15. Let u = 3ax + bx3 . Then du = (3a + 3bx2 ) dx = 3(a + bx2 ) dx, so
a + bx2
√
dx =
3ax + bx3
1
3
du
1
=
3
u1/2
u−1/2 du =
1
3
√
dt
1
17. Let u = t. Then du = √ and √ dt = 2 du, so
2 t
t
19. Let u = sin θ. Then du = cos θ dθ, so
u−1/3
1
3
du =
1
3
1
3
2
3
3ax + bx3 + C.
√
cos t
√ dt =
t
cos θ sin6 θ dθ =
21. Let u = 1 + z 3 . Then du = 3z 2 dz and z 2 dz =
z2
√
dz =
3
1 + z3
· 2u2 + C =
cos u (2 du) = 2 sin u + C = 2 sin
u6 du = 17 u7 + C =
1
7
√
t + Ċ.
sin7 θ + C.
du, so
· 32 u2/3 + C = 12 (1 + z 3 )2/3 + C.
23. Let u = cot x. Then du = − csc2 x dx and csc2 x dx = −du, so
√
cot x csc2 x dx =
√
u3/2
+ C = − 23 (cot x)3/2 + C.
u (−du) = −
3/2
25. Let u = sec x. Then du = sec x tan x dx, so
sec3 x tan x dx =
sec2 x (sec x tan x) dx =
27. Let u = sin x. Then du = cos x dx, so
u2 du = 13 u3 + C =
cos x
dx =
sin2 x
1
du =
u2
1
3
sec3 x + C.
u−2 du =
1
1
u−1
+C =− +C =−
+C
−1
u
sin x
[or −csc x + C ].
29. Let u = x + 2. Then du = dx, so
x
√
dx =
4
x+2
u−2
√
du =
4
u
(u3/4 − 2u−1/4 ) du = 47 u7/4 − 2 · 43 u3/4 + C
= 47 (x + 2)7/4 − 83 (x + 2)3/4 + C
204
¤
CHAPTER 5
INTEGRALS
In Exercises 31–34, let f (x) denote the integrand and F (x) its antiderivative (with C = 0).
31. f (x) = x(x2 − 1)3 .
u = x2 − 1 ⇒ du = 2x dx, so
x(x2 − 1)3 dx =
u3
1
2
du = 18 u4 + C = 18 (x2 − 1)4 + C
Where f is positive (negative), F is increasing (decreasing). Where f
changes from negative to positive (positive to negative), F has a local
minimum (maximum).
33. f (x) = sin3 x cos x. u = sin x
sin3 x cos x dx =
Note that at x =
π
,
2
⇒ du = cos x dx, so
u3 du = 14 u4 + C =
1
4
sin4 x + C
f changes from positive to negative and F has a local
maximum. Also, both f and F are periodic with period π, so at x = 0 and
at x = π, f changes from negative to positive and F has local minima.
35. Let u = x − 1, so du = dx. When x = 0, u = −1; when x = 2, u = 1. Thus,
2
(x
0
− 1)25 dx =
1
−1
u25 du = 0 by
Theorem 6(b), since f (u) = u25 is an odd function.
37. Let u = 1 + 2x3 , so du = 6x2 dx. When x = 0, u = 1; when x = 1, u = 3. Thus,
1
0
x2 1 + 2x3
5
39. Let u = t/4, so du =
π
0
41.
1
4
π/4
0
sec2 (t/4) dt =
π/6
−π/6
3
1
dx =
u5
1
6
du =
1 6 3
u 1
6
1
6
=
1
(36
36
− 16 ) =
1
(729
36
728
36
− 1) =
=
182
.
9
dt. When t = 0, u = 0; when t = π, u = π/4. Thus,
sec2 u (4 du) = 4 tan u
π/4
0
= 4 tan π4 − tan 0 = 4(1 − 0) = 4.
tan3 θ dθ = 0 by Theorem 6(b), since f (θ) = tan3 θ is an odd function.
