MIDTERM #1 (KEY) JONES, FALL 2001

The solutions are included for non-multiple choice problems, except for 3(b).

Problem 1(a)(i). Evaluate the following integral:

∫_{0}^{3} x^{2}\sqrt{9 − x^{2}} dx.

Solution. (Note, some exams had 16 in place of 9 for this problem.) We substitute x = 3 sin θ, so \sqrt{9 − x^{2}} = \sqrt{9 cos^{2} θ} = 3 cos θ, and dx =

3 cos θ dθ. We get

∫ x^{2}\sqrt{9 − x^{2}} dx = ∫ (3 sin θ)^{2}(3 cos θ)(3 cos θ dθ) = 81 ∫ cos^{2} θ sin^{2} θ dθ.

Now we note that sin θ cos θ = \frac{1}{2} sin(2θ), so this becomes

81 ∫ (1/2 sin(2θ))^{2} dθ = \frac{81}{4} ∫ sin^{2}(2θ) dθ.

Now we substitute sin^{2}(2θ) = (1 − sin(4θ))/2 (double the double-angle formula!) to get

\frac{81}{4} ∫ \frac{1 − sin(4θ)}{2} dθ = \frac{81}{8} (θ + cos(4θ)) + C.

(We could have also replaced cos^{2} θ = 1 − sin^{2} θ, and then used the doubleangle formula twice.)

Now replacing the limits of integration, we have x = 0 at sin θ = 0, or θ = 0 and x = 3 at sin θ = 1 or θ = π/2, so we have

∫_{0}^{3} x^{2}\sqrt{9 − x^{2}} dx = \frac{81}{8} (θ + cos(4θ))_{0}^{π/2} = \frac{81}{8}(π/2 + 1 − 1) = \frac{81π}{16} .

Problem 1(a)(ii). Evaluate the integral

∫ \frac{x}{x^{2} − x + 6} dx.

Solution. Note that the denominator does not factor: (x − 3)(x + 2) = x^{2} − x − 6 ≠ x^{2} − x + 6. Therefore the method of partial fractions does not apply because the denominator is irreducible qudratic.

Instead, we would like to substitute u = x^{2} − x + 6, so that du = (2x − 1) dx, hence we let

∫ \frac{x}{x^{2} − x + 6} dx = \frac{1}{2} ∫ \frac{(2x − 1) + 1}{x^{2} − x + 6} dx

= \frac{1}{2} ∫ \frac{2x − 1}{x^{2} − x + 6} dx + \frac{1}{2} ∫ \frac{1}{x^{2} − x + 6} dx.

The first integral is now just

\frac{1}{2} ∫ \frac{du}{u} = \frac{1}{2} ln |u| + C = \frac{1}{2} ln |x^{2} − x + 6| + C.

For the second integral, we must complete the square:

x^{2} − x + 6 = (x − 1/2)^{2} − 1/4 + 6 = (x − 1/2)^{2} + 23/4.

Hence

\frac{1}{x^{2} − x + 6} = \frac{1}{(x − 1/2)^{2} + 23/4} = \frac{4/23}{(4/23)(x − 1/2)^{2} + 1}.

Substituting u = (2/\sqrt{23})(x − 1/2), we have du = 2/\sqrt{23} dx, and

∫ \frac{1}{x^{2} − x + 6} dx = ∫ \frac{(4/23)(\sqrt{23}/2)}{u^{2} + 1} du = \frac{2}{\sqrt{23}} ∫ \frac{1}{u^{2} + 1} du

= \frac{2}{\sqrt{23}} tan^{−1} u + C = \frac{2}{\sqrt{23}} tan^{−1} (\frac{2}{\sqrt{23}}(x − 1/2)) + C.

Putting these together (remembering the 1/2), we get

∫ \frac{x}{x^{2} − x + 6} dx = \frac{1}{2} ln |x^{2} − x + 6| + \sqrt{23} tan^{−1} (\frac{1}{\sqrt{23}}(x − 1/2)) + C.

Problem 1(b). Evaluate

∫_{0}^{∞} \frac{dx}{x^{2} − 5}

or show that it is divergent.

Solution. The function 1/(x^{2}−5) is discontinuous at x = \sqrt{5}, therefore we must write

∫_{0}^{∞} \frac{dx}{x^{2} − 5} = lim_{t→\sqrt{5}^{+}} ∫_{0}^{t} \frac{dx}{x^{2} − 5} + lim_{t→\sqrt{5}^{−}} ∫_{t}^{∞} \frac{dx}{x^{2} − 5}.

