Practice Exam 1
Techniques of Integration
Do all ten cold, timed, no notes, before opening a single solution. Then grade yourself and send every miss back to the section it came from — each question is tagged with its topic for exactly that reason. Reading solutions without attempting the problems produces the feeling of understanding and none of the substance.
Evaluate \(\displaystyle\int x\ln x\,dx\).
Solution
LIATE: L beats A, so \(u = \ln x\) and \(dv = x\,dx\).
Then \(du = dx/x\) and \(v = x^2/2\).
Why the other choice fails: taking \(u = x\) requires integrating \(\ln x\) to get \(v\) — which is itself a parts problem. LIATE exists to stop you walking into that.
Evaluate \(\displaystyle\int \sin^2 x\cos^3 x\,dx\).
Solution
Cosine carries the odd power, so peel one off and let \(u = \sin x\), \(du = \cos x\,dx\).
Evaluate \(\displaystyle\int \frac{dx}{x^2\sqrt{9 - x^2}}\).
Solution
Form \(\sqrt{a^2 - x^2}\) with \(a = 3\), so \(x = 3\sin\theta\), \(dx = 3\cos\theta\,d\theta\), \(\sqrt{9 - x^2} = 3\cos\theta\).
From the sine triangle: opposite = \(x\), hypotenuse = 3, adjacent = \(\sqrt{9 - x^2}\), so \(\cot\theta = \sqrt{9 - x^2}/x\).
Common loss of points: stopping at \(-\tfrac19\cot\theta\). The answer must be in terms of \(x\).
Evaluate \(\displaystyle\int \frac{5x - 3}{x^2 - 2x - 3}\,dx\).
Solution
Degree 1 over degree 2 — proper, no long division. Factor: \(x^2 - 2x - 3 = (x - 3)(x + 1)\). Distinct linear factors, so use cover-up.
- \(A\): cover \((x - 3)\), evaluate \((5x - 3)/(x + 1)\) at \(x = 3 \rightarrow 12/4 =\) \(3\)
- \(B\): cover \((x + 1)\), evaluate \((5x - 3)/(x - 3)\) at \(x = -1 \rightarrow (-8)/(-4) =\) \(2\)
Check: recombining \(3/(x - 3) + 2/(x + 1)\) gives \((3x + 3 + 2x - 6)/\big((x - 3)(x + 1)\big) = (5x - 3)/(x^2 - 2x - 3)\). ✓
Evaluate \(\displaystyle\int \frac{x^2 + x + 2}{(x + 1)(x^2 + 1)}\,dx\).
Solution
\(x^2 + 1\) is irreducible, so it gets a linear numerator:
Clearing denominators: \(x^2 + x + 2 = A(x^2 + 1) + (Bx + C)(x + 1)\).
- \(x = -1\): \(1 - 1 + 2 = 2 = 2A \Rightarrow A = 1\)
- \(x^2\) coefficient: \(1 = A + B \Rightarrow B = 0\)
- constant term: \(2 = A + C \Rightarrow C = 1\)
The lesson: irreducible quadratics produce arctangents (and sometimes logarithms, when the numerator supplies the derivative of the denominator). Writing a bare constant over \(x^2 + 1\) instead of \(Bx + C\) is the standard error — here it happened to work out because \(B = 0\), but you cannot know that in advance.
Evaluate \(\displaystyle\int \frac{dx}{x^2 + 4x + 13}\).
Solution
The quadratic is irreducible (discriminant \(16 - 52 < 0\)), so partial fractions does not apply. Complete the square:
Now it matches \(\int du/(u^2 + a^2)\) with \(u = x + 2\) and \(a = 3\):
Do not lose the \(1/a\). Dropping the 1/3 is the most common error on this exact problem type.
Use Simpson’s rule with \(n = 4\) to estimate \(\displaystyle\int_0^1 e^{-x^2}\,dx\), to five decimal places. Why can this integral not be done exactly?
Solution
\(\Delta x = 1/4 = 0.25\). Evaluate at the five nodes:
| \(x\) | 0.00 | 0.25 | 0.50 | 0.75 | 1.00 |
|---|---|---|---|---|---|
| \(e^{-x^2}\) | 1.00000 | 0.93941 | 0.77880 | 0.56978 | 0.36788 |
| weight | 1 | 4 | 2 | 4 | 1 |
The true value is 0.74682, so four intervals already gives five-digit accuracy — that is the \(1/n^4\) convergence earning its keep.
Why no exact answer: \(e^{-x^2}\) has no elementary antiderivative. This is a proven theorem, not a gap in anyone’s technique. The function is important enough that its integral was given its own name — the error function, \(\operatorname{erf}(x)\) — which you will meet again in noise and probability analysis.
Determine whether \(\displaystyle\int_1^\infty \frac{\ln x}{x^2}\,dx\) converges, and evaluate it if so.
Solution
Type I. Write the limit first, then integrate by parts with \(u = \ln x\), \(dv = x^{-2}\,dx\), so \(v = -1/x\):
Now take \(T \to \infty\). By L’Hôpital, \(\ln T/T \to 0\), and \(1/T \to 0\):
The idea worth keeping: \(\ln x\) grows, but any positive power of \(x\) beats it. The \(1/x^2\) decay wins comfortably, so the logarithm in the numerator does not save the integrand from convergence.
A damped signal has instantaneous value \(v(t) = e^{-2t}\cos(3t)\). Evaluate \(\displaystyle\int_0^\infty v(t)\,dt\) and explain why the integral converges.
Solution
Use the cyclic-parts result with \(a = -2\), \(b = 3\):
Here \(a^2 + b^2 = 4 + 9 = 13\), so the antiderivative is
At the upper limit, \(e^