SECTION 4.2
4.2
■ Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places.
1.
3.
5.
y
y
2
0
1
y
s1 x 2 dx,
10
0
■
2.
n 10
4.
n5
sin sx dx,
■
7– 8
n5
x dx,
■
■
6.
■
y
1
dx, n 4
2x 7
3
1
y
4
0
y
n4
■
■
■
19–22
■
2x* i
n l i1
2
5x*i x,
sx* x,
i
n l i1
■
■
■
20.
y
21.
y (1 x ) dx
22.
y 3x 5 dx
2
2
■
■
■
■
■
■
9.
23.
y
3
24.
y
8
25.
y
10
11.
13.
15.
17.
Copyright © 2013, Cengage Learning. All rights reserved.
■
b
y
4
y
1
a
1
0
y
c dx
x 2 dx
12.
ax b dx
14.
2
1
1
y
b
0
10.
t 3 t 2 1 dt
16.
x 3 4x dx
■
■
18.
■
■
■
y
7
y
5
y
0
2
1
26.
y
5
1
5
y
b
a
y
5
2
■
6
f x dx y f x dx 3
■
■
■
■
■
y
12
f x dx
6
5
f x dx y f x dx
0
2
7
f x dx y f x dx
2
3
■
f x dx y
0
■
■
3
■
f x dx y
■
■
6
5
f x dx
■
■
■
■
■
27. If x28 f x dx 1.7 and x58 f x dx 2.5, find x25 f x dx.
6 2x dx
28. If x01 f t dt 2, x04 f t dt 6, and x34 f t dt 1,
find x13 f t dt.
2 3x x dx
2
29–31
3
■
■
■
y
■
3
0
■
Use the form of the definition of the integral given in
Theorem 4 to evaluate the integral.
9–18
■
2 x dx
■ Write the given sum or difference as a single integral in
the form xab f x dx .
0, 1
■
■
1
23–26
1, 4
■
3
1 2x dx
1
n
8. lim
Evaluate the integral by interpreting it in terms of areas.
y
n
7. lim
3
■
19.
■
secx3 dx, n 6
0
■
tan x dx,
Express the limit as a definite integral on the given interval.
■
1
S Click here for solutions.
1–6
3
■
THE DEFINITE INTEGRAL
A Click here for answers.
5
THE DEFINITE INTEGRAL
2x 2 3x 4 dx
29.
y
0
31.
y
1
3
Px 2 Qx R dx
t 3 2t 3 dt
■
■
■
■
■
■
1
■
Use Property 8 to estimate the value of the integral.
x 2 2x dx
30.
3
y
4
cos x dx
s1 x 4 dx
■
■
■
■
■
■
■
■
■
2
■
SECTION 4.2
THE DEFINITE INTEGRAL
4.2
ANSWERS
E Click here for exercises.
1.
153.125
2.
−0.7873
3.
1.8100
4.
0.3450
5.
6.4643
S Click here for solutions.
3.9379
1 2
7. 0 2x − 5x dx
4√
8. 1
x dx
6.
9.
(b − a) c
10.
9
11.
15
12. 83
a
+b
2
14. 19.5
13.
4
15. 3
16.
P
a3
b3
−
3
3
+Q
b2
a2
−
2
2
b4
+ 2b2
4
18. 140.25
17.
19.
10
20.
4
21.
0
22. 41
6
23.
12
1
8
f (x) dx
f (x) dx
0
10
25. 7 f (x) dx
6
26. 0 f (x) dx
Copyright © 2013, Cengage Learning. All rights reserved.
24.
27.
−0.8
28.
−9
29.
−3 ≤
0 −3
x2 + 2x dx ≤ 9.
√
π/3
2π
≤ π/4 cos x dx ≤ 24
1 √
√
31. 2 ≤ −1 1 + x4 dx ≤ 2 2
π
30. 24
+ R (b − a)
SECTION 4.2
4.2
THE DEFINITE INTEGRAL
■
3
SOLUTIONS
E Click here for exercises.
1.
The width of the intervals is ∆x = (5 − 0) /5 = 1 so the
partition points are 0, 1, 2, 3, 4, 5 and the midpoints are
0.5, 1.5, 2.5, 3.5, 4.5. The Midpoint Rule gives
5
0
x3 dx ≈
5
i=1
5.
∆x = (10 − 0) /5 = 2, so the endpoints are 0, 2, 4, 6, 8, and
10, and the midpoints are 1, 3, 5, 7, and 9. The Midpoint
Rule gives
10
f (xi ) ∆x
0
5
√
sin x dx ≈
f (xi ) ∆x
i=1
√
√
= 2 sin 1 + sin 3
= (0.5)3 + (1.5)3 + (2.5)3 + (3.5)3 + (4.5)3
√
√
√ + sin 5 + sin 7 + sin 9
= 153.125
≈ 6.4643
2.