43. Let u = 1 + 2x, so du = 2 dx. When x = 0, u = 1; when x = 13, u = 27. Thus,
13
27
dx
3
0
=
2
u−2/3
1
(1 + 2x)
1
2
1
2
du =
45. Let u = x2 + a2 , so du = 2x dx and x dx =
2a2
a
x
x2 + a2 dx =
u1/2
a2
0
1
2
du =
1
2
· 3u1/3
27
1
= 32 (3 − 1) = 3.
du. When x = 0, u = a2 ; when x = a, u = 2a2 . Thus,
2 3/2
u
3
1
2
2a2
a2
=
1 3/2
u
3
2a2
a2
=
1
3
(2a2 )3/2 − (a2 )3/2 =
47. Let u = x − 1, so u + 1 = x and du = dx. When x = 1, u = 0; when x = 2, u = 1. Thus,
2
x
1
√
x − 1 dx =
1
√
(u + 1) u du =
0
1
(u3/2 + u1/2 ) du =
0
49. Let u = x−2 , so du = −2x−3 dx. When x =
1
1/2
cos(x−2 )
dx =
x3
1
cos u
4
du
−2
=
1
2
1
,
2
+ 23 u3/2
1
=
0
2
5
+
u = 4; when x = 1, u = 1. Thus,
4
cos u du =
1
2 5/2
u
5
1
sin u
2
4
1
=
1
(sin 4 − sin 1).
2
2
3
=
16
.
15
1
3
2
√
2 − 1 a3
SECTION 5.5
THE SUBSTITUTION RULE
¤
205
51. From the graph, it appears that the area under the curve is about
1 + a little more than 12 · 1 · 0.7 , or about 1.4. The exact area is given by
1√
A = 0 2x + 1 dx. Let u = 2x + 1, so du = 2 dx. The limits change to
2 · 0 + 1 = 1 and 2 · 1 + 1 = 3, and
√
3
3
A = 1 u 12 du = 12 23 u3/2 =
1
1
3
3
√
√
3−1 = 3−
1
3
≈ 1.399.
53. First write the integral as a sum of two integrals:
I=
2
(x
−2
f (x) = x
√
+ 3) 4 − x2 dx = I1 + I2 =
2
−2
x
√
4 − x2 dx +
2
−2
3
√
4 − x2 dx. I1 = 0 by Theorem 6(b), since
√
4 − x2 is an odd function and we are integrating from x = −2 to x = 2. We interpret I2 as three times the area of
a semicircle with radius 2, so I = 0 + 3 ·
1
2
π · 22 = 6π.
55. The volume of inhaled air in the lungs at time t is
V (t) =
t
0
=
5
4π
f(u) du =
− cos v
t 1
0 2
2πt/5
0
57. Let u = 2x. Then du = 2 dx, so
=
2
0
2π
5
sin
5
4π
2πt/5 1
2
0
u du =
− cos
2π
t
5
4
0
f(2x) dx =
+1 =
f (u)
1
2
5
2π
sin v
5
4π
1 − cos
du =
4
0
1
2
substitute v =
dv
2π
t
5
2π
5 u,
dv =
liters
f (u) du = 12 (10) = 5.
59. Let u = −x. Then du = −dx, so
b
a
f (−x) dx =
−b
−a
f (u)(−du) =
−a
−b
f (u) du =
−a
−b
f (x) dx
From the diagram, we see that the equality follows from the fact that we are
reflecting the graph of f , and the limits of integration, about the y-axis.
61. Let u = 1 − x. Then x = 1 − u and dx = −du, so
1
0
63.
xa (1 − x)b dx =
π/2
0
f (cos x) dx =
=
0
1
(1 − u)a ub (−du) =
π/2
0
0
π/2
f sin
π
2
−x
1
0
ub (1 − u)a du =
dx
f(sin u)(−du) =
u=
π/2
0
π
2
1
0
xb (1 − x)a dx.
− x, du = −dx
f (sin u) du =
π/2
0
f (sin x) dx
Continuity of f is needed in order to apply the substitution rule for definite integrals.
65. Let u = 5 − 3x. Then du = −3 dx and dx = − 13 du, so
dx
=
5 − 3x
1
− 13 du = − 13 ln |u| + C = − 13 ln |5 − 3x| + C.
u
67. Let u = ln x. Then du =
dx
, so
x
(ln x)2
dx =
x
69. Let u = 1 + ex . Then du = ex dx, so
u2 du = 13 u3 + C = 13 (ln x)3 + C.
