2
Areas and the Fundamental Theorem of Calculus
Area Under the Curves
2.1
Estimate the area under the graph of f (x) = x2 + 2 from x = −1 to x = 2 in each of the
following ways, and sketch the graph and the rectangles in each case.
(a) By using three rectangles and left endpoints.
(b) Improve your estimate in (a) by using 6 rectangles.
(c) Repeat part (a) using midpoints.
(d) Repeat part (b) using midpoints.
(e) From your sketches in parts (a), (b), (c) and (d), which appears to be the best estimate?
Solution
–1
–1
–0.5
–0.5
6
6
5
5
4
4
3
3
2
2
1
1
0
0.5
1
x
1.5
2
–1
–0.5
0
6
6
5
5
4
4
3
3
2
2
1
1
0
0.5
1
x
1.5
2
–1
–0.5
0
0.5
1
x
1.5
2
0.5
1
x
1.5
2
Figure 1: For problem 2.1 from left to right:(a), (b), (c) (d)
(a) Each of the rectangular strips has a base width of 1. Using left endpoints, we find the
height of each strip from the values of f (x) at x = −1, 0, 1. The combined area of the strips
is therefore: A = 1[((−1)2 + 2) + (02 + 2) + (12 + 2)] = 8
1
(b) A = 0.5[((−1)2 + 2) + ((−0.5)2 + 2) + (02 + 2) + (0.52 + 2) + (12 + 2) + (1.52 + 2)] =
0.5[(−1)2 + (−0.5)2 + 02 + 0.52 + 12 + 1.52 + 6(2)] = 8.375
(c) A = 1[(−0.5)2 + 0.52 + 1.52 + 3(2)] = 8.75
(d) A = 0.5[(−0.75)2 + (−0.25)2 + 0.252 + 0.752 + 1.252 + 1.752 + 6(2)] = 8.9375
(e) Midpoints using 6 rectangles.
2.2
Consider the function y = f (x) = ex on the interval [0, 1].
Subdivide the interval [0, 1] into 4 equal subintervals of width 0.25 and find an approximation
to the area of this region using four rectangular strips.
(a)Use the left endpoints of each interval to obtain the heights of the rectangular strips.
(b) Use the right endpoints of each interval to obtain the heights of the rectangular strips.
(c) Explain the reason why your answer in (a) is different from your answer in (b).
Solution
(a)Using the left hand end points of the subintervals (“lower sums”) we get the approximation
Area ≈
1 0
e + e0.25 + e0.5 + e0.75 ≈ 1.51.
4
(b)Using the right hand end points of the subintervals (“upper sums”) we get the approximation
1 0.25
Area ≈
e
+ e0.50 + e0.75 + e1 ≈ 1.942.
4
(c) The answer in (a) is an under-estimation of the actual area. The answer in (b) is an
over-estimation of the actual area. In this case, answer (b) > answer (a). Note: The rightend point method does not always under-estimate the actual area. Consider y = e−x . The
area obtained using right-endpoints < the area obtained using left-endpoints.
2.3
Find the area between the x axis and the graph of the function y = f (x) = 2 − x between
x = 0 and x = 2.
(a) By using your knowledge about the area of a triangular region.
(b) By setting up the problem as a sum of the areas of n rectangular strips, using the
appropriate summation formula, and letting the number of strips (n) get larger and larger
to arrive at the result. Show that your answer in (b) is then identical to the answer in (a).
Solution
2
(a) The intercepts of the line y = f (x) = 2 − x on the x and y axes are both 2 so that the
height and the base of the triangle have lengths 2. Thus its area is A = (1/2)2 × 2 = 2.
(b)We now subdivide the interval [0, 2] on the x axis into n equal pieces (each has width
h = 2/n) and consider the strips that are so defined. The coordinate of the right endpoint
of the k’th interval is xk = 2(k/n). The y coordinate on the straight line corresponding to
xk (which is also the height of the k’th strip is
yk = f (xk ) = 2 − xk = 2 − 2(k/n) = 2(1 − k/n).
Thus the area of the (approximately rectangular) k’th strip is
Ak = width × height = 2/n × 2(1 − k/n)
The total area of n strips is
A(n) =
n
X
k=1
(4/n)(1 − k/n) = (4/n)
"
n
X
k=1
1 − (1/n)
n
X
k=1
k
#
A(n) = (4/n)n − (4/n2 )(n(n + 1)/2) = 4 − 2(n + 1)/n.