The width of the interval ∆x = (3 − 1) /4 = 0.5 so the
partition points are 1.0, 1.5, 2.0, 2.5, 3.0 and the midpoints
are 1.25, 1.75, 2.25, 2.75.
3
1
6.
4
1
dx ≈
f (xi ) ∆x
2x − 7
i=1
1
1
= 0.5
+
2 (1.25) − 7 2 (1.75) − 7
+
∆x = (π − 0) /6 =
π
6,
so the endpoints are 0, π6 ,
4π 5π
, 6 , and 6π
, and the midpoints
6
6
11π
and 12 . The Midpoint Rule gives
π
1
1
+
2 (2.25) − 7 2 (2.75) − 7
0
sec (x/3) dx ≈
=
6
i=1
π
6
π 3π
, ,
12 12
f (xi ) ∆x
π
sec 36
+ sec 3π
+ sec 5π
36
36
+ sec
≈ −0.7873
are
2π 3π
6 , 6 ,
5π 7π 9π
, , ,
12 12 12
7π
36
+ sec
9π
36
+ sec 11π
36
≈ 3.9379
3.
∆x = (2 − 1) /10 = 0.1 so the partition points are
1.0, 1.1, . . . , 2.0 and the midpoints are 1.05, 1.15, . . . , 1.95.
2
10
1 + x2 dx ≈
f (xi ) ∆x
1
i=1
= 0.1
1 + (1.05)2 +
+ ··· +
1 + (1.15)2
∆x =
1
4
π
4
−0 =
π
16
π
so the partition points are 0, 16
,
n→∞ i=1
8.
Copyright © 2013, Cengage Learning. All rights reserved.
On [1, 4], lim
n n→∞ i=1
x∗i ∆x =
4
1
√
x dx.
2π
,
16
3π 4π
,
16 16
π 3π 5π 7π
and the midpoints are 32
, 32 , 32 , 32 . The Midpoint
Rule gives
π/4
π π
tan x dx ≈ 16
tan 32
+ tan 3π
32
0
5π
+ tan 32 + tan 7π
32
≈ 0.3450
On [0, 1],
n 1
2 (x∗i )2 − 5x∗i ∆x = 0 2x2 − 5x dx.
lim
1 + (1.95)2
≈ 1.8100
4.
7.
9.
b
a
c dx = lim
n→∞
n
b−a b−a
nc
c = lim
n→∞
n i=1
n
= lim (b − a) c = (b − a) c
n→∞
4
■
SECTION 4.2
10.
THE DEFINITE INTEGRAL
n
9i
9 6 − 2 −2 +
n→∞ n i=1
n
n
9 18i
= lim
10 −
n→∞ n i=1
n
n
n
162 90
= lim
1− 2
i
n→∞
n i=1
n i=1
162 n(n + 1)
90
n− 2
= lim
n→∞
n
n
2
1
= lim 90 − 81 · 1 1 +
n→∞
n
7
(6 − 2x) dx = lim
−2
= 90 − 81 = 9
11. 1 x − 2 dx
2
n
3 3i
= lim
−2
1+
n→∞ n i=1
n
2
n
6i
3 9i
−
1
+
= lim
n→∞ n i=1 n2
n
n
n
n
18 3 27
2
i
+
i
−
1
= lim
n→∞ n3 i=1
n2 i=1
n i=1
18 n (n + 1)
3
27 n (n + 1) (2n + 1)
+
−
n
= lim
n→∞ n3
6
n2
2
n
1
1
1
9
·1 1+
2+
+9·1 1+
−3
= lim
n→∞ 2
n
n
n
9 = 2 · 2 + 9 − 3 = 15
5
2
12. 1 2 + 3x − x dx
2 n
4i
4i
4 − 1+
= lim
2+3 1+
n→∞ n i=1
n
n
2
n
4i
16i
4 +4
= lim
− 2 +
n→∞ n i=1
n
n
n
n
n
16 16 64 = lim − 3
i2 + 2
i+
1
n→∞
n i=1
n i=1
n i=1
64 n (n + 1) (2n + 1)
16 n (n + 1)
16
+ 2
+ n
= lim − 3
n→∞
n
6
n
2
n
1
1
1
32
2+
+8·1 1+
+ 16
= lim − · 1 1 +
n→∞
3
n
n
n
4
2
= − 64
+ 8 + 16 =
3
Copyright © 2013, Cengage Learning. All rights reserved.