√
ex 1 + ex dx =
√
u du = 23 u3/2 + C = 23 (1 + ex )3/2 + C.
√
Or: Let u = 1 + ex . Then u2 = 1 + ex and 2u du = ex dx, so
√
ex 1 + ex dx = u · 2u du = 23 u3 + C = 23 (1 + ex )3/2 + C.
2π
5
du
206
¤
CHAPTER 5 INTEGRALS
71. Let u = tan x. Then du = sec2 x dx, so
etan x sec2 x dx =
eu du = eu + C = etan x + C.
73. Let u = 1 + x2 . Then du = 2x dx, so
1+x
dx =
1 + x2
1
dx +
1 + x2
= tan−1 x +
2I = −2
ln 1 + x2 + C = tan−1 x +
du
= tan−1 x +
u
ln|u| + C
[since 1 + x2 > 0].
ln 1 + x2 + C
1
2
1
2
sin x cos x
dx = 2I. Let u = cos x. Then du = − sin x dx, so
1 + cos2 x
sin 2x
dx = 2
1 + cos2 x
75.
1
2
1
2
x
dx = tan−1 x +
1 + x2
u du
= −2 ·
1 + u2
1
2
ln(1 + u2 ) + C = − ln(1 + u2 ) + C = − ln(1 + cos2 x) + C.
Or: Let u = 1 + cos2 x.
77.
cot x dx =
cos x
dx. Let u = sin x. Then du = cos x dx, so
sin x
e
dx
√
=
x ln x
4
1
du = ln |u| + C = ln |sin x| + C.
u
dx
. When x = e, u = 1; when x = e4 ; u = 4. Thus,
x
79. Let u = ln x, so du =
e4
cot x dx =
u−1/2 du = 2 u1/2
4
1
1
= 2(2 − 1) = 2.
81. Let u = ez + z, so du = (ez + 1) dz. When z = 0, u = 1; when z = 1, u = e + 1. Thus,
1
0
83.
ez + 1
dz =
ez + z
e+1
1
1
du = ln |u|
u
e+1
= ln |e + 1| − ln |1| = ln(e + 1).
1
sin x
t
x sin x
=x·
. By Exercise 62,
= x f (sin x), where f(t) =
1 + cos2 x
2 − t2
2 − sin2 x
π
0
x sin x
dx =
1 + cos2 x
π
x f (sin x) dx =
0
π
2
π
f (sin x) dx =
0
π
2
π
0
sin x
dx
1 + cos2 x
Let u = cos x. Then du = − sin x dx. When x = π, u = −1 and when x = 0, u = 1. So
π
2
π
0
sin x
π
dx = −
1 + cos2 x
2
=
−1
1
du
π
=
1 + u2
2
1
−1
du
π
tan−1 u
=
1 + u2
2
π
π
π π
[tan−1 1 − tan−1 (−1)] =
− −
2
2 4
4
5 Review
1. (a)
n
i=1
f (x∗i ) ∆x is an expression for a Riemann sum of a function f .
x∗i is a point in the ith subinterval [xi−1 , xi ] and ∆x is the length of the subintervals.
(b) See Figure 1 in Section 5.2.
(c) In Section 5.2, see Figure 3 and the paragraph beside it.
=
1
−1
π2
4
CHAPTER 5 REVIEW
¤
207
2. (a) See Definition 5.2.2.
(b) See Figure 2 in Section 5.2.
(c) In Section 5.2, see Figure 4 and the paragraph by it (contains “net area”).
3. See the Fundamental Theorem of Calculus after Example 8 in Section 5.3.
4. (a) See the Net Change Theorem after Example 5 in Section 5.4.
(b)
t2
t1
r(t) dt represents the change in the amount of water in the reservoir between time t1 and time t2 .
5. (a)
120
60
v(t) dt represents the change in position of the particle from t = 60 to t = 120 seconds.
(b)
120
60
|v(t)| dt represents the total distance traveled by the particle from t = 60 to 120 seconds.
(c)
120
60
a(t) dt represents the change in the velocity of the particle from t = 60 to t = 120 seconds.
6. (a)
f (x) dx is the family of functions {F | F 0 = f }. Any two such functions differ by a constant.
(b) The connection is given by the Net Change Theorem:
b
a
f(x) dx =
f (x) dx
b
a
if f is continuous.
