REVIEW, MIDTERM #1 (Key): MATH 1B

Warning: I have not seen a copy of the midterm examination. Concepts and problems reviewed here should not be taken as an exclusive list! It should give you a good sense, though, of what will be asked.

Problem 1. The integral

∫_{a}^{∞} \frac{1}{x^{\sqrt{3}}} dx

converges:

(a) For all values of a > 0;

(b) For all values of a ≥ 0;

(c) For a = 1 only;

(d) For no value of a.

Solution. The integral

∫_{1}^{∞} \frac{1}{x^{p}} dx

converges for p > 1 and diverges for p ≤ 1. Since \sqrt{3} > 1, the integral converges for any a > 0. However, the integral

∫_{0}^{1} \frac{1}{x^{p}} dx

converges for p < 1 and diverges for p ≥ 1, so the integral does not converge for a = 0—in fact,

∫_{0}^{∞} \frac{1}{x^{p}} dx

converges for no value of p. The answer is (a).

Problem 2. Evaluate the integral

∫ \frac{4x^{2} + 4x + 1}{4x^{2} − 4x + 1} dx.

Solution. The degree of the numerator is not less than that of the denominator, so we use long division:

\frac{4x^{2} + 4x + 1}{4x^{2} − 4x + 1} = 1 + \frac{8x}{4x^{2} − 4x + 1}.

Therefore

∫ \frac{4x^{2} + 4x + 1}{4x^{2} − 4x + 1} dx = ∫ dx + ∫ \frac{8x}{4x^{2} − 4x + 1} dx = x + ∫ \frac{8x}{4x^{2} − 4x + 1} dx.

To evaluate the second term, we use partial fractions. First we factor the denominator:

4x^{2} − 4x + 1 = (2x − 1)^{2}.

Therefore we have the partial fractions decomposition

\frac{8x}{(2x − 1)^{2}} = \frac{A}{2x − 1} + \frac{B}{(2x − 1)^{2}} .

Multiplying through, we obtain

8x = A(2x − 1) + B = 2Ax + (B − A);

we see that A = 4 and B = A = 4; therefore

∫ \frac{8x}{(2x − 1)^{2}} dx = ∫ \frac{4}{2x − 1} dx+∫ \frac{4}{(2x − 1)^{2}} dx = 2 ln(2x−1)− \frac{2}{2x − 1} +C;

together this implies that our integral is

x + 2 ln(2x − 1) − \frac{2}{2x − 1} + C.

Problem 3. The area of the surface obtained by rotating the curve y = f(x) from a ≤ x ≤ b about the x-axis is given by:

(a) ∫_{a}^{b} πf(y)^{2} dy;

(b) ∫_{f(a)}^{f(b)} πf(x)^{2} dy;

(c) ∫_{a}^{b} 2πy\sqrt{1 + (\frac{df}{dx})^{2}} dx;

(d) ∫_{f(a)}^{f(b)} 2πx\sqrt{1 + (\frac{df}{dx})^{2}} dy.

Solution. Drawing a picture, we see that the radius of the circles we wish to sum up is y, hence

A = ∫ 2πy ds = ∫_{a}^{b} 2πy\sqrt{1 + (\frac{dy}{dx})^{2}} dx

which is (b).

Problem 4. Evaluate the integral

∫ \sqrt{x} ln x dx.

Solution. We use integration by parts, with u = ln x, so du = 1/x, and dv = \sqrt{x} dx, so v = 2/3x^{3/2}, and

∫ \sqrt{x} ln x dx = ∫ u dv = uv − ∫ v du = \frac{2}{3}x^{3/2} ln x − ∫ \frac{2}{3}x^{1/2} dx

= \frac{2}{3}x\sqrt{x} ln x − \frac{4}{9}x\sqrt{x} + C.

Problem 5. Which of the following does not have an elementary antiderivative?

(a) x^{2} ln x;

(b) sinh x;

(c) xe^{x^{2}};

(d) e^{x^{2}}.