If we now increase the number of strips, n → ∞, we find that this area approaches the true
area of the triangle, A = 4 − 2 = 2. (This uses the fact that the ratio (n + 1)/n approaches
1 as n gets large.
2.4
Find the area between the graphs of the following two functions: y = f (x) = x2 and
y = g(x) = 2 − x2 . (Hint: what do we mean by “between”? Where does this region begin
and where does it end?) This problem should be set up in the form of a sum of areas of
rectangular strips. You should NOT use previous familiarity with integration techniques to
solve it.
Solution These curves intersect when x2 = 2 − x2 which occurs when x = ±1. The region
is symmetric about the x axis, so we can compute the area in the first quadrant and then
double the answer obtained. Splitting up the region into n pieces, we obtain rectangles with
base of width 1/n. The kth rectangle has an x-coordinate, xk = k/n. The height of each of
the rectangles that fit right between the curves is the difference between the two curves:
hk = 2 − x2k − x2k = 2 − 2x2k = 2(1 − (k/n)2 ).
The sum to calculate is thus:
n−1
X
k
2(1 −
n
k=0
!2
)
X
X
1
2 n−1
2 n−1
=
1− 3
k2 .
n
n k=0
n k=0
We find, using the summation formulae, that the result is
(n − 1)(2n − 1)
2n2 − 3n + 1
2 (n − 1)(n)(2n − 1)
=
2−
=
2−
= 2−(2/3)+(1/n)−(1/3n2 )
3
2
2
n
6
3n
3n
In the limit as n → ∞ this converges to the area A = 2 − (2/3) = 4/3. The entire area is
double this, i.e. 8/3.
2−
3
4
3
y=x^2
y2
1
0
–1
–2
1
x
2
y=2-x^2
–1
Figure 2: For problem 2.4 solution
2.5
Determine a function f (x) and an interval on the x-axis such that the expression shown
below is equal to the area under the graph of f (x) over the given interval:
s
n
X
3
3i
1+
lim
n→∞
n
i=0 n
What do the single summands that appear in this expression represent?
Solution
q
In the expression ni=0 n3 1 + 3i
we find the quantity n3 that acts as a (constant) width, and
n
3i
that plays the role of the x coordinate xi (which is i steps of size 3/n from the origin.)
n
The expression after the summation represents the
√ area of a rectangle with base ∆x = 3/n
and height f (xi ) where f is the function f (x) = 1 + x. The sum extends over n pieces of
size 3/n so that the entire interval covered by those pieces is 0 < x < 3.
P
When we take the limit of the sum as n increases, (and as ∆x → dx) we arrive at the definite
integral
Z 3√
1 + x dx.
0
The general form will be
Z
a+3
a
q
1 + (x − a) dx.
2.6
Consider the function y = x3 . Using the spreadsheet, create one plot which contains all of
the following superimposed:
(a) a graph of this function for 0 < x < 1.
4
1.0
1.0
bar graph for y=x^3
bar graph for y=x^3
area fn
area function
y=(x^4)/4
y=(x^4)/4
0.0
0.0
0.0
1.0
0.0
1.0
Figure 3: solution to spreadsheet assignment
(b) a bar-graph showing 20 rectangular strips, whose top right corner lies on the graph of
the function.
(c) The function A(x) which adds up the areas of the first strip, the first two strips, the first
three strips.. etc. (See Figure 2.4 of the lecture notes for an example).
(d) The function g(x) = x4 /4 which is the anti-derivative of the original function.
Solution
Two acceptable answers are given in Figure 3. We see that the curve g(x) = x4 /4 is close
to but not identical to the computed area A(x) because the width of the strips is finite. (We
get the true area only in the limit as the width of the strips goes to zero.)
2.7
Determine the values of each of the following definite integrals. Each of these can be done
using simple geometry, and need no ”integration” techniques. It will be helpful to sketch
the regions and functions involved. Recall that we have formulae for areas of rectangles,
triangles,
and circles. Z
Z 2 √
Z 1
1
(1 − x)dx, (c)
2xdx, (b)
(a)
4 − x2 dx
0
−2
−1
Solution
(a) The graph of the function y = 2x is a straight line, and the interval is [0,Z1]. This is a
1
2xdx = 1.
triangular region with base 1 and height 2. Its area is A = (1/2)·1·2 = 1. Thus
0
(b) This is also a triangular region, with
base 2 and height 2. (The height is the value of y
Z 1
at x = −1.) Its area is A = 2. Thus
(1 − x)dx = 2.
