MA1S11 (Dotsenko) Solutions to Tutorial/Exercise Sheet 10
Week 12, Michaelmas 2013
1. Evaluate the indefinite integrals
√
Z
(2 + 3 x)20
√
a)
dx
x
b)
Z
√
x3 1 + 4x dx
√
Solution. a) Let us do the u-substitution u = 2 + 3 x, so that
3
du
= √ ,
dx
2 x
and
3
du
dx = √ dx,
dx
2 x
du =
making the integral become
√
√
Z
Z
(2 + 3 x)20
2u21
2
2(1 + x)21
20
√
u du =
dx =
+C =
+C
x
3
63
63
For b), we use u = 1 + 4x so du = 4 dx, and dx = 14 du. Plus, we have to substitute
x = 41 (u − 1) back in to the equation, the equation has x3 in it and
3
x =
1
(u − 1)
4
3
=
1 3
(u − 3u2 + 3u − 1)
64
(1)
giving
√
Z
√
1
(u3 − 3u2 + 3u − 1) u du =
x 1 + 4x dx =
256
Z
1
u9/2
3u7/2 3u5/2 u3/2
=
(u7/2 − 3u5/2 + 3u3/2 − u1/2 ) du =
−
+
−
+C =
256
1152
896
640
384
(1 + 4x)9/2 3(1 + 4x)7/2 3(1 + 4x)5/2 (1 + 4x)3/2
−
+
−
+C
=
1152
896
640
384
Z
3
2. Evaluate the integrals
a)
Z
0
π/2
sin x dx
b)
Z
1
2
2
y −y
1
−3
dy
and
Z
2
1
y 2 − y −3 dy
a) is fairly obvious: we know an antiderivative of sin x, that is − cos x
Solution.
hence
π/2
Z
0
sin xdx = − cos x]π/2
= − cos
0
π
+ cos 0 = −0 + 1 = 1.
2
For b), we know an antiderivative of y 2, it is y 3/3 and of y −3, it is − y 2 , hence
3
2
Z 2
y
1
8 1 1 1
47
2
−3
y − y dy =
+ 2
= + − − =
3
2y
3 8 3 2
24
1
1
−2
and
Z
1
2
2
y −y
−3
dy =
1
y3
+ 2
3
2y
1
2
3. Evaluate the integrals
Z 5
a)
(3 + 2w)(3w + w 2 )5 dw
=
1 1 8 1
47
+ − − =−
3 2 3 8
24
b)
−1
Z
π/2
cos xesin x dx
−π
Solution. For a), let us consider the u-substitution u = 3w + w 2 , so that du =
(3 + 2w) dw. When w = −1 by substituting back in we have u = −2 and when
w = 5, u = 40, hence
6 40
Z 5
Z 40
u
406 − 64
2 5
5
(3 + 2w)(3w + w ) dw =
u du =
=
6 −2
6
−1
−2
For b) let u = sin x, so that du = cos x dx. When x = −π we have u = 0 and when
x = π/2, we have u = 1, giving
Z π/2
Z 1
sin x
cos xe
dx =
eu du = eu ]10 = e − 1.
0
−π
4. Evaluate the integral
Z
2
−1
Solution. We note that the function
integral as
Z
2
−1
Z
p
2 + |x| dx =
=
0
−1
√
p
2 + |x| dx.
p
2 + |x| is piecewise defined, and evaluate our
0
−1
Z
p
2 + |x| dx +
Z
0
2
p
2 + |x| dx =
0
2
2
2
3/2
3/2
2 − x dx +
2 + x dx = − (2 − x)
+ (2 + x)
=
3
3
0
−1
0
√
2
2
2
2
16 8 √
2 + 2 3.
= − 23/2 + 33/2 + 43/2 − 23/2 =
−
3
3
3
3
3
3
Z
2
√
2
5. Show that
Z
2
1/2
1
1
dx = 0
sin x −
x
x
Solution. Let us consider the u-substitution u = 1/x, so that du = − x12 dx, that
is dx = − u12 du. We note that for x = 1/2, we have u = 2, and for x = 2, we have
u = 1/2, so
Z
2
1/2
Z 1/2
Z 2
1
1
1
1
1
1
dx =
du,
u sin
sin x −
−u
− 2 du = −
sin u −
x
x
u
u
u
2
1/2 u
so this integral is equal to its negative, and therefore vanishes.
3