MATH 101 HOMEWORK 3 – SOLUTIONS
1. We know that the average value of f (x) = x3 + 1 on some interval [a, b] is 9. Prove
that b ≥ 2. Can we say anything about a?
By the Mean Value Theorem, the average value of f is equal to f (c) for some c in [a, b].
That is, c3 + 1 = 9 for some c in [a, b]. But if c3 + 1 = 9 then c = 2, hence 2 belongs to
the interval [a, b], i.e. a ≤ 2 ≤ b.
Z
0
sin x
2
ex
2. Find F (x), if F (x) =
+t2
dt.
0
Z
Z sin x
2
d x2
e
dt =
et dt
e ·
dx
0
0
Z sin x
Z sin x
2
2
2
2
d
= (ex )0 ·
et dt + ex ·
et dt
dx 0
0
Z sin x
2
2
2
x2
= 2x e ·
et dt + ex · esin x · cos x.
d
F (x) =
dx
0
sin x
x2 +t2
0
3. Evaluate the integrals:
Z 3π/2
Z π
Z
5
5
(a)
| sin x| dx =
sin x dx −
0
0
3π/2
sin5 x dx. We use the substitution u = cos x,
π
du = − sin x dx to integrate
Z
Z
Z
5
4
sin x dx = sin x sin x dx = (1 − cos2 x)2 sin x dx
Z
=−
Z
(1 − u ) du = −
2 2
(1 − 2u2 + u4 ) du = −u +
= − cos x +
u5
2u3
−
+C
3
5
2 cos3 x cos5 x
−
+ C.
3
5
Therefore
Z 3π/2
2 cos3 x cos5 x π
2 cos3 x cos5 x 3π/2
5
−
) + (cos x −
+
)
| sin x| dx = (− cos x +
3
5
3
5
0
π
0
= (1 −
Z
(b)
1
2
x2 − x + 1
√
dx =
x
2 1
2 1
8
2 1
+ ) − (−1 + − ) + 0 − (−1 + − ) = .
3 5
3 5
3 5
5
Z
2
x
1
3/2
−x
1/2
+x
−1/2
2
2 5/2 2 3/2
1/2 dx = ( x − x + 2x )
5
3
1
√
34 2 − 11
≈ 2.4722.
=
15
1
Z
Z
dx
(c)
=
x
e + e−x
x
stitution u = e .
ex dx
=
e2x + 1
Z
u dx
= tan−1 u + C = tan−1 (ex ) + C, using the sub2
u +1
4. Evaluate the areas of the following planar regions:
(a) the region bounded from below by the lines y = x and y = −x, and from above by the
parabola y = 2 − x2 :
Observe first that the region is symmetric with respect to the y axis. We need to find the
intersection points. Set 2 − x2 = x, then x2 + x − 2 = 0, x = 1 (the other root x = −2 is
negative). Thus the area in question is
Z
1
(2 − x2 − x) dx = (2x −
2
0
x2 1
7
x3
1 1
− ) = 2 − − = .
3
2 0
3 2
6
(b) the region bounded by the half-parabola y = x2 , x ≥ 0; the half-parabola y = (x − 2)2 ,
x ≥ 2; and the horizontal lines y = a2 and y = b2 :
This can be done without any calculus at all! Cut off a piece on the right and move it to
the left of the region to get a perfect rectangle of sidelengths 2 and b2 − a2 (we assume
that b > a). Thus the area is 2(b2 − a2 ).
If you do want to use calculus, it is better to interchange the roles played by the x and y
√
√
variables – so that the two parabolas have equations x = y and x = 2 + y. The area is
then given by
Z b2
2
√
√
(2 + y − y)dy = 2y|ba2 = 2(b2 − a2 ).
a2
2