7. The precise version of this statement is given by the Fundamental Theorem of Calculus. See the statement of this theorem and
the paragraph that follows it at the end of Section 5.3.
8. See the Substitution Rule (5.5.4). This says that it is permissible to operate with the dx after an integral sign as if it were a
differential.
1. True by Property 2 of the Integral in Section 5.2.
3. True by Property 3 of the Integral in Section 5.2.
5. False.
For example, let f (x) = x2 . Then
1
0
√
x2 dx =
1
0
x dx = 12 , but
1
0
x2 dx =
1
3
=
√1 .
3
7. True by Comparison Property 7 of the Integral in Section 5.2.
9. True.
11. False.
The integrand is an odd function that is continuous on [−1, 1], so the result follows from Theorem 5.5.6(b).
The function f (x) = 1/x4 is not bounded on the interval [−2, 1]. It has an infinite discontinuity at x = 0, so it is
not integrable on the interval. (If the integral were to exist, a positive value would be expected, by Comparison
Property 6 of Integrals.)
13. False.
15. False.
For example, the function y = |x| is continuous on R, but has no derivative at x = 0.
b
a
f (x) dx is a constant, so
d
dx
b
a
f (x) dx = 0, not f (x) [unless f (x) = 0]. Compare the given statement
carefully with FTC1, in which the upper limit in the integral is x.
208
¤
CHAPTER 5 INTEGRALS
6
1. (a)
L6 =
[∆x =
f (xi−1 ) ∆x
i=1
6−0
6
= 1]
= f (x0 ) · 1 + f (x1 ) · 1 + f(x2 ) · 1 + f(x3 ) · 1 + f(x4 ) · 1 + f (x5 ) · 1
≈ 2 + 3.5 + 4 + 2 + (−1) + (−2.5) = 8
The Riemann sum represents the sum of the areas of the four rectangles
above the x-axis minus the sum of the areas of the two rectangles below the
x-axis.
6
(b)
M6 =
f(xi ) ∆x
[∆x =
i=1
6−0
6
= 1]
= f(x1 ) · 1 + f(x2 ) · 1 + f(x3 ) · 1 + f (x4 ) · 1 + f (x5 ) · 1 + f (x6 ) · 1
= f(0.5) + f (1.5) + f (2.5) + f (3.5) + f(4.5) + f (5.5)
≈ 3 + 3.9 + 3.4 + 0.3 + (−2) + (−2.9) = 5.7
3.
1
0
x+
√
1 − x2 dx =
1
0
x dx +
1
0
√
1 − x2 dx = I1 + I2 .
I1 can be interpreted as the area of the triangle shown in the figure
and I2 can be interpreted as the area of the quarter-circle.
Area = 12 (1)(1) + 14 (π)(1)2 =
5.
6
0
f (x) dx =
4
0
6
4
f (x) dx +
1
2
+ π4 .
6
4
f (x) dx ⇒ 10 = 7 +
7. First note that either a or b must be the graph of
x
0
f (x) dx ⇒
f (t) dt, since
0
0
6
4
f(x) dx = 10 − 7 = 3
f (t) dt = 0, and c(0) 6= 0. Now notice that b > 0 when c
is increasing, and that c > 0 when a is increasing. It follows that c is the graph of f (x), b is the graph of f 0 (x), and a is the
graph of
x
0
f (t) dt.
9.
2
1
8x3 + 3x2 dx = 8 · 14 x4 + 3 · 13 x3
11.
1
0
1 − x9 dx = x −
9
13.
1
√
u − 2u2
du =
u
1 10 1
x 0
10
9
1
= 1−
2
1
1
10
y(y 2 + 1)5 dy =
2
1
−0 =
2
1
u5
1
2
du =
1
2
1
2
1 6 2
u 1
6
= 2 · 24 + 23 − (2 + 1) = 40 − 3 = 37
9
10
(u−1/2 − 2u) du = 2u1/2 − u2
15. Let u = y 2 + 1, so du = 2y dy and y dy =
1
0
= 2x4 + x3
9
1
= (6 − 81) − (2 − 1) = −76
du. When y = 0, u = 1; when y = 1, u = 2. Thus,
=
1
(64
12
− 1) =
63
12
=
21
.
