Download Math133 Midterm Exam Practice Solutions: Diff. Equations, Sequences, Integrals and more Exams Calculus in PDF only on Docsity!
Third Midterm Exam–Practice Test, Math
- Solve the following differential equation
dy
dx
y cos
y
with the initial condition y(0) = 1.
- State whether the following sequences of numbers {an}n≥ 1 converge or
diverge. If they converge, find the limit.
(a) an =
√n 2 n + 1
(b) an = (−3)n
- Compute the integral and simplify as much as possible
∫ √ 9 − w^2
w^2
dw
(Hint: Use the identity 1 + cot^2 θ = 1/ sin 2 θ and compute also (cot θ)′)
- Determine whether the following improper integral converges. State the
method you have used. ∫ (^) ∞
1
ex^ − x
dx
- Compute the integral ∫ (^1)
0
ln x dx
Solutions:
- Separating the variables yields
∫ dy √ y cos^2
y
dx
Substituting u =
y, du = 1 2 √ y dy^ we obtain
∫ dy √ y cos^2
y
sec 2 u du = 2 tan u + C = 2 tan
y + C
and 2 tan
y = x + C.
Solving for y we get
y =
tan
− 1 (
x
2
+ C)
The initial condition y(0) = 1 leads to
±1 = tan − 1 (C) and C = ± tan(1)
so we get two solutions.
- (a) We may consider the function f (x) =
√x 2 x + 1 and compute its limit
for x → +∞. We first take the logarithm g(x) = ln f (x) =
ln (2x+1) x and use l’Hopital’s rule. We have
lim x→+∞
2 x + 1
and therefore limx→+∞ g(x) = 0, i.e. limx→+∞ f (x) = 1. Hence
lim n→+∞
√ n 2 n + 1 = 1.
(b) This sequence has no limit since |an| → +∞ and signs of the numbers an alternate.
- Substitute w = 3 sin θ, − π 2 ≤ θ ≤ π 2 so that dw = 3 cos θ dθ and
9 − w^2 = 3 | cos θ| = 3 cos θ. We obtain
∫ √ 9 − w^2
w^2
dw =
3 cos θ
9 sin 2 θ
· 3 cos θ dθ =
cot 2 θ dθ.
Note that sin
2 θ + cos 2 θ = 1 implies
1 + cot 2 θ =
sin 2 θ
and therefore 1 √ ex^ − x
ex/^2
2 e −x/ 2 .
Hence ∫ (^) ∞
1
ex^ − x
dx ≤
1
2 e −x/ 2 dx
2 lim b→+∞
∫ (^) b
1
e −x/ 2 dx
2 lim b→+∞
(e − 1 / 2 − e −b/ 2 )
2 e − 1 / 2 < +∞.
This means that the integral
1
√^1 ex^ −x dx converges.
- We first compute the integral of the logarithm function using integration
by parts: (^) ∫
ln x dx =
1 · ln x dx = x ln x − x + C.
Then ∫ (^1)
0
ln x dx = lim a→ 0
a
ln x dx
= lim a→ 0
(x ln x − x)| 1 a
= − 1 − lim a→ 0
(a ln a)
= − 1 − lim a→ 0
ln a 1 a
= − 1 − lim a→ 0
1 a − 1 a^2 = − 1