13.
1
0
8
3
n
1 i
+b
a
n→∞ n i=1
n
n
n
b a
= lim
i
+
1
n→∞ n2 i=1
n i=1
b
a n (n + 1)
+
n
= lim
n→∞ n2
2
n
a
1
a
+b = +b
= lim
·1 1+
n→∞ 2
n
2
(ax + b) dx = lim
14.
0 2x2 − 3x − 4 dx
2
n
3i
3i
3 = lim
2 −3 +
−4
− 3 −3 +
n→∞ n i=1
n
n
2
n
45i
3 18i
= lim
+ 23
−
n→∞ n i=1
n2
n
n
n
n
135 69 54 2
= lim
i − 2
i+
1
n→∞ n3 i=1
n i=1
n i=1
135 n (n + 1)
69
54 n (n + 1) (2n + 1)
−
+
n
= lim
n→∞ n3
6
n2
2
n
1
135
1
1
2+
−
+ 69
= lim 9 · 1 1 +
·1 1+
n→∞
n
n
2
n
−3
= 9 · 2 − 135
+ 69 = 19.5
2
1 3
2
15. −1 t − t + 1 dt
3 2
n
2i
2 2i
= lim
− −1 +
+1
−1 +
n→∞ n i=1
n
n
3
n
12i2
6i
2 8i
−
+
= lim
−
1
n→∞ n i=1
n3
n2
n
2
4i
4i
−
+
1
+
1
−
n2
n
3
2
n
16i
10i
2 8i
−1
− 2 +
= lim
n→∞ n i=1 n3
n
n
n
n
n
n
32 20 2 16 3
2
= lim
i
−
i
+
i
−
1
n→∞ n4 i=1
n3 i=1
n2 i=1
n i=1
2
2
32 n (n + 1) (2n + 1)
16 n (n + 1)
= lim
− 3
n→∞ n4
4
n
6
20 n (n + 1)
2
+ 2
− n
n
2
n
2
1
1
16
1
·1 1+
2+
−
= lim 4 · 12 1 +
n→∞
n
3
n
n
1
+ 10 · 1 1 +
−2
n
= 4 − 32
+ 10 − 2 = 43
3
b
16. a P x2 + Qx + R dx
2
n
b−a
b−a = lim
P a+
i
n→∞
n i=1
n
b−a
i +R
+Q a+
n
2
n
(b − a) 2
b−a = lim
i
P
n→∞
n i=1
n2
b−a
i + P a2 + Qa + R
+ (2P a + Q)
n
3 n
n
(b − a) 2
(b − a)2 i
+
(2P
a
+
Q)
i
= lim P
2
n→∞
n3
n
i=1
i=1
n
b−a + P a2 + Qa + R
1
n i=1
SECTION 4.2
= lim
n→∞
P
(b − a)3 n (n + 1) (2n + 1)
n3
6
(b − a)2 n (n + 1)
+ (2P a + Q)
n2
2
b−a
2
n
+ P a + Qa + R
n
3 1
1
P (b − a)
2+
= lim
1 1+
n→∞
6
n
n
1
Q
(b − a)2 1 1 +
+ Pa +
2
n
2
+ P a + Qa + R (b − a)
(b − a)3
Q
(b − a)2
+ Pa +
3
2
+ P a2 + Qa + R (b − a)
3
a3
b
=P
− b2 a + ba2 −
+ ab2 − 2a2 b + a3 + a2 b − a3
3
3
2
b
a2
+Q
− ab +
+ ab − a2 + R (b − a)
2
2
3
2
3
2
b
a
b
a
=P
−
+Q
−
+ R (b − a)
3
3
2
2
b 3
17. 0 x + 4x dx
3
n
b bi
bi
= lim
+4
n→∞ n i=1
n
n
4 n
2
n
b b 3
= lim
i
+
4
i
n→∞ n4 i=1
n2 i=1
4 2
4b2 n (n + 1)
b n (n + 1)2
= lim
+
n→∞ n4
4
n2
2
2
4
1
1
b
2
2
= lim
+ 2b · 1 1 +
·1 1+
n→∞
4
n
n
= lim
n→∞
=
19.
3
1
81
4
THE DEFINITE INTEGRAL
■
5
2
81 2
1
1
1
·1 1+
2+
+ 27 · 1 1 +
4
n
n
n
1
+ 21
+ 45 · 1 1 +
n
+ 54 + 45 + 21 = 140.25
(1 + 2x) dx can be interpreted as the area under the graph
of f (x) = 1 + 2x between x = 1 and x = 3. This is equal
to the area of the rectangle plus the area of the triangle, so
3
(1 + 2x) dx = A = 2 · 3 + 12 · 2 · 4 = 10.