−1
5
√
(c) The function is y = 4 − x2 , which can be written as y 2 +x2 = 4, (with y positive). Thus,
the region is a semi-circle of radius 2, and its area is (1/2)πr 2 = (1/2)4π = 2π. Therefore,
Z 2 √
4 − x2 dx = 2π
−2
2.8
Find the area under the function shown below between x = 1 and x = 5 by your favorite
method. You may want to use simple geometrical dissection, or to describe the functions
that make up this shape.
4
3
(2,3)
(4,2)
y2
y=f(x)
1
(3,1)
0
1
2
3
x
5
4
6
7
Figure 4: For problem 2.8
Solution
See Figure 5.
Method 1:
A1 = 1 · 3 = 3
y
(2,3)
y=−2x+7
3 y=x−2
(4,2)
2 (3,1)
1 A1
A2A3
A4
1
2
3
4
5
x
Figure 5: Problem 2.8 solution
6
A2 =
Z
3
(−2x + 7) dx = (−x
2
2
3
+ 7x)
2
= (−9 + 21) − (−4 + 14) = 12 − 10 = 2
A3 =
=
Z
4
3
(x − 2) dx = (
4
x2
− 2x)
2
3
9
16
−8 −
− 6 = 0 − (−1.5) = 1.5
2
2
A4 = 1 · 2 = 2
Area under y = f (x) from x = 1 to x = 5 is sum of four areas : A1 + A2 + A3 + A4 =
3 + 2 + 1.5 + 2 = 8.5.
Method 2:
Using the formula for the area of trapezoid Atrap = h(t + b)/2, we obtain
A2 =
1
× 1 × (1 + 3) = 2.
2
and
1
× 1 × (1 + 2) = 1.5.
2
Thus, A = A1 + A2 + A3 + A4 = 7.5 + 3 = 3 + 2 + 1.5 + 2 = 8.5.
A3 =
7
The Fundamental Theorem of Calculus
2.9
(a) Give a concise statement of the Fundamental Theorem of Calculus.
(b) Why is it a useful practical tool?
Solution
(a) The
Fundamental Theorem of Calculus states that if F (x) is any anti-derivative of f (x)
R b
then a f (x)dx = F (b) − F (a)
(b) the Fundamental Theorem of Calculus is useful since it gives us a shortcut for calculating
integrals. Without it, we would have to use tedious summation and limits to compute areas,
volumes, etc. of irregular regions.
2.10
Use the Fundamental Theorem of Calculus to compute each of the following integrals. The
last few are a little more challenging, and will require special care. Some of these integrals
may not exist. Explain why.
(a)
(d)
(g)
Z
(m)
(p)
(s)
sin(x) dx
Z
(b)
0
Z
Z
Z
(j)
π
π
4
− π4
1
2 cos(x) dx
2
(x + 1) dx
3x
1/2
dx
(k)
0
Z
Z
x1/2 dx
(q)
−1
2
(1/x) dx
(t)
0
Z
(c)
3
2
(x − 2) dx
4
2
(x + 1) dx
Z
(i)
√
(1 + x) dx
Z
(l)
1
1
2ex dx
(o)
0
1
x−2 dx
(r)
−1
Z
Z
Z
π
2
cos(x) dx
0
Z
(f)
0
4
Z
(n)
1
0
Z
Z
3
(2/x) dx
Z
Z
(h)
2 sin(x) dx
Z
0
(e)
−1
1
π
4
1
(x − 1) dx
−1
4
x1/2 dx
0
2
(3/x) dx
1
−3
x1/3 dx
−2
1
2|x| dx
−1
1
(2/x) dx
−1
Solution
(a)
Z
π
0
π
sin(x) dx = − cos(x) = − cos(π) + cos(0) = −(−1) + 1 = 2
0
8
Z
(b)
(c)
Z
(e)
Z
(f)
Z
− π4
x2
2
(x − 2) dx =
x2
2
(x − 1) dx =
1
(x2 + 1) dx = (
−1
4
(j)
Z
(k)
Z
0
0
1
0
4
1
2
(o)
Z
3
1
1
0
2 cos(x) dx =
0
π/4
4 sin(x)
0
!
