Calculus I, Fall 2014

Brief solutions to Midterm Exam 2

1. (10 pts) Solve for y(x) on x < 0 from

y′′(x) = x^{−2}, y(−1) = 1, y′(−1) = 2

Ans:

y(x) = − ln |x| + x + 2

2. (12 pts) Graph f(x) = x^{1/3}(x − 4). Indicate all critical points and points of inflection. Ans: f′(x) = \frac{4}{3}x^{− \frac{2}{3}} (x − 1) and f′′(x) = \frac{4}{9}x^{− \frac{5}{3}} (x + 2). We have

\{ \begin{matrix}Critical points: x=0, 1 \\ Points of inflection: x=0, -2 \end{matrix}

3. (16 pts) Find the limits of the following expressions:

(a) lim_{x→0^{+}} x^{x} (b) lim_{x→0} \frac{x^{2} cos \frac{1}{x}}{sin x}

Ans:

(a) lim_{x→0^{+}} x^{x} = 1 (b) lim_{x→0} \frac{x^{2} cos \frac{1}{x}}{sin x} = lim_{x→0} \frac{x}{sin x}x cos \frac{1}{x} = 0

4. (16 pts) Evaluate

(a) ∫_{1}^{2} \frac{1}{x(1 + ln^{2} x)}dx (b) ∫_{0}^{4} x\sqrt{2x + 1} dx

Ans: (a) Let u = ln x

∫_{1}^{2} \frac{1}{x(1 + ln^{2} x)}dx = ∫_{0}^{ln 2} \frac{1}{1 + u^{2}} du = tan^{−1}(ln 2)

(b) Let z = 2x + 1

∫_{0}^{4} x\sqrt{2x + 1} dx = ∫_{1}^{9} \frac{1}{4}(z − 1)\sqrt{z}dz = \frac{121}{5} − \frac{13}{3} = \frac{298}{15}

5. (10 pts) Evaluate

lim_{n→∞} ∑_{k=n}^{2n} \frac{n}{k^{2}}

Ans:

lim_{n→∞} ∑_{k=n}^{2n} \frac{n}{k^{2}} = lim_{n→∞}(\frac{n}{n^{2}} + ∑_{k=n+1}^{2n} \frac{1}{n} \frac{1}{(\frac{k}{n})^{2}} ) = ∫_{1}^{2} \frac{1}{x^{2}} dx = \frac{1}{2}

6. (12 pts) Let f be a real valued function defined on {x ≥ 0} satisfying (a): f(0) = −1, (b): f′(x) ≥ 1/2 for all x ≥ 0. Prove that f(x) = 0 has \underline{one and only one} solution on \{x ≥ 0\}. Ans: Step 1: f(x) = 0 has \underline{one} solution on \{x ≥ 0\}: Since

f(2) = f(0) + ∫_{0}^{2} f′(x)dx ≥ −1 + (2 − 0) min_{0≤x≤2} f′(x) ≥ −1 + 2 × \frac{1}{2} = 0

and f(0) = −1 < 0. Moreover, f′(x) exist for all x ≥ 0, which implies f(x) is continuous on x ≥ 0. By Intermediate Value Theorem, there exists c between 0 and 2 such that f(c) = 0. Step 2: f(x) = 0 has \underline{only one} solution on \{x ≥ 0\}: If there exist c_{1} ≥ 0, c_{1} ≠ c such that f(c_{1}) = 0, by Rolle’s Theorem, there exist c_{2} between c and c_{1} such that

f′(c_{2}) = 0

which is contradict to f′(x) ≥ \frac{1}{2} for all x ≥ 0. Hence f(x) = 0 has only one solution on x ≥ 0.

7. (12 pts) State both parts of Fundamental Theorem of Calculus, prove that part 1 implies part 2. Ans: Proof of part 1 implies part 2: Let G(x) be any antiderivative of f(x), which is

\frac{d}{dx}G(x) = f(x)

2 Since \frac{d}{dx} ∫_{a}^{x} f(t)dt = f(x) = \frac{d}{dx}G(x), we have

∫_{a}^{x} f(t)dt + c = G(x)

for some constant c ∈ R. Then G(b) − G(a) = ∫_{a}^{b} f(t)dt + c − (∫_{a}^{a} f(t)dt + c) = ∫_{a}^{b} f(t)dt.

8. (12 pts) Suppose f(x) satisfies ∫_{0}^{x^{2}} e^{−t}f(t)dt = x for all x ≥ 0. Find f(4) and f′(4). Ans: Taking derivative on both sides twice gives us

2xe^{−x^{2}}f(x^{2}) = 1

(2 − 4x^{2})e^{−x^{2}}f(x^{2}) + 2xe^{−x^{2}}f′(x^{2})2x = 0

which leads us to get

f(4) = \frac{e^{4}}{4}

f′(4) = \frac{7}{32}e^{4}