4
CHAPTER 5 REVIEW
5
17.
1
¤
209
dt
1
does not exist because the function f (t) =
has an infinite discontinuity at t = 4;
(t − 4)2
(t − 4)2
that is, f is discontinuous on the interval [1, 5].
19. Let u = v 3 , so du = 3v 2 dv. When v = 0, u = 0; when v = 1, u = 1. Thus,
1
0
1
0
v 2 cos(v 3 ) dv =
π/4
21.
−π/4
cos u
1
3
1
3
du =
sin u
1
0
= 13 (sin 1 − 0) =
1
3
sin 1.
t4 tan t
t4 tan t
dt = 0 by Theorem 5.5.6(b), since f (t) =
is an odd function.
2 + cos t
2 + cos t
23. Let u = sin πt. Then du = π cos πt dt, so
sin πt cos πt dt =
u
1
π
=
1
2
du =
1
π
· 12 u2 + C =
1
(sin πt)2
2π
+ C.
25. Let u = 2θ. Then du = 2 dθ, so
π/8
0
sec 2θ tan 2θ dθ =
π/4
0
1
2
sec u tan u
du =
1
2
sec u
π/4
0
sec π4 − sec 0 =
1
2
√
2−1 =
1
2
√
2 − 12 .
27. Since x2 − 4 < 0 for 0 ≤ x < 2 and x2 − 4 > 0 for 2 < x ≤ 3, we have x2 − 4 = −(x2 − 4) = 4 − x2 for 0 ≤ x < 2 and
x2 − 4 = x2 − 4 for 2 < x ≤ 3. Thus,
3
0
2
x2 − 4 dx =
0
(4 − x2 ) dx +
= 8−
8
3
3
2
(x2 − 4) dx = 4x −
− 0 + (9 − 12) −
8
3
−8 =
16
3
In Exercises 29 and 30, let f(x) denote the integrand and F (x) its antiderivative (with C = 0).
x3
3
−3+
29. Let u = 1 + sin x. Then du = cos x dx, so
cos x dx
√
=
1 + sin x
u−1/2 du = 2u1/2 + C = 2
√
1 + sin x + C.
31. From the graph, it appears that the area under the curve y = x
√
x between x = 0
and x = 4 is somewhat less than half the area of an 8 × 4 rectangle, so perhaps
about 13 or 14. To find the exact value, we evaluate
4
0
x
√
x dx =
x
33. F (x) =
0
4
0
d
dx
2 5/2
x
5
4
0
= 25 (4)5/2 =
t2
d
dt ⇒ F 0 (x) =
1 + t3
dx
35. Let u = x4 . Then
g 0 (x) =
x3/2 dx =
x4
0
x
0
64
5
= 12.8.
t2
x2
dt =
3
1+t
1 + x3
dg du
du
dg
= 4x3 . Also,
=
, so
dx
dx
du dx
cos(t2 ) dt =
d
du
u
0
cos(t2 ) dt ·
du
du
= cos(u2 )
= 4x3 cos(x8 ).
dx
dx
2
+
0
16
3
=
x3
− 4x
3
32
3
−
9
3
3
2
=
23
3
210
¤
CHAPTER 5 INTEGRALS
x
x
1
cos θ
cos θ
cos θ
dθ =
dθ + √
dθ =
√
θ
θ
θ
x
1
x
√
√
cos x cos x 1
2 cos x − cos x
0
√ =
− √
y =
x
2x
x 2 x
x
37. y =
1
cos θ
dθ −
θ
√
x
1
cos θ
dθ
θ
⇒
√
√
√
√
√
12 + 3 ≤ x2 + 3 ≤ 32 + 3 ⇒ 2 ≤ x2 + 3 ≤ 2 3, so
√
√
3√
3√
2(3 − 1) ≤ 1 x2 + 3 dx ≤ 2 3(3 − 1); that is, 4 ≤ 1 x2 + 3 dx ≤ 4 3.
39. If 1 ≤ x ≤ 3, then
41. 0 ≤ x ≤ 1
⇒ 0 ≤ cos x ≤ 1 ⇒ x2 cos x ≤ x2
43. ∆x = (3 − 0)/6 =
1
2,
⇒
1
0
x2 cos x dx ≤
1
0
x2 dx =
1
3
x3
1
0
=
1
3
so the endpoints are 0, 12 , 1, 32 , 2, 52 , and 3, and the midpoints are 14 , 34 , 54 , 74 , 94 , and
[Property 7].
