Calculus Final Exam

Calculus Final Exam

1. Find the following limit.

(a) (8 points) lim

x→0

· 1

ln(1 + x)−1 x

¸ .

Solution: lim

x→0

x − ln(1 + x) xln(1 + x) = lim

x→0

1 − 1 1 + x ln(1 + x) + x

1 + x

= lim

x→0

x

(1 + x)ln(1 + x) + x

= lim_x→0 1

1 + ln(1 + x) + 1 =1 2.

(b) (8 points) lim

x→∞

¡e^x+ 1¢_1/x .

Solution: lim

x→∞

¡e^x+ 1¢_1/x

= lim

x→∞eln^(ex+1)1/x= e^limx→∞

ln(e^x+ 1)

x = e^limx→∞

e^x e^x+ 1

= e^limx→∞ e^x e^x = e¹.

2. (8 points) Let f (x) = x^xx, for x > e. Find f⁰(x).

Solution: ln f (x) = x^xlnx and lnln f (x) = xlnx + lnlnx. Therefore, 1

ln f (x)· f⁰(x)

f (x) = lnx +1+ 1 lnx·1

x, and f⁰(x) = x^xx· x^x· lnx · [1 + 1

xlnx+ lnx].

3. Determine whether the improper integral converges; explain it.

(a) (6 points) Z _∞

1

lnx x² dx

Solution: (Method 1)Since Z _∞

1

lnx

x² dx = −lnx x |^∞₁ +

Z _∞

1

1

x²dx = −lnx x −1

x|^∞₁ = 1, the integral converges.

(Method 2) Since lim

x→∞

lnx/x² 1/x^3/2 = lim

x→∞

lnx

x^1/2 = lim

x→∞

1 x ·1

2x^−1/2

= lim

x→∞

1

2x^1/2 = 0, and Z _∞

0

1

x^3/2dx = −2

x^1/2|^∞₁ = 2 <∞, Z _∞

1

lnx

x² dx converges by the limiting comparison test.

(b) (6 points) Z _∞

1

sin(1/x) dx

[Hint: Note that lim

x→∞

sin(1/x)

1/x = lim

x→0⁺

sin x x .]

Calculus Final Exam

Solution: Since lim

x→∞

sin(1/x) 1/x = lim

x→0⁺

sin x

x = 1, and^R₁^∞1

xdx = lnx|^∞₁ =∞, Z _∞

1

sin(1/x) dx =∞ diverges by the limiting comparison test.

(c) (6 points) Z ₁

0

sin x x dx

[Hint: Note that the function sin x

x defined on (0, 1] can be extended to a continuous function g(x) =

(

sin x/x, x 6= 0

1, x = 0 on [0, 1].]

Solution: Since lim

x→0g(x) = lim

x→0

sin x

x = 1 = g(0), g(x) is continuous on [0, 1], and the function G : [0, 1] → R defined by G(a) =^R_a¹g(x)dx is continuous on [0, 1] by the Fundamental Theo- rem of Calculus.

Now Z ₁

0

sin x

x dx = lim

a→0⁺

Z ₁

a

sin x

x dx = lim

a→0⁺

Z ₁

a

g(x) dx = lim

a→0⁺G(a) = G(0) exists, the im- proper integral

Z ₁

0

sin x

x dx converges.

4. Let f (x) = (

e^−1/x2, x 6= 0

0, x = 0.

(a) (4 points) Show that f is continuous at 0.

Solution: Since lim

x→0f (x) = lim

x→0e^−1/x2 = 0 = f (0), f is continuous at x = 0.

(b) (4 points) Show that f is differentiable at 0.

Solution: Since lim

x→0⁺

f (x) − f (0)

x − 0 = lim

x→0⁺

e^−1/x2

x = lim_y→∞e^−y2

1/y = lim_y→∞ y e^y2

= limy→∞ 1

2ye^y2 = 0, and lim

x→0⁻

f (x) − f (0)

x − 0 = lim

x→0⁻

e^−1/x2

x = limy→−∞e^−y2

1/y = limy→−∞ y e^y2

= lim_y→−∞ 1

2ye^y2 = 0, we have lim

x→0

f (x) − f (0)

x − 0 = 0 and f is differentiable at x = 0.