3
− 2x =
2
!
1
− x −1
9
4
1
−6− +4=
2
2
2
!
1 3
(−1)2
1
+ 1 = − − = −2
= −1−
2
2
2 2
−1
+1
+ x)
1
1
8
= ( + 1) − (− − 1) =
3
3
3
3 4
2
3
(2/x) dx = 2 ln(x) = 2 ln(3)
1
1
2ex dx = 2ex = 2(e − 1)
x
0
1/3
−3
3
3
(−3)4/3 − (−2)4/3 = 1.36
dx = x4/3 =
4
4
−2
x1/2 dx does not exist as x1/2 is not defined for x < 0
−1
1
x−2 dx does not exist. In the course notes we showed that
−1
(it diverged to +∞.) In exactly the same way
+∞.
(r)
Z
√
4
= √ =2 2
2
0
−2
Z
=1−0=1
53 13
124
(x + 1) =
−
=
3
3
3
3
0
4
2
2
16
x1/2 dx = x3/2 = (4)3/2 − 0 =
3
3
3
0
1
2
3 x1/2 dx = 3 · x3/2 = 2(13/2 − 0) = 2
3
0
!
√
16
2
23
2x3/2 4
=
4
+
−
1
+
=
(1 + x) dx = x +
3
3
3
3
1
−3
Z
!
1
Z
(n)
π
4
√
1
π
= −2 cos( ) − cos(0) = −2 √ − 1 = 2 − 2
4
2
(3/x) dx = 3 ln x = 3 ln 2
1
Z
x3
3
(x + 1)2 dx =
4
(i)
Z
Z
2 cos(x) dx = 2
−1
(h)
(q)
π
4
1
Z
(p)
0
2
(g)
(m)
cos(x) dx =
π/2
sin(x)
3
Z
π/4
−2 cos(x)
0
0
(d)
(l)
2 sin(x) dx =
0
π
2
Z
Z
π
4
1
2|x|dx =
−1
Z
0
(−2x) dx +
−1
Z
x−2 dx and
0
0
2
1
0
R1
2x dx = −x −1
1
2
R0
−1
R2
0
x−2 dx did not exist
x−2 dx both diverge to
+x =1+1=2
0
(s) The problem here Ris that 1/x is not defined
at x = 0. If theR given integral existed,
R
then we would have 02 (1/x)dx = limr→0 r2 (1/x)dx. However, r2 (1/x)dx = ln(x)|2r =
9
ln(2) − ln(r),R and the limit limr→0 (ln(2) − ln(r)) does not exist (i.e., is ∞)). Hence the
limit limr→0 r2 (1/x)dx does not exist. Therefore, the given integral does not exist either.
(t)
Z
1
(2/x) dx = 2
−1
Z
1
(1/x) dx. 1/x is discontinuous at x = 0, and 1/x is unbounded on
−1
the interval [−1, 1]. Therefore this integral does not exist.
2.11
Find the following integrals using the Fundamental theorem of Calculus.
(a)
(e)
(h)
Z
Z
x
a
e
T
(b)
sec (5x) dx
1
T
a
dt
Z
2
c
Z
kt
x1/2
dx
x
A cos(ks) ds
(c)
0
(f)
(i)
Z
Z
x
Z
x
1
2
dt
1 + t2
(sin(3y)) dy
0
x
b
Ct dt
Z
(g)
(j)
m
Z
x
b
x
b
(d)
Z
x
0
1
dq
aq
3
ds
s2
3 dt
Solution
kt
kt
dt = ek |xa = k1 (ekx − eka )
a e
R
(b) 0x A cos(ks) ds = Ak sin(ks)|x0 = Ak sin(kx)
R
C m+1 x
C
(c) bx Ctm dt = m+1
t
|b = m+1
(xm+1 − bm+1 ), m 6= −1
R
Rx
x dt
m
x
b Ct dt = C b t = C ln(t)|b = C(ln(x) − ln(b)), m = −1, b 6= 0
R 1
R 1
(d) 0x aq
dq - does not exist since 0x aq
dq = a1 ln |q||x0 = a1 (ln |x|
(a)
Rx
undefined.
