Calculus 2, Quiz 1
40 Mininutes. Please check your answers before you turn in your answer sheet. No partial credit if the answer is wrong.
(1) (6 Points) Evaluate
n→∞lim
1
n + 1+ 1
n + 2+ · · · + 1 n + n
.
Sol:
The sum inside the limit is the Riemann sum of 1/(1 + x) with respect to the partition {k/n : 1 ≤ k ≤ n} :
1
n + 1+ 1
n + 2+ · · · + 1 n + n =
n
X
k=1
1 n· 1
1 +nk. Hence the limit is convergent to the Riemann integral
Z 1 0
dx
1 + x = ln(1 + x)|10= ln 2.
(2) (11 Points) Evaluate Z
p(x − 1)(x − 3)dx.
Sol:
After completing the square, we have x2− 4x + 3 = (x − 2)2− 1. We let p(x − 1)(x − 3) = tan θ.
Then x − 2 = sec θ or x = 2 + sec θ. Thus dx = sec θ tan θdθ. The original integral becomes R tan2θ sec θdθ. We have seen in class (think aboutR sec3θdθ) that
Z
tan2θ sec θdθ = 1
2tan θ sec θ −1
2ln | tan θ + sec θ| + C.
Thus Z
p(x − 1)(x − 3)dx = 1
2(x − 2)p
(x − 1)(x − 3) −1 2ln |p
(x − 1)(x − 3) + (x − 2)| + C.
(3) (7 Points) Evaluate
Z ln x x
2 dx.
Sol:
Write
Z ln x x
2 dx =
Z
ln2x ·dx x2 =
Z ln2xd
−1 x
= −ln2x
x +
Z 1 xd ln2x
= −ln2x
x +
Z 1
x· 2 ln x · 1 xdx
= −ln2x x + 2
Z ln x x2 dx.
1
2
Now we only need to computeR ln x/x2. Similarly, Z ln x
x2 dx = Z
ln xd
−1 x
= −ln x x +
Z 1 xd ln x
= −ln x x +
Z 1 x· 1
xdx
= −ln x x −1
x+ C.
(4) (12 Points) Find the partial fraction expansion of the following given rational functions.
(a) x3+ 1
x3− 5x2+ 6x = A +B x + C
x − 2+ D x − 3. We know
x3+ 1
x3− 5x2+ 6x = 1 + 5x2− 6x + 1 x3− 5x2+ 6x Hence A = 1. We see that
5x2− 6x + 1 = B(x − 2)(x − 3) + Cx(x − 3) + Dx(x − 2).
Plugging x = 0, we obtain 6B = 1. Hence B = 1/6. Plugging x = 2, we obtain −2C = 9.
Hence C = −9/2. Plugging x = 3, we obtain 3D = 28. Thus D = 28/3.
(b)
x
x2− 3x + 2
2
= A
x − 1+ B
(x − 1)2 + C
(x − 2)+ D (x − 2)2 We write
x
x2− 3x + 2 = a
x − 1+ b x − 2.
Hence a(x − 2) + b(x − 1) = x. Plugging b = 1, we find a = −1 and plugging x = 2, we find b = 2. Hence
x
x2− 3x + 2 = −1 x − 1+ 2
x − 2. Taking the square of both side of the equation, we have
x
x2− 3x + 2
2
= 1
(x − 1)2 + 4
(x − 2)2+ −4 (x − 1)(x − 2). Now,
−4
(x − 1)(x − 2) = c
x − 1 + d x − 2.
We find c(x − 2) + d(x − 1) = −4. Plugging x = 1, we find c = 4 and plugging x = 2, we find d = −4. Thus
−4
(x − 1)(x − 2) = 4
x − 1 + −4 x − 2. Therefore
x
x2− 3x + 2
2
= 4
x − 1+ 1
(x − 1)2 + −4
x − 2+ 4 (x − 2)2. (c) x2+ 5x + 4
x4+ 5x2+ 4 = Ax + B
x2+ 1 +Cx + D x2+ 4 .
Observe that (x2+ 4) − (x2+ 1) = 3. Hence 1
3
3
(x2+ 1)(x2+ 4) = 1 3
(x2+ 4) − (x2+ 1) (x2+ 1)(x2+ 4) = 1
3
1
x2+ 1 − 1 x2+ 4
.
3
In other words,
1
(x2+ 1)(x2+ 4) =1 3
1
x2+ 1 − 1 x2+ 4
. Multiplying both side of the equation by x2+ 5x + 4, we see that
x2+ 5x + 4 x4+ 5x2+ 4 =1
3
x2+ 5x + 4
x2+ 1 −x2+ 5x + 4 x2+ 4
. Notice that
x2+ 5x + 4
x2+ 1 = 1 +5x + 3
x2+ 1, x2+ 5x + 4
x2+ 4 = 1 + 5x x2+ 4. Thus
x2+ 5x + 4 x4+ 5x2+ 4 =1
3
5x + 3 x2+ 1 − 5x
x2+ 4
. (5) (14 Points)Evaluate the following improper integrals.
(a) (7 Points) Z ∞
2
dx x2+ x − 2. Write
1
x2+ x − 2 =1 3
1
x − 1− 1 x + 2
. (We know (x + 2) − (x − 1) = 3.) Thus
Z dx
x2+ x − 2 =1
3(ln |x − 1| − ln |x + 2|) + C =1 3ln
x − 1 x + 2
+ C.
For M > 2, we have Z M 2
dx
x2+ x − 2 =1 3
lnM − 1 M + 2− ln1
4
. Thus
Z ∞ 2
dx
x2+ x − 2 = lim
M →∞
Z M 2
dx
x2+ x − 2 = 1 3ln 4 (b) (7 Points)
Z 1 0
dx (2 − x)√
1 − x. (Hint: use t =√ 1 − x.) Write t =√
1 − x. Then 2 − x = 1 + t2and dt = −dx 2√
1 − x. Thus Z b
0
dx (2 − x)√
1 − x = Z
√1−b
1
−2dt 1 + t2
= 2 Z 1
√1−b
dt
1 + t2 = 2 tan−1t|1√1−b
= 2π
4 − tan−1√ 1 − b
. Hence
Z 1 0
dx (2 − x)√
1 − x = lim
b→1
Z b 0
dx (2 − x)√
1 − x =π 2. (6) (Bonus 15 Points) Evaluate
Z 1 0
tan−1x 1 + x dx.