11
4 .
The Midpoint Rule gives
3
0
sin(x3 ) dx ≈
6
f(xi ) ∆x =
i=1
1
2
sin
1 3
4
3 3
4
+ sin
+ sin
5 3
4
+ sin
7 3
4
+ sin
9 3
4
+ sin
11 3
4
≈ 0.280981.
45. Note that r(t) = b0 (t), where b(t) = the number of barrels of oil consumed up to time t. So, by the Net Change Theorem,
8
0
r(t) dt = b(8) − b(0) represents the number of barrels of oil consumed from Jan. 1, 2000, through Jan. 1, 2008.
47. We use the Midpoint Rule with n = 6 and ∆t =
24
0
24 − 0
6
= 4. The increase in the bee population was
r(t) dt ≈ M6 = 4[r(2) + r(6) + r(10) + r(14) + r(18) + r(22)]
≈ 4[50 + 1000 + 7000 + 8550 + 1350 + 150] = 4(18,100) = 72,400
49. Let u = 2 sin θ. Then du = 2 cos θ dθ and when θ = 0, u = 0; when θ =
π/2
0
x
51.
2
0
f (2 sin θ) cos θ dθ =
x
f (t) dt = x sin x +
0
f (x) 1 −
0
1
1 + x2
f (u)
1
2
du =
1
2
2
0
f (u) du =
1
2
2
0
π
,
2
u = 2. Thus,
f(x) dx = 12 (6) = 3.
f (t)
f(x)
dt ⇒ f(x) = x cos x + sin x +
[by differentiation] ⇒
1 + t2
1 + x2
= x cos x + sin x ⇒ f (x)
53. Let u = f (x) and du = f 0 (x) dx. So 2
55. Let u = 1 − x. Then du = −dx, so
1
0
b
a
x2
1 + x2
f (x)f 0 (x) dx = 2
f (1 − x) dx =
0
1
= x cos x + sin x ⇒ f (x) =
f (b)
f (a)
u du = u2
f(u)(−du) =
1
0
f (b)
f (a)
1 + x2
(x cos x + sin x)
x2
= [f (b)]2 − [f (a)]2 .
f (u) du =
1
0
f (x) dx.
PROBLEMS PLUS
1. Differentiating both sides of the equation x sin πx =
U x2
0
f (t) dt (using FTC1 and the Chain Rule for the right side) gives
sin πx + πx cos πx = 2xf (x2 ). Letting x = 2 so that f (x2 ) = f (4), we obtain sin 2π + 2π cos 2π = 4f (4), so
f (4) = 14 (0 + 2π · 1) =
π
.
2
Ux
3. Differentiating the given equation,
0
f (t) dt = [f (x)]2 , using FTC1 gives f (x) = 2f (x) f 0 (x) ⇒
f (x)[2f 0 (x) − 1] = 0, so f (x) = 0 or f 0 (x) = 12 . Since f (x) is never 0, we must have f 0 (x) =
f (x) = 12 x + C. To find C, we substitute into the given equation to get
1 2
x
4
U x 1
0
2
and f 0 (x) =
2
t + C dt = 12 x + C
1
2
⇒
⇔
+ Cx = 14 x2 + Cx + C 2 . It follows that C 2 = 0, so C = 0, and f(x) = 12 x.
5. f (x) =
]
g(x)
0
1
√
dt, where g(x) =
1 + t3
]
cos x
[1 + sin(t2 )] dt. Using FTC1 and the Chain Rule (twice) we have
0
1
1
f 0 (x) = s
g0 (x) = s
[1 + sin(cos2 x)](− sin x). Now g π2 =
3
3
1 + [g(x)]
1 + [g(x)]
f0
1
2
π
2
]
0
[1 + sin(t2 )] dt = 0, so
0
1
= √
(1 + sin 0)(−1) = 1 · 1 · (−1) = −1.