Solution. We can integrate (a) using integration by parts; we in fact obtain

∫ x^{2} ln x dx = \frac{1}{3}x^{3} ln x − \frac{1}{9}x^{3},

but this is not important. For (b), the antiderivative of sinh x is cosh x. For (c), we may make the substitution u = x^{2} to find

∫ xe^{x^{2}} dx = \frac{1}{2}e^{x^{2}}.

Therefore the answer is (d).

Problem 6. Evaluate the integral

∫ csc^{4} x dx.

Solution. We first substitute 1 + cot^{2} x = csc^{2} x, so

∫ csc^{4} x dx = ∫ (csc^{2} x)(1 + cot^{2} x) dx = ∫ csc^{2} x dx + ∫ csc^{2} x cot^{2} x dx.

Since (d/dx)(cot x) = − csc^{2} x, the first term is

∫ csc^{2} x dx = − cot x + C.

For the second, make the substitution u = cot x, so then du = − csc^{2} x dx, hence

∫ csc^{2} x cot^{2} x dx = − ∫ u^{2} du = −\frac{u^{3}}{3} + C = −\frac{1}{3} cot^{3} x + C.

Putting these together,

∫ csc^{4} x dx = − cot x − \frac{1}{3} cot^{3} x + C.

Problem 7. If the function f(x) is continuous on −1 ≤ x ≤ 1 except at the point x = a for some −1 < a < 1, then the improper integral

∫_{−1}^{1} f(x) dx

is written:

(a) lim_{t→a^{+}} ∫_{−1}^{t} f(x) dx + lim_{t→a^{−}} ∫_{t}^{1} f(x) dx;

(b) lim_{x→a} ∫_{−1}^{x} f(x) dx;

(c) (df/dx)(a) − lim_{t→a} ∫_{−t}^{t} f(x) dx;

(d) lim_{a→∞} ∫_{−1}^{1} f(x) dx.

Solution. The answer is (a); the rest are nonsense.

Problem 8. Evaluate the integral

∫ \sqrt{x^{2} − 1} dx.

Solution. This is a problem involving trigonometric substitution: we substitute x = sec θ, so that dx = sec θ tan θ dθ, and

\sqrt{x^{2} − 1} = \sqrt{sec^{2} θ + 1} = tan θ

hence

∫ \sqrt{x^{2} − 1} dx = ∫ tan θ(sec θ tan θ) dθ
= ∫ sec θ tan^{2} θ dθ.

Now integrate by parts, with u = tan θ, du = sec^{2} θ dθ, dv = sec θ tan θ dθ, v = sec θ, so

∫ sec θ tan^{2} θ dθ = ∫ u dv = uv − ∫ v du
= sec θ tan θ − ∫ sec^{3} θ dθ
= sec θ tan θ − ∫ sec θ(sec^{2} θ) dθ
= sec θ tan θ − ∫ sec θ(tan^{2} θ + 1) dθ
= sec θ tan θ − ∫ sec θ tan^{2} θ dθ − ∫ sec θ dθ.

Since ∫ sec θ dθ = ln | sec θ + tan θ| + C, bringing the integral to the other side, we obtain

sec θ tan^{2} θ dθ = \frac{1}{2} sec θ tan θ − \frac{1}{2} ln | sec θ + tan θ|.

Replacing this with the values of x, we get

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

Problem 9. The sequence

a_{n} = \frac{2n^{2}}{n^{2} + 1}

is:

(a) Bounded, monotonic, and convergent;

(b) Unbounded, monotonic, and not convergent;

(c) Bounded, not monotonic, and convergent;

(d) Bounded, not monotonic, and not convergent.

Solution. The sequence is bounded above by 2 (n^{2}+1 > n^{2}) and below by 0, say, so it is bounded. It is monotonic increasing, and convergent (it has limit 1, since

lim_{n→∞} a_{n} = lim_{n→∞} \frac{2}{1 + 1/n^{2}} = 2.

Problem 10. Find the area of the surface obtained by rotating the curve

y = x^{2} + 1

about the y-axis for 0 ≤ x ≤ 1.

Solution. Looking at the graph, we note that the radius of our circles is obtained with radius x, hence we have

A = ∫ 2πx ds = ∫_{0}^{1} 2πx\sqrt{1 + (\frac{dy}{dx})^{2}} dx = 2π ∫_{0}^{1} x\sqrt{1 + 4x^{2}} dx

since dy/dx = 2x.

