MIDTERM #1 (KEY) JONES, SPRING 1998

Problem 1. Which of the following formulas is correct...?

Solution. The correct answer is (c). We have from the product formula

(u(x)v(x))′ = u(x)v′(x) + u′(x)v(x).

Integrating both sides, we get (c).

Problem 2. If you wanted to expand

\frac{2x + 7}{(x + 1)^{2}(x^{2} + x + 19)^{2}}

in partial fractions you would use the sum...:

Solution. The answer is (b). Recall that we put a linear factor in the numerator for each irreducible quadratic factor, and a simple constant for the others.

Problem 3. Which of the following functions cannot be integrated in terms of elementary functions...?

Solution. For (a), we can use integration by parts: in fact, we have

∫ x^{2} ln x dx = \frac{1}{3}x^{3} ln x − \frac{1}{9}x^{3} + C.

For (c), we can use integration by parts twice, then bring the unknown integral to the other side:

∫ e^{x} sin x dx = −e^{x} cos x+∫ e^{x} cos x dx = −e^{x} cos x+e^{x} sin x−∫ e^{x} sin x dx

so that

∫ e^{x} sin x dx = \frac{1}{2}e^{x}(sin x − cos x) + C.

For (d), we make the substitution u = ln x, so that du = 1/x dx and

∫ \frac{1}{x ln x} dx = ∫ \frac{1}{u} du = ln u + C = ln(ln x) + C.

For (e), we make the substitution u = x^{2}, so that du = 2x dx and

∫ x sin(x^{2}) dx = \frac{1}{2} ∫ sin u du = −\frac{1}{2} cos u + C = −\frac{1}{2} cos x^{2} + C.

The answer is (b).

Problem 4. Which of the following statements is always correct for a function f(x) with 0 ≤ f(x) ≤ C...?

Solution. We just test examples to see which statement is true. Take f(x) = 1/x^{2}, for example. Then

∫_{1}^{∞} \frac{1}{x^{2}} dx

converges by the p-series test (2 > 1), but

∫_{1}^{∞} \sqrt{\frac{1}{x^{2}}} dx = ∫_{1}^{∞} \frac{1}{x} dx

diverges by the same test. Therefore (a) is out. Similarly,

∫_{1}^{∞}(1/x^{2})^{−2} dx = ∫_{1}^{∞} x^{4} dx

obviously diverges, so (b) is out.

For (c), do the same analysis above with f(x) = 1/\sqrt{x}; this shows that (c) is out. For (e), try the function f(x) = 1/x; this shows that (e) is out.

The answer is (d). You can also see this by the comparison test, since

f(x) ≥ \frac{f(x)}{1 + x}

for x ≥ 1, so if ∫_{1}^{∞} f(x) dx converges, so does the smaller integral ∫_{1}^{∞} f(x)/(1+ x) dx.

Problem 5. Which of the following statements is correct...?

Solution. The answer is (a). We have

|E_{S}| ≤ \frac{K(b − a)^{5}}{180n^{4}} ,

so if we double n, the term on the right is changed by a factor of 1/16.

Statement (b) is false: it is Simpson’s rule which is exact on quadratics. Statement (c) is false, certainly there exist functions (like sin x^{2}) which do not have elementary antiderivatives, so we have to approximate. The error for the midpoint rule is

|E_{M}| ≤ \frac{K(b − a)^{3}}{24n^{2}}

where K = max_{a≤x≤b} |f′′(x)|, so (d) is false. Statement (e) is false: in Simpson’s rule, we draw quadratic curves fitting the points (it is the trapezoidal rule which uses a linear approximation).

Problem 6. Find the arc-length function for the curve y = x^{2}/8 − ln x starting at (1, 1/8).

Solution. The arc length function is

s(x) = ∫_{1}^{x} \sqrt{1 + f′(t)^{2}} dt

where f(t) = t^{2}/8 − ln t. We have f′(t) = t/4 − 1/t, so

s(x) = ∫_{1}^{x} \sqrt{1 + (t/4 − 1/t)^{2}} dt = ∫_{1}^{x} \sqrt{1 + t^{2}/16 − 1/2 + 1/t^{2}} dt
= ∫_{1}^{x} \sqrt{t^{2}/16 + 1/2 + 1/t^{2}} dt = ∫_{1}^{x} \sqrt{(t/4 + 1/t)^{2}} dt
= ∫_{1}^{x}(t/4 + 1/t) dt = t^{2}/8 + ln t|_{1}^{x} = x^{2}/8 + ln x − 1/8.

Problem 7(i). Evaluate the following indefinite integral:

∫ \frac{1}{(x + 3)(x − 2)} dx.

Solution. We use partial fractions. We write

\frac{1}{(x + 3)(x − 2)} = \frac{A}{x + 3} + \frac{B}{x − 2}

and then multiply through to get

1 = A(x − 2) + B(x + 3)

With x = 2 we see that 5B = 1, so B = 1/5. With x = −3 we get −5A = 1, so A = −1/5. Therefore

∫ \frac{1}{(x + 3)(x − 2)} dx = −\frac{1}{5} ∫ \frac{1}{x + 3} dx + \frac{1}{5} ∫ \frac{1}{x − 2} dx

= −\frac{1}{5} ln |x + 3| + \frac{1}{5} ln |x − 2| + C.

Problem 7(ii). Evaluate the following indefinite integral:

∫ (ln x)^{2} dx.

Solution. We use integration by parts, with u = (ln x)^{2} so du = (2 ln x)/x dx and dv = dx, so v = x. We get

∫ (ln x)^{2} dx = x(ln x)^{2} − ∫ 2 ln x dx.

Repeating this with now u = ln x so du = 1/x dx and dv = dx, v = x, we get

∫ ln x dx = x ln x − ∫ dx = x ln x − x + C.

Together this gives

∫ (ln x)^{2} dx = x(ln x)^{2} − 2x ln x + 2x + C.

Problem 7(iii). Evaluate the following indefinite integral:

∫ \frac{1}{\sqrt{1 − 4x^{2}}} dx.

Solution. We substitute x = 1/2 sin θ, so that \sqrt{1 − 4x^{2}} = \sqrt{1 − sin^{2} θ} = cos θ, and dx = 1/2 cos θ dθ, so that

∫ \frac{1}{\sqrt{1 − 4x^{2}}} dx = ∫ \frac{1}{cos θ} (1/2) cos θ dθ

= \frac{1}{2} ∫ dθ = \frac{1}{2}θ + C

= \frac{1}{2} sin^{−1} 2x + C.