1+0
7. f (x) = 2 + x − x2 = (−x + 2)(x + 1) = 0
else. The integral
Ub
a
⇔ x = 2 or x = −1. f (x) ≥ 0 for x ∈ [−1, 2] and f (x) < 0 everywhere
(2 + x − x2 ) dx has a maximum on the interval where the integrand is positive, which is [−1, 2]. So
a = −1, b = 2. (Any larger interval gives a smaller integral since f (x) < 0 outside [−1, 2]. Any smaller interval also gives a
smaller integral since f (x) ≥ 0 in [−1, 2].)
9. (a) We can split the integral
Un
0
n kU
S
i
[[x]] dx into the sum
i=1
i−1
l
[[x]] dx . But on each of the intervals [i − 1, i) of integration,
[[x]] is a constant function, namely i − 1. So the ith integral in the sum is equal to (i − 1)[i − (i − 1)] = (i − 1). So the
original integral is equal to
n
S
(i − 1) =
i=1
(b) We can write
Now
Ub
0
Ub
a
[[x]] dx =
[[x]] dx =
U [[b]]
0
Ub
0
[[x]] dx −
[[x]] dx +
Ub
[[b]]
n−1
S
i=
i=1
Ua
0
(n − 1)n
.
2
[[x]] dx.
[[x]] dx. The first of these integrals is equal to 12 ([[b]] − 1) [[b]],
by part (a), and since [[x]] = [[b]] on [[[b]] , b], the second integral is just [[b]] (b − [[b]]). So
Ub
0
[[x]] dx = 12 ([[b]] − 1) [[b]] + [[b]] (b − [[b]]) =
Therefore,
Ub
a
[[x]] dx =
1
2
[[b]] (2b − [[b]] − 1) −
1
2
1
2
[[b]] (2b − [[b]] − 1) and similarly
[[a]] (2a − [[a]] − 1).
Ua
0
[[x]] dx =
1
2
[[a]] (2a − [[a]] − 1).
211
212
¤
CHAPTER 5 PROBLEMS PLUS
11. Let Q(x) =
]
x
b
c
d
b
c
d
P (t) dt = at + t2 + t3 + t4 = ax + x2 + x3 + x4 . Then Q(0) = 0, and Q(1) = 0 by the
2
3
4
2
3
4
0
x
0
given condition, a +
c d
b
+ + = 0. Also, Q0 (x) = P (x) = a + bx + cx2 + dx3 by FTC1. By Rolle’s Theorem, applied to
2 3 4
Q on [0, 1], there is a number r in (0, 1) such that Q0 (r) = 0, that is, such that P (r) = 0. Thus, the equation P (x) = 0 has a
root between 0 and 1.
More generally, if P (x) = a0 + a1 x + a2 x2 + · · · + an xn and if a0 +
a2
an
a1
+
+ ··· +
= 0, then the equation
2
3
n+1
P (x) = 0 has a root between 0 and 1. The proof is the same as before:
] x
a 1 2 a2 3
an
Let Q(x) =
x +
x + ··· +
xn . Then Q(0) = Q(1) = 0 and Q0 (x) = P (x). By
P (t) dt = a0 x +
2
3
n+1
0
Rolle’s Theorem applied to Q on [0, 1], there is a number r in (0, 1) such that Q0 (r) = 0, that is, such that P (r) = 0.
13. Note that
d
dx
]
0
x
]
0
u
]
f(t) dt du =
15. lim
n→∞
Ux
0
f (t) dt by FTC1, while
0
d
dx
Hence,
x
]
x
0
f (u)(x − u) du =
] x
] x
d
d
x
f (u)(x − u) du =
f(u) du −
f(u)u du
dx
dx 0
0
Ux
Ux
= 0 f(u) du + xf(x) − f(x)x = 0 f(u) du
U x U u
0
0
f (t) dt du + C. Setting x = 0 gives C = 0.
1
1
1
+√ √
+··· + √ √
√ √
n n+n
n n+1
n n+2
u
u
u
1
n
n
n
= lim
+
+ ··· +
n→∞ n
n+1
n+2
n+n
$
#
1
1
1
1
s
+s
+··· + √
= lim
n→∞ n
1+1
1 + 1/n
1 + 2/n
n
i
1[
f
n→∞ n
n
i=1
= lim
=
]
0
1
1
where f (x) = √
1+x
√
√
1
1
√
dx = 2 1 + x 0 = 2 2 − 1
1+x