We now make a trigonometric substitution, x = 1/2 tan θ, so that dx = 1/2 sec^{2} θ dθ, and

\sqrt{1 + 4x^{2}} = \sqrt{1 + tan^{2} θ} = sec θ

so that

∫ x\sqrt{1 + 4x^{2}} dx = ∫ \frac{1}{2} tan θ sec θ \frac{1}{2} sec^{2} θ dθ = \frac{1}{4} ∫ sec^{3} θ tan θ dθ.

Now let u = sec θ; then du = sec θ tan θ dθ, so

\frac{1}{4} ∫ sec^{2} θ(sec θ tan θ) dθ = \frac{1}{4} ∫ u^{2}du = \frac{u^{3}}{12} + C

= \frac{sec^{3} θ}{12} + C = \frac{(1 + 4x^{2})^{3/2}}{12} + C.

So

A = 2π \frac{(1 + 4x^{2})^{3/2}}{12} |_{0}^{1} = \frac{5\sqrt{5} − 1}{6}π.

Problem 11. The partial fractions decomposition of

\frac{1}{(x − 3)^{2}(x^{2} + 3)}

is:

(a) \frac{A}{x − 3} + \frac{Bx + C}{(x − 3)^{2}} + \frac{Dx + E}{x^{2} + 3} ;

(b) \frac{A}{(x − 3)^{2}} + \frac{B}{x^{2} + 3};

(c) \frac{A}{x − 3} + \frac{B}{(x − 3)^{2}} + \frac{Cx + D}{x^{2} + 3} ;

(d) \frac{Ax + B}{(x − 3)^{2}} + \frac{Cx + D}{x^{2} + 3} ;

(e) \frac{A}{x − 3} + \frac{B}{(x − 3)^{2}} + \frac{C}{x^{2} + 3}.

Solution. The answer is (c). Note we only put a linear factor in the numerator when we have an irreducible quadratic factor in the denominator. Problem 12. Evaluate the integral

∫_{0}^{π} sin^{5} x cos^{2} x dx.

Solution. We let u = cos x, so du = − sin x dx, hence

∫_{0}^{π} cos^{2} x sin^{4} x sin x dx = − ∫_{0}^{π} u^{2}(1 − u^{2})^{2} du

= − ∫_{0}^{π}(u^{2} − 2u^{4} + u^{6}) du

= −\frac{1}{3}u^{3} + \frac{2}{5}u^{5} − \frac{1}{7}u^{7} + C

= −\frac{1}{3} cos^{3} x + \frac{2}{5} cos^{5} x − \frac{1}{7} cos^{7} x + C.

Problem 13. For the integral

∫_{0}^{2} e^{x} dx,

given the formula

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

with |f^{(4)}(x)| ≤ K for a ≤ x ≤ b, the error in approximating the integral using Simpson’s rule for n = 2 is:

(a) |E_{S}| ≤ 1/90;

(b) |E_{S}| ≤ e/90;

(c) |E_{S}| ≤ e^{2}/90;

(d) |E_{S}| ≤ e^{2}/2880.

Solution. We have f^{(4)}(x) = e^{x}, which obtains its maximum on 0 ≤ x ≤ 2 at x = 2, so |f^{(4)}(x)| ≤ e^{2}. Therefore

|E_{S}| ≤ \frac{e^{2}(2^{5})}{180(2^{4})}

which is (c).

Problem 14. Determine if the integral

∫_{1}^{2} \frac{1}{x\sqrt{ln x}} dx

is convergent, and evaluate it if so.

Solution. Make the substitution u = ln x; then du = 1/x dx, so

∫ \frac{1}{x\sqrt{ln x}} dx = ∫ \frac{1}{\sqrt{u}} du = 2\sqrt{u} + C = 2\sqrt{ln x} + C.

The function 1/(x\sqrt{ln x}) has a discontinuity at x = 1, so we have

∫_{1}^{2} \frac{1}{x\sqrt{ln x}} dx = lim_{t→1^{−}} ∫_{t}^{2} \frac{1}{x\sqrt{ln x}} dx = lim_{t→1^{−}} 2\sqrt{ln x}|_{t}^{2} = 2\sqrt{ln 2}.