1
Or: Use the formula for the area of a trapezoid:
A = 12 (2) (3 + 7) = 10.
=P
b4
+ 2b2
4
5 3
18. 2 t − 2t + 3 dt
3
n
3i
3 3i
= lim
+3
−2 2+
2+
n→∞ n i=1
n
n
n
6i
54i2
36i
3 27i3
+
8
−
4
−
+
3
+
+
= lim
n→∞ n i=1
n3
n2
n
n
3
2
n
54i
30i
3 27i
+ 2 +
= lim
+7
n→∞ n i=1
n3
n
n
n
n
n
n
162 90 21 81 3
2
i
+
i
+
i
+
1
= lim
n→∞ n4 i=1
n3 i=1
n2 i=1
n i=1
2
2
162 n (n + 1) (2n + 1)
81 n (n + 1)
+ 3
= lim
n→∞ n4
4
n
6
90 n (n + 1) 21
+ n
+ 2
n
2
n
20.
3
−1
(2 − x) dx can be interpreted as A1 − A2 , where A1 and
A2 are the areas of the triangles shown. Thus,
3
(2 − x) dx = 12 · 3 · 3 − 12 · 1 · 1 = 4.
−1
21.
2
−2
(1 − |x|) dx can be interpreted as the area of the middle
triangle minus the areas of the outside ones, so
2
(1 − |x|) dx = 12 · 2 · 1 − 2 · 12 · 1 · 1 = 0.
−2
Copyright © 2013, Cengage Learning. All rights reserved.
=
22.
3
0
|3x − 5| dx can be interpreted as the area under the graph
of the function f (x) = |3x − 5| between x = 0 and x = 3.
This is equal to the sum of the areas of the two triangles, so
3
|3x − 5| dx = 12 · 53 · 5 + 12 3 − 53 4 = 41
.
0
6
23.
3
1
f (x) dx +
6
12
f (x) dx + 6 f (x) dx
12
12
6
= 1 f (x) dx + 6 f (x) dx = 1 f (x) dx
3
6
■
SECTION 4.2
24.
25.
26.
27.
28.
29.
8
THE DEFINITE INTEGRAL
f (x) dx +
5
f (x) dx
8
8
5
= 0 f (x) dx + 5 f (x) dx = 0 f (x) dx
7
10
f (x) dx − 2 f (x) dx
2
= 27 f (x) dx + 710 f (x) dx − 27 f (x) dx = 710 f (x) dx
5
0
6
f (x) dx − −3 f (x) dx + 5 f (x) dx
−3
5
0
6
0
= −3 f (x) dx + 0 f (x) dx − −3 f (x) dx + 5 f (x) dx
6
= 0 f (x) dx
5
8
8
f (x) dx + 5 f (x) dx = 2 f (x) dx ⇒
2
5
5
f (x) dx + 2.5 = 1.7 ⇒ 2 f (x) dx = −0.8
2
1
3
4
4
f (t) dt + 1 f (t) dt + 3 f (t) dt = 0 f (t) dt
0
3
⇒ 2 + 1 f (t) dt + 1 = −6 ⇒
3
f (t) dt = −6 − 2 − 1 = −9
1
5
0
If f (x) = x2 + 2x, −3 ≤ x ≤ 0, then f (x) = 2x + 2 = 0
when x = −1, and f (−1) = −1. At the endpoints,
f (−3) = 3, f (0) = 0. Thus the absolute minimum is
m = −1 and the absolute maximum is M = 3. Thus
0 2
−1 [0 − (−3)] ≤ −3
x + 2x dx ≤ 3 [0 − (−3)] or
0 −3 ≤ −3 x2 + 2x dx ≤ 9.
Copyright © 2013, Cengage Learning. All rights reserved.
If
π
4
√
≤ x ≤ π3 , then 12 ≤ cos x ≤ 22 , so
√ π/3
1 π
− π4 ≤ π/4 cos x dx ≤ 22 π3 − π4 or
2 3
√
π/3
π
2π
≤ π/4 cos x dx ≤ 24
.
24
√
√
4
31. For −1 ≤ x ≤ 1, 0 ≤ x ≤ 1 and 1 ≤ 1 + x4 ≤ 2, so
√
1 √
1 [1 − (−1)] ≤ −1 1 + x4 dx ≤ 2 [1 − (−1)] or
√
1 √
2 ≤ −1 1 + x4 dx ≤ 2 2.
30.