Calculus Midterm Exam II

Calculus Midterm Exam II

1. (10 points) On the curve x = sin²y, evaluate dy/dx and d²y/dx²at the point¡1 2,π/4¢

.

Solution: Differentiating twice both sides with respect to x, we get 1 = 2 sin y cos ydy

dx, and 0 = 2 sin y cos yd²y

dx²+ 2(cos²y − sin²y)(dy

dx)². Evaluating at¡1 2,π/4¢

, we get dy

dx= 1, and d²y dx² = 0.

2. Let f (x) =



 x sin(1

x), x 6= 0

0, x = 0.

(a) (5 points) Show that f is continuous at 0.

Solution: Since 0 ≤ |x sin(1

x)| ≤ |x|, and lim_x→0|x| = 0, we have lim_x→0f (x) = 0 = f (0), and f is continuous at 0.

(b) (5 points) Show that f is not differentiable at 0.

Solution: Since lim_x→0 f (x) − f (0)

x − 0 = lim_x→0sin(1

x), and lim_n→∞sin( 1

1/nπ) = 0 while lim_n→∞sin( 1

1/(π^{/2 + 2n}π^{)) = 1, limx→0sin(}

1

x) does not exist. Therefore, f is not differen- tiable at 0.

3. (10 points) Use the Mean Value Theorem to show that sin x ≤ x for any x > 0.

Solution: Mean Value Theorem implies that sin x − sin 0 = cos c(x − 0) where c is a number satis- fying 0 < c < x. Therefore, we have sin x = x cos c ≤ |x|| cos c| ≤ |x| = x.

4. (10 points) Let f (x) = 1

2cos x for x ∈ [0,π/2]. Use the Intermediate Value Theorem to show that there exists a c ∈ (0,π/2) such that f (c) = c. [Such number c is called a fixed point of f .]

Solution: Consider the function g(x) = f (x) − x. Since g(0) = 1

2 > 0, and g(π/2) = −π/2 < 0, the Intermediate Value Theorem implies that there is a number c ∈ (0,π/2) such that g(c) = 0, i.e.

f (c) = c.

5. (5 points) Is it true that d dx

h^Z _x

a

f (t)dt i

= Z _x

a

d

dt[ f (t)]dt? Explain your answer.

Calculus Midterm Exam II

Solution: Since d dx

h^Z x

a

f (t)dt i

= f (x), while Z _x

a

d

dt[ f (t)]dt = f (x) − f (a), and f (a) may not be zero, d

dx h^Z _x

a

f (t)dt i

, in general, is not equal to Z _x

a

d

dt[ f (t)]dt.

6. (a) (5 points) Find d dx

£^{Z x2+1}

cos x

f (t)dt¤

Solution: d dx

£^{Z x2+1}

cos x

f (t)dt¤

= 2x f (x²+ 1) + sin x f (cos x).

(b) (5 points) Let f be a continuous function and set F(x) =^R₀^xx f (t)dt. Find F⁰(x).

Solution: F⁰(x) = x f (x) +^R₀^xf (t)dt.

7. Assume that f and g are continuous, that a < b, and that^R_a^bf (x)dx >^R_a^bg(x)dx.

(a) (5 points) Show that L_g(P) < U_f(P) for all partitions P of [a, b].

Solution: Since U_f(P) >^R_a^bf (x)dx >^R_a^bg(x)dx > L_g(P) for all partitions P of [a, b], we have U_f(P) > L_g(P).

(b) (5 points) Is it true that L_g(P) < L_f(P) for all partitions P of [a, b]? Explain your answer.

[Hint: Consider f (x) = x for x ∈ [−1, 2], g(x) = 0 for x ∈ [−1, 2], and note that Z ₂

−1f (x)dx >

Z ₂

−1g(x)dx. ]

Solution: Choose P = {−1, 2}, then L_f(P) = (−1) · 3 = −3 and Lg(P) = 0. Therefore, in general, the claim is not true.

(c) (5 points) Is it true that U_g(P) < U_f(P) for all partitions P of [a, b]? Explain your answer.

[Hint: Consider f (x) = (

2x, if x ∈ [0, 1]

2, if x ∈ [1, 3/2], g(x) =







4, if x ∈ [0, 1/4]

−16x + 8, if x ∈ [1/4, 1/2]

0, if x ∈ [1/2, 3/2]

,

and note that^R₀^3/2f (x)dx >^R₀^3/2g(x)dx. ]

Solution: Choose P = {0, 3/2}, then Uf(P) = (2) · 3/2 = 3 and Ug(P) = 4 · 3/2 = 6. There- fore, in general, the claim is not true.

8. (10 points) Calculate the indefinite integral Z

(x − 1)²⁰x²dx.

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Calculus Midterm Exam II

Solution: Setting u = x − 1, we get x = u + 1, and dx = du. Thus Z

(x − 1)²⁰x²dx = Z

u²⁰(u + 1)²du =

Z ¡

u²²+ 2u²¹+ u²⁰¢

du = u²³/23 + u²²/11 + u²¹/21 +C = (x − 1)²³/23 + (x − 1)²²/11 + (x − 1)²¹/21 +C

9. (10 points) Determine the centroid of the region bounded between the curves y = x², y = 0, x = 1, x = 2.

Solution: The centroid ( ¯x, ¯y) can be found by ¯x = R₂

1x · x²dx R₂

1x²dx and ¯y = R₂

1

x² 2 · x²dx R₂

1x²dx . Therefore, ( ¯x, ¯y) = (45/28, 93/70)

10. (10 points) The region between the graph of y = sin x and the x−axis, 0 ≤ x ≤π, is revolved about the line y = 1. Find the volume of the solid that is generated.

Solution: The volume V =^R₀^ππ^£1 − (1 − sin x)²¤

dx =π²−π^¡x + 2 cos x +x

2−sin 2x 4

¢|^π₀ = 4π− π^2/2.

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