5. Find the integral.

(a) (6 points) Z

cos(lnx)dx

Solution: Since Z

cos(lnx)dx = x cos(lnx) + Z

sin(lnx)dx = x cos(lnx) + x sin(lnx)

−^Rcos(lnx)dx, we have Z

cos(lnx)dx = 1

2[x cos(lnx) + x sin(lnx)] +C.

(b) (6 points)

Z e^x

e^2x+ 5e^x+ 6dx

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Calculus Final Exam

Solution: Setting u = e^x, we have

Z e^x

e^2x+ 5e^x+ 6dx =

Z 1

u²+ 5u + 6du

=

Z ¡ 1

u + 2− 1 u + 3

¢du = ln|u + 2| − ln|u + 3| +C = ln(e^x+ 2) − ln(e^x+ 3) +C,

or = ln¡e^x+ 2 e^x+ 3

¢+C.

(c) (6 points)

Z 1

(x²+ 1)²dx

Solution: Setting x = tanθ^{, we get}

Z 1

(x²+ 1)²dx =

Z sec²θ

(1 + tan²θ)²dθ ⁼

Z 1

sec²θ^dθ

= Z

cos²θ^dθ ⁼

Z 1 + cos 2θ

2 dθ ⁼ θ

2 +sin 2θ

4 + C = 1

2tan⁻¹x + x

2(1 + x²)+ C, where we have use that sin 2θ ^{= 2 sin}θ^{· cos}θ ^{= 2√ x}

1 + x²· 1

√1 + x².

6. (8 points) Find the critical numbers and determine where f achieves its maximum values and where f achieves its minimum values. DO NOT NEED TO FIND EXTREME VALUES.

f (x) =^R₀^xt(t − 1)²(t + 1)³dt.

Solution: Since f⁰(x) = x(x − 1)²(x + 1)³, x = 0, 1, −1 are critical numbers.

Since f⁰(x) is

(> 0 in (−∞− 1) ∪ (0, 1) ∪ (1,∞),

< 0 in (−1, 0).

Hence, f (−1) is a local maximum value of f , and f (0) is a local minimum value of f .

7. (8 points) Determine the centroid of the region bounded above by y = x and below by y = x².

Solution: Setting x = x², we get x = 0, or 1. Then the area A of the region is A =^R₀¹(x − x²)dx = x²

2 −x³ 3|¹₀=1

6, Ax =^R₀¹x(x − x²)dx = x³

3 −x⁴

4|¹₀= 1 12, and Ay =^R₀¹x + x²

2 (x − x²)dx =x³ 6 − x⁵

10|¹₀= 1 15. Hence, the centroid is at (x, y) = (1

2,2 5).

8. (8 points) Let {an}^∞_n=1 be a bounded increasing sequence of real numbers. Prove that lim

n→∞anexists.

[Hint: Consider c = lub{an|n ∈ Z+} = the least upper bound of the set {an|n ∈ Z+}.]

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Calculus Final Exam

Solution: Since {a_n}^∞₁ is bounded, c = lub{a_n|n ∈ Z₊} exists. For any ε > 0, there exists an m ∈ Z+such that c −ε^{< a}m≤ a_nfor all n ≥ m, where we have used the fact that {a_n} is increasing.

Therefore, we have 0 ≤ c − an<ε for all n ≥ m, where the first inequality holds since c is an upper bound of {a_n}. Hence, |c − a_n| <εfor all n ≥ m, and lim

n→∞a_n= c.

9. (8 points) Show that tan x > x for any x ∈ (0,π 2).

Solution: Consider the function f (x) = tan x − x. Note that f (0) = 0, and f⁰(x) = sec²x − 1 > 0 for each x ∈ (0,π

2). For each x ∈ (0,π

2), Mean Value Theorem implies that there is a c ∈ (0, x) such that f (x) = f (x) − f (0) = f⁰(c)(x − 0) = f⁰(c)x > 0. Hence, f (x) > 0 for each x ∈ (0,π

2), or tan x > x for any x ∈ (0,π

2).

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