Download Final Exam Review Solutions - Calculus II | MATH 1132 and more Exams Calculus in PDF only on Docsity!
Math 1132Q: Fall 2009 Final Exam Review Solutions
- Use the definition of the integral to evaluate
∫ (^2) 0
(3 − x) dx By the Theorem 4 of section 5.2 on page 368, ∫ (^2) 0
(3 − x) dx = limn→∞
∑^ n i=
(3 − xi)∆x
where ∆x =
n
and xi = 0 + i∆x = 2 i n
. Thus we have ∫ (^2) 0
(3 − x) dx = (^) nlim→∞
∑^ n i=
2 i n
n = limn→∞
n
∑^ n i=
2 i n ) = limn→∞(
n
∑^ n i=
n^2
∑^ n i=
i)
= (^) nlim→∞(
n n −
n^2
n(n + 1) 2 ) = lim n→∞(6 − 2(n^2 + n) n^2 ) = lim n→∞(6 − 2 −
n
We can check our result as follows ∫ (^2) 0 (3 − x) dx = 3x − x^2 2
∣∣ ∣∣ ∣
2
0
- Find dgdx , where g(x) =
∫ (^0) 3 x
u^2 − 1 u^2 + 1 du
Note that g(x) = −
∫ (^3) x 0
u^2 − 1 u^2 + 1 du, so
dg dx
(3x)^2 − 1 (3x)^2 + 1
27 x^2 − 3 9 x^2 + 1
∫ sin^3 (θ) cos^5 (θ) dθ ∫ sin^3 (θ) cos^5 (θ) dθ =
∫ sin^3 (θ) cos^4 (θ) d(sin(θ)) =
∫ sin^3 (θ)(1 − sin^2 (θ))^2 d(sin(θ))
=
∫ u^3 (1 − u^2 )^2 du =
∫ (u^3 − 2 u^5 + u^7 ) du = u^4 4
u^6 3
u^8 8
+ C
sin^4 (θ) 4
sin^6 (θ) 3
sin^8 (θ) 8
+ C
∫ x sin^2 x dx
∫ x sin^2 x dx =
∫ x · 1 − cos 2x 2 dx =
∫ (x − x cos(2x)) dx
=
x^2 2 − (x · sin(2x) 2
∫ (^) sin(2x) 2 dx)) (integration by parts)
= x^2 4
x sin(2x) 4
cos(2x) 8
+ C
∫ x^5 e−x 3 dx Let u = x^3 , so that du = 3x^2 dx, ∫ x^5 e−x 3 dx =
∫ ue−u^ du =
(−ue−u^ +
∫ e−u^ du) =
(−ue−u^ − e−u) + C
=
(−x^3 e−x 3 − e−x 3 ) + C
∫ (^3) x (^2) − 2 x^2 − 2 x − 8 dx
∫ (^3) x (^2) − 2 x^2 − 2 x − 8
dx =
∫ 3 + 6 x + 22 x^2 − 2 x − 8
dx = 3x +
∫ (^6) x + 22 (x − 4)(x + 2)
dx
= 3 x +
3(x − 4)
3(x + 2)
dx = 3x +
ln(x − 4) −
ln(x + 2) + C
- Find the volume of the solid obtained by rotating the region bounded by y = x and y =
x about the x-axis. The curve y = x and y =
x intersect at the points (0, 0) and (1, 1). The curve y =
x is on the top of the region and the curve y = x is on the bottom of the region.
V =
∫ (^1) 0
A(x) dx =
∫ (^1) 0
π(x − x^2 ) dx = π ( x^2 2
x^3 3
∣∣ ∣∣ ∣
1
0
π 6
- A heavy rope, 50ft long, weighs 0.5lb/ft and hangs over the edge of a building 120ft high. How much work is done in pulling half the rope to the top of the building?