Now x^{2} − 5 = (x − \sqrt{5})(x + \sqrt{5}), so by partial fractions, we have

\frac{1}{x^{2} − 5} = \frac{A}{x − \sqrt{5}} + \frac{B}{x + \sqrt{5}}

so

1 = A(x + \sqrt{5}) + B(x − \sqrt{5}).

Letting x = \sqrt{5} we see that A = 1/(2\sqrt{5}); letting x = −\sqrt{5} we get x = −1/(2\sqrt{5}). Hence

∫ \frac{dx}{x^{2} − 5} = ∫ \frac{1/(2\sqrt{5})}{x − \sqrt{5}} dx + ∫ \frac{−1/(2\sqrt{5})}{x + \sqrt{5}} dx

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

Therefore, for the first integral, say, we have

lim_{t→\sqrt{5}^{+}} ∫_{0}^{t} \frac{dx}{x^{2} − 5} = lim_{t→\sqrt{5}^{+}} \frac{1}{2\sqrt{5}} ln |\frac{x − \sqrt{5}}{x + \sqrt{5}}| → ln 0 = −∞

so the integral is divergent.

Problem T.he error in estimating ∫_{a}^{b} f(x) dx using Simpson’s rule with n

intervals is at most K(b − a)^{5}180n^{4} when |f^{(4)}(x)| ≤ K for a ≤ x ≤ b. How large should n be to sure the error is less than 10^{−5} in estimating

∫_{1}^{3} \frac{ln x}{2} dx

using Simpson’s rule?

Solution. We have f(x) = 3/2 ln x, so f′(x) = 3/(2x) = 3/2x^{−1}, f′′(x) = −3/2x^{−2}, f′′′(x) = 3x^{−3}, and f^{(4)}(x) = −9x^{−4} = −9/x^{4}. Therefore |f^{(4)}(x)| = 9/x^{4}. This function is clearly decreasing on 1 ≤ x ≤ 3, so it reaches its maximum at x = 1, namely, |f^{(4)}(x)| ≤ 9, so we may take K = 9.

Therefore

|E_{S}| ≤ \frac{9(3 − 1)^{2}}{180n^{4}} < 10^{−5}

so

n^{4} > \frac{9(32)(10^{5})}{180} = 16(10^{4}),

i.e. n > \sqrt[4]{16(10^{4})} = 20. Since n must be even in applying Simpson’s rule, we take n = 22, or any even integer thereafter.

Problem 3(a). Is the series

∑_{n=1}^{∞} \frac{1}{n^{2} + 3n + 2}

convergent? If so find its sum.

Solution. The terms of the series a_{n} = 1/(n^{2} + 3n + 2) are positive and decreasing, so the integral test applies. We have

∫_{1}^{∞} \frac{1}{x^{2} + 3x + 2} dx.

Since x^{2} + 3x + 2 = (x + 2)(x + 1), we use partial fractions:

\frac{1}{x^{2} + 3x + 2} = \frac{A}{x + 2} + \frac{B}{x + 1}

and so 1 = A(x + 1) + B(x + 2). Letting x = −1 we get B = 1, x = −2 we get A = −1, hence

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

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

Therefore

∫_{1}^{∞} \frac{1}{x^{2} + 3x + 2} dx = lim_{t→∞} ln |\frac{x + 1}{x + 2}|_{1}^{t} .

We now have:

lim_{t→∞} ln \frac{t + 1}{t + 2} = ln lim_{t→∞} \frac{t + 1}{t + 2} = ln 1 = 0

so the original series is convergent.

From the above we see

∑_{n=1}^{∞} \frac{1}{n^{2} + 3n + 2} = ∑_{n=1}^{∞} \frac{1}{n + 1} − \frac{1}{n + 2}.

This is a telescoping series: we have

∑_{n=1}^{N} = \frac{1}{2} − \frac{1}{3} + \frac{1}{3} − \frac{1}{4} + ⋯ + \frac{1}{N + 2} = \frac{1}{2} + \frac{1}{N + 2}.

Therefore

∑_{n=1}^{∞} \frac{1}{n^{2} + 3n + 2} = lim_{N→∞} (\frac{1}{2} + \frac{1}{N + 2}) = \frac{1}{2}.