(e)
(f)
(g)
(h)
(i)
(j)
RT
1
1
2
T
c sec (5x) dx = 5 tan(5x)|c = 5 (tan(5T ) − tan(5c))
Rx 2
−1
x
−1
−1
1 1+t2 dt = 2 tan (t)|1 = 2 tan (x) − 2π/4 = 2 tan (x)
Rx
3
b s2 ds
RT
− ln(0)), and ln(0) is
− π/2
= 3 bx s−2 ds = 3 s−1 |xb = 3(1/b − 1/x)
√
√
√
= 2 x|Ta = 2( T − a), T ≥ a > 0
R
−1
1
a x1/2
Rx
1
x
0 sin(3y) dy = − 3 cos(3y)|0
Rx
x
b 3 dt = 3t|b = 3x − 3b
= − 31 cos(3x) +
1
3
2.12
Determine the values of the integrals shown below, this time using the Fundamental Theorem
of Calculus : (i.e. find the anti-derivative of each of the functions and evaluate at the two
endpoints.)
(a)
Z
1
2xdx
0
10
(b)
Z
1
−1
(1 − x)dx
Solution
(a)
Z
1
0
"
#
x2 1 h 2 i 1
2xdx = 2
| = x |0 = 1.
2 0
"
#
"
#
"
#
12
(−1)2
x2 1
(b)
|−1 = 1 −
− −1 −
= 2.
(1 − x)dx = x −
2
2
2
−1
Z
1
2.13
Consider the function y = f (x) = ex on the interval [0, 1].
Find the area under the graph of this function over this interval using the Fundamental
Theorem of Calculus.
Solution
A=
Z
1
0
ex dx = ex |10 = e − 1.
2.14
Use the F.T.C. to calculate 2.4. what can you say about the relationship between sigma
summation and integration?
Solution
Z
1
−1
2 − x − x2 dx = 2
Z
1
−1
1 − x2 dx
x3 1
1
1
8
= 2(x − ) = 2(1 − + 1 − ) =
3 −1
3
3
3
The value of the integral is the same as the total area of N strips where N → ∞.
2.15
(a) Use an integral to estimate the sums
√
k.
k=1
PN
(b) For N = 4 draw a sketch in which the value computed by summing the four terms is
compared to the value found by the integration. (Your sketch should show the graph of the
appropriate function and a set of steps that represent the above sum.)
Solution
√
We will take f (x) = x. Figure 6 shows N = 4 and N = 16 rectangles approximating the
area under this curve.
Let S be the sum
11
2.0
4.0
0.0
0.0
0.0
4.0
0.0
16.0
Figure 6: For problem 2.15 solution
S=
We could rewrite this as
S =1·
√
√
√
1+ 2+···+ N
√
1+1·
√
√
2 + · · · + 1 · N,
making the sum look like the total area contained
√ in N rectangular strips (of width 1) with
heights specified by the function y = f (x) = x. (See red rectangles in the Figure.) The
area under the curve would be an approximation for this sum, so that
Z
√
√
√
S = 1+ 2+···+ N ≈
N
0
Z
√
x dx =
N
0
x
1/2
x3/2 N
2 3/2
dx =
= N .
3/2 0
3
2.16
Find the area under the graphs of these functions:
(a) f (x) = 1/x between x = 1 and x = 3.
(b) v(t) = at between t = 0 and t = T , where a, T are fixed constants and a > 0.
(c) h(u) = u3 between u = 1/2 and u = 2.
Solution
3
1
dx = ln x = ln 3 − ln 1 = ln 3
(a) Area =
1 x
1
Z T
t2 T
aT 2
(b) Area =
at dt = a · =
2 0
2
0
4 !
Z 2
1 4
1
1
255
u4 2
3
2 −
=4−
=
u du = =
(c) Area =
4 1/2 4
2
64
64
1/2
Z
3
12
2.17
Find the area between the two curves y = 1 − x and
y = x2 − 1 for x >R 0. Explain the
R1
relationship of your answer to the two integrals I1 = 0 (1 − x)dx and I2 = 01 (x2 − 1)dx.