W = lim n→∞
∑^ n i=
- 5 x∗ i ∆x + 0. 5 ·
∫ (^25) 0
- 5 x dx +
x^2 4
∣∣ ∣∣ ∣
25
0
ft-lb
- What is the average value of h(x) = cos^4 (x) sin(x) on the interval [0, π]?
havg =
π
∫ (^) π 0
cos^4 (x) sin(x) dx = −
π
∫ (^) π 0
cos^4 (x) d(cos(x))
= −
π
cos^5 (x) 5
∣∣ ∣∣ ∣
π
0
5 π
∫ (^1) 0
x^2 + 1 dx Let x = tan θ, ∫ (^1) 0
x^2 + 1 dx =
∫ π 4 0
√ tan^2 (θ) + 1 sec^2 (θ) dθ =
∫ π 4 0 sec^3 (θ) dθ
sec(θ) tan(θ) + ln | sec(θ) + tan(θ)| 2
∣∣ ∣∣ ∣
π 4
0
2 + ln(
- Does the sequence converge? If so, find its limit. an = cos^2 n 2 n Note that 0 ≤ cos^2 n 2 n^
2 n and limn→∞
2 n^ = 0, by the Squeeze Theorem (of the section 11.1 on page 679), the sequence converges to 0.
Do the following series converge? If you can find the sum, do so.
∑^ ∞ n=
1 + 2n 3 n =
∑^ ∞ n=
)n
∑^ ∞ n=
)n
∑^ ∞ n=
(−1)n^ n ln n
diverges by the Test for Divergence
- ∑∞ n=1 ne−n^ converges by the Integral Test
∑^ ∞ n=
n^2 + 4 converges by the Comparison Test with the p-series (p = 2)
∑^ ∞ n=
n(n + 1) converges by the Comparison Test with the p-series (p = 2)
∑^ ∞ n=
n^2 + 1
diverges by the Comparison Test with the harmonic series (p = 1)
∑^ ∞ n=
(−1)n−^1 2 n n^4 diverges by the Test for Divergence
∑^ ∞ n=
sin
n
) diverges by the Limit Comparison Test with the harmonic series
∑^ ∞ n=
( n^2 + 1 2 n^2 + 1
)n converges by the Root Test
Find the interval of convergence of the power series.
∑^ ∞ n=
(n!)(2x − 1)n
Let an = (n!)(2x − 1)n. Then ∣∣ ∣∣^ an+ an
∣∣ ∣∣ = (n + 1)| 2 x − 1 | → ∞, as n → ∞
By the Ratio Test, the series diverges when x 6 = 1/2. Thus the series converges only when x = 1/2.
∑^ ∞ n=
(x − 2)n nn Let an = (x − 2)n nn^
. Then √ n |an| = |x − 2 | n
→ 0 , as n → ∞ By the Root Test, the series converges for all values of x. So the interval of convergence is (−∞, ∞)
∑^ ∞ n=
(x − 2)n n^2 + 1 Let an = (x − 2)n n^2 + 1
. Then ∣∣ ∣∣^ an+ an
∣∣ ∣∣ = n
(n + 1)^2 + 1 |x − 2 | → |x − 2 |, as n → ∞
By the Ratio Test, the series converges when |x − 2 | < 1 ⇔ 1 < x < 3. When x = 1, the series is convergent by the Alternating Series Test, and when x = 3, the series is also convergent by the Comparison Test. So the interval of convergence is [1, 3].
- Find the Taylor series for f (x) = ex^ centered at a = 3.
∑^ ∞ n=
f (n)(a) n! (x − a)n^ =
∑^ ∞ n=
f (n)(3) n! (x − 3)n^ =
∑^ ∞ n=
e^3 n! (x − 3)n
- Find a power series representation for f (x) = ln(5 − x)
ln(5 − x) = −
5 − x dx = −
1 − x/ 5 dx = −
∫ (
∑^ ∞ n=
x 5 )n) dx
∑^ ∞ n=
∫ (^) xn 5 n^ dx = −
5 n+
∑^ ∞ n=
xn+ n + 1
5 n+
∑^ ∞ n=
xn n
Put x = 0 in this equation and obtain C = ln 5.Thus
f (x) = ln(5 − x) = −
5 n+
∑^ ∞ n=
xn n
- Find the Maclaurin series for f (x) = cos(3x) Note that f ′(x) = −3 sin(3x), f ′′(x) = − 32 cos(3x), f (3)(x) = 3^3 sin(3x) and f (4)(x) = 3^4 cos(3x), so f (0) = 1,f ′(0) = 0, f ′′(0) = − 32 , f (3)(0) = 0 and f (4)(0) = 3^4 , so the Maclaurin series is ∑^ ∞ n=
(−1)n^ 32 nx^2 n (2n)!