The integral is convergent.

Problem 15. The sum of the series

1 − \frac{1}{2} + \frac{1}{4} − \frac{1}{8} + …

is:

(a) 1/2;

(b) 1;

(c) 2/3;

(d) 3/2;

(e) The series is divergent.

Solution. This series is

∑_{n=1}^{∞} (\frac{−1}{2})^{n−1}

which is a geometric series: in general,

∑_{n=1}^{∞} ar^{n−1} = \frac{a}{1 − r} , |r| < 1

so for us, a = 1 and r = −1/2, so the sum is 1/(1 + 1/2) = 2/3. The answer is (c).

Problem 16. Evaluate

∫_{e}^{e^{2}} \frac{ln(ln x)}{x} dx.

Solution. Substitute t = ln x, then dt = 1/x dx, so

∫ \frac{ln(ln x)}{x} dx = ∫ ln t dt.

Now do integration by parts, with u = ln t, du = 1/t dt, dv = dt, v = t, so

∫ ln t dt = ∫ u dv = uv − ∫ v du = t ln t − ∫ t(1/t) dt = t ln t − t + C.

Hence

∫_{e}^{e^{2}} \frac{ln(ln x)}{x} dx = ((ln x)(ln(ln x)) − ln x)_{e}^{e^{2}} = 2 ln 2 − 2 + 1 = 2 ln 2 − 1.

Problem 17. For the integral

∫_{a}^{b}(2x^{2} + 3x − 4) dx,

which of the following gives the best approximation for fixed n?

(a) Left endpoint approximation;

(b) Right endpoint approximation;

(c) Midpoint approximation;

(d) The Trapezoidal rule;

(e) Simpson’s rule.

Solution. Simpson’s rule is exact on quadratics (recall, we are fitting a quadratic each time around!). The answer is (e).

Problem 18. Set up an integral to compute the length of the curve y = e^{x} for 1 ≤ y ≤ e. Suggest a substitution, but do not evaluate completely.

Solution. We have

L = ∫ ds = ∫_{1}^{e} \sqrt{1 + (\frac{dx}{dy})^{2}} dy.

Writing x = ln y, we see that dx/dy = 1/y, so

L = ∫_{1}^{e} \sqrt{1 + (1/y)^{2}} dy = ∫_{1}^{e} \frac{\sqrt{y^{2} + 1}}{y} dy.

We substitute y = tan θ, so \sqrt{y^{2} + 1} = sec θ, and dy = sec^{2} θ, so
∫ \frac{\sqrt{y^{2} + 1}}{y} dy = ∫ \frac{sec θ}{tan θ} sec^{2} θ dθ
= ∫ \frac{sec θ}{tan θ} (1 + tan^{2} θ) dθ
= ∫ csc θ dθ + ∫ sec θ tan θ dθ.

From here, we could finish the evaluation.

Problem 19. If f is continuous, then

∫_{−∞}^{∞} f(x) dx = lim_{t→∞} ∫_{−t}^{t} f(x) dx.

(a) True;

(b) False;

(c) Cannot be determined from the information given.

Solution. The answer is (b). Recall that ∫_{−∞}^{∞} x^{3} dx, for example, is undefined (divergent), but

lim_{t→∞} ∫_{−t}^{t} x^{3} = lim_{t→∞} 0 = 0.

Problem 20. Determine if the integral

∫_{0}^{1} \begin{matrix}√ & \overline{1 + \sqrt{x}} \\ √ & \overline{x} \end{matrix} dx

is convergent, and evaluate it if so.

Solution. Use the comparison test; since

\frac{1}{\sqrt{x}} ≤ \begin{matrix}√ & \overline{1 + \sqrt{x}} \\ √ & \overline{x} \end{matrix}

(we just make the numerator a little larger), and

∫_{1}^{∞} \frac{1}{\sqrt{x}} dx

diverges (by the p-test, with p = 1/2 ≤ 1), the original integral diverges as well.