Solution
The area between the curves y = x2 − 1 and y = 1 − x for x > 0 is the area of the shaded
region (see Figure 7). Points of intersection of the curves: x2 − 1 = 1 − x ⇒ x2 + x − 2 = 0 ⇒
x = 1, x = −2. The second point does not belong to the specified region x > 0. Therefore,
the area can be calculated as follows:
S=
Z
1
0
(1 − x − (x2 − 1))dx =
Z
1
0
(−x2 − x + 2)dx
S = (−x3 /3 − x2 /2 + 2x)|10 = −1/3 − 1/2 + 2 = 7/6.
y
y=1−x
y=x2−1
1
1
x
Figure 7: For problem 2.17
Remark:
We can write the anser in the form S = 01 (1 − x)dx − 01 (x2 − 1)dx = I1 − I2 . I1 is the
area under y = 1 − x. I2 is a negative quantity (because y = x2 − 1 is under the x axis for
0 < x < 1). −I2 is the area between the x axis and this curve.
R
R
2.18
Find the area between the graphs of y = f (x) = 2 − 3x2 and y = −x2 .
Solution
The points of intersection of these two curves satisfy 2 − 3x2 = −x2 , so that 2 = 2x2 , x2 =
1, x = ±1. The region is symmetric about the y axis so the area is
2
Z
1
0
2
2
((2 − 3x ) − (−x ))dx = 2
Z
1
0
!
1
8
x3 1
=
4
1
−
= .
(2 − 2x )dx = 4 x −
3 0
3
3
2
13
2.19
(a) Find the area enclosed between the graphs of the functions y = f (x) = xn and the
straight line y = x in the first quadrant. (Note that we are considering positive values of n
and that for n = 1 the area is zero.)
(b) Use your answer in part (a) to find the area between the graphs of the functions y =
f (x) = xn , y = g(x) = x1/n in the first quadrant. (Hint: what is the relationship between
these two functions and what sort of symmetry do their graphs satisfy?)
Solution
(a) For n = 1, 2, . . . the graphs y = xn and y = x intersect at x = 0, 1. Thus the area is
Z
1
0
n
(x − x )dx =
x2
xn+1 1 1
1
n−1
−
=
.
= −
2
n+1 0 2 n+1
2(n + 1)
!
(b) The graphs y = xn and y = x1/n intersect at x = 0, 1. They form an inverse function pair
and therefore are symmetric about the line y = x. Thus the area between them is twice the
answer in (a).
2.20
A piece of tin shaped like a leaf blade is to be cut from a square sheet of tin. The shape of
one of the sides is given by the function y = x2 and the shape is to be symmetric about the
line y = x. How much tin goes into making the shape? How much is left over? Assume that
the thickness of the sheet is such that each square cm weighs one gm, and that y and x are
in meters.
1.4
y = x^2
1.2
1
y 0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
x
1
1.2
1.4
Figure 8: For problem 2.20
Solution
14
The lower edge is given by y = x2 . The shape is symmetric about y = x. Therefore, the
upper edge is given by y = x1/2 (inverse function to y = x2 ).
Amount of tin = area of shaded region=A.
A=
Z
1
x
1/2
0
dx −
Z
1
0
x2 dx = 2 ·
2−1
1
=
2(2 + 1)
3
(from problem 2.19, with n = 2).
Thus the amount of tin used is
1 2
g
10, 000
m · 10000 2 =
g.
3
m
3
The amount of tin left over is
1 m2 · 10000
g
10, 000
20, 000
−
g=
g
2
m
3
3
2.21
Find the area between the graphs of the functions y = f (x) = 2x, y = g(x) = 1 + x2 and
the y axis.
Solution
The graphs intersect tangentially at x = 1, and so the area is
Z
1
0
2
(1 + x − 2x)dx =
Z
1
0
2.22
(x − 1)3 1 1
(x − 1) dx =
= .
3
3
0
2
Find the area of the finite plane region bounded by the parabola y = 6 − x2 and y = x2 − 4x.
[from the April 97 Final Exam].
Solution
The intersection points of the two parabolas are found by setting 6 − x2 = x2 − 4x, i.e. by
solving 2x2 − 4x − 6 = 0. This equation has the two solutions x = −1 and x = 3. It follows
that the region we are looking for is bounded by the parabola y = 6 − x2 on the top, by the
parabola y = x2 − 4x on the bottom, and by two vertical lines at x = −1 and at x = 3 on
the sides. The area of this region is given by the difference
A=
Z
3
−1
2
(6 − x )dx −
Z
3
−1
(x2 − 4x)dx
A = (6x − x3 /3)|3−1 − (x3 /3 − 2x2 )|3−1
A = [(18 − 9) − (−6 + 1/3)] − [(9 − 18) − (−1/3 − 2)] = 64/3.
15
3.5
3
y = 3x^2
2.5
2
y
1.5
1
0.5
0
0.2
0.4
0.6
0.8
x
1
1.4
1.2
Figure 9: For problem 2.23
2.23
Find the area of the shape shown in Figure 9.
Solution
Area of shape shown = (Area of rectangle with vertices (0, 0), (1, 0), (0, 3), (1, 3)) − (Area
under curve y = 3x2 ) = 3 −
R1
0
2.24
Let g(x) =
Rx
0
1
3x2 dx = 3 − x3 = 3 − (1 − 0) = 2 sq. units
0
f (t)dt, where f (t) is of the function whose graph is shown.
5
4
3
y
2
f (x)
1
0
1
2
3
x
4
5
6
–1
–2
–3
Figure 10: For problem 2.24
(a) Evaluate g(0), g(1), g(2), g(3) and g(6).
(b) On what intervals is g(x) increasing?
(c) Where does g(x) have a maximum value?
16
7
(d) Sketch a rough graph of g(x).
Solution
(a)
g(0) = 0,
g(2) = g(1) +
g(3) = g(2) +
Z
Z
g(1) =
1
2dx =
0
1
2
= 2,
2 2xdx = 2 + x = 2 + 4 − 1 = 5,
(−4(x − 3))dx = 5 + (−2x
g(6) = g(3) +
1
2x
0
2
3
2
Z
Z
5
2
1
3
+ 12x)
2
Z
(−(x − 3))dx +
3
= 5 − 18 + 36 + 8 − 24 = 7,
6
(−2)dx
5
6
5
25
x2
9
⇒ g(6) = 7 + (− + 3x) − 2x = 7 −
+ 15 + − 9 − 12 + 10 = 3.
2
2
2
5
3
(b) g(x) is increasing on [0,3].
(c) g(x) has a maximum value at x = 3.
(d) See Figure 11.
7
6
5
g(x)
4
3
2
1
0
0
1
2
3
x
4
5
6
Figure 11: For problem 2.24(d)
2.25
Consider the functions shown in Figure 12(a) and (b). In each case, use the sketch
of this
Rx
function y = f (x) to draw a sketch of the graph of the related function F (x) = 0 f (t)dt.
(Assume that F (−1) = −4 in (a), and F (x) = −0.5 at the left end of the interval in (b).)
Solution
See Figure 13
17
5
1
4
x
0
3
2
–1
1
–2
–1
0
1
x
2
–3
–1
Figure 12: For problem 2.25
2.26
Consider the functions shown in Figure 14(a) and (b). In each case, use the sketch
of this
Rx
function y = f (x) to draw a sketch of the graph of the related function F (x) = 0 f (t)dt.
Solution
See Figure 15
2.27
Consider the functions shown in Figure 16(a) and (b). In each case, use the sketch
of this
R
function y = f (x) to draw a sketch of the graph of the related function F (x) = 0x f (t)dt.
(Assume that F (0) = 0 in both (a) and (b).)
Solution
See Figure 17
18
1
4
x
0
2
–1
–1
2
1
x
–2
–2
–3
–4
Figure 13: For solution to problem 2.25
0.6
0.4
0.4
0.2
0.2
–1
–0.5
0.5
x
1
–0.2
–1
–0.5
0
–0.4
–0.2
Figure 14: For problem 2.26
19
0.5
x
1
0.6
0.4
0.4
0.2
0.2
–0.5
–1
0.5
1
x
–0.2
0
–0.5
–1
0.5
1
x
–0.4
–0.2
Figure 15: For solution to problem 2.26
8
2
6
4
1
2
0
0
2
4
x
6
8
10
–1
–2
–4
–2
Figure 16: For problem 2.27
20
2
4
x
6
8
3
8
2
6
4
1
2
0
0
2
4
x
6
8
2
10
–2
–1
–4
–2
–6
Figure 17: For solution to problem 2.27
21
4
x
6
8
10