MATH 101 MIDTERM 2 – SOLUTIONS
1. Evaluate the following integrals:
Z
Z 3
A
C B
(a)
+
dx,
dx
=
+
(x − 1)2 (x + 2)
x − 1 (x − 1)2 x + 2
A(x + 2)(x − 1) + B(x + 2) + C(x − 1)2 = 3,
A(x2 + x − 2) + B(x + 2) + C(x2 − 2x + 1) = 3,
(A + C)x2 + (A + B − 2C)x + (−2A + 2B + C) = 3,
hence A + C = 0, C = −A, A + B − 2C = 3A + B = 0, B = −3A, and
−2A + 2B + C = −2A − 6A − A = −9A = 3. Hence A = −1/3, B = 1,
C = 1/3, and
Z
Z 3
−3
1
1
dx
=
+
+
dx
(x − 1)2 (x + 2)
3(x − 1) (x − 1)2 3(x + 2)
1
1
1
= − ln |x − 1| −
+ ln |x + 2| + C.
3
x−1 3
Z
Z
Z
dx
dx
du
=
=
, where we sub2
3/2
2
3/2
2
(x + 4x + 8)
((x + 2) + 4)
(u + 4)3/2
stituted u = x + 2, du = dx. Now let u = 2 tan θ, du = 2 sec2 θ dθ, then
u2 + 4 = 4(tan2 θ + 1) = 4 sec2 θ, and
(b)
Z
du
=
2
(u + 4)3/2
=
Z
2 sec2 θ dθ
=
(4 sec2 θ)3/2
Z
dθ
=
4 sec θ
Z
cos θ
dθ
4
sin θ
u
x+2
+C = √
+ C.
+C = √ 2
2
4
4 4+u
4 x + 4x + 8
(c) Integrate by parts with f = x, f 0 = 1, g 0 = sec2 x, g = tan x:
Z
x sec2 x dx = x tan x −
Z
tan x dx = x tan x − ln | sec x| + C.
Z
2. Determine whether the integral
0
divergent.
π/2
sin
3/2
cos x
dx is convergent or
x + sin1/2 x
Substitute u = sin x, du = cos x dx, then the integral above becomes
Z
1
0
Z 1
du
du
≤
.
3/2
1/2
u +u
0 u1/2
Since the second integral is convergent (1/2 < 1), the first one is also convergent.
(If we substitute u2 = sin x instead, then with just a little more effort we can
compute the exact value of the integral. It turns out to be π/2.)
Z
3. Prove that the integral
0
∞
e−2x
dx is convergent and has value less than
x2 + 1
1/2.
We have x2 + 1 > 1 for all x > 0, so
Z
∞
0
e−2x
dx <
x2 + 1
Z
∞
0
e−2x dx = −
e−2x ∞ 1
= .
2 0
2
R
4. We want to evaluate numerically the integral 01 f (x) dx, given that f (0) =
1.0, f (1/6) = 1.1, f (2/6) = 1.2, f (3/6) = 1.4, f (4/6) = 1.7, f (5/6) = 1.8,
f (1) = 1.8. Write out (but do not evaluate) the Simpson’s approximation S6 .
We have n = 6, h = 1/6. Hence
S6 =
1
(1.0 + 4 · 1.1 + 2 · 1.2 + 4 · 1.4 + 2 · 1.7 + 4 · 1.8 + 1.8).
3·6
5. We do not know the exact formula for the function f , but we can compute
numerically its values as needed.
We know that |f 00(x)| ≤ x2 + x + 2 for all x
Z
2
f (x)dx using the midpoint approximation
in [0, 2]. We want to evaluate
0
Mn , with accuracy 10−4 or better. How large does n need to be?
We can take K = 22 + 2 + 2 = 8. Thus we need to have
8 · 23
8
≤ 10−4 , n2 ≥ · 104 , n ≥
2
24n
3
s
8
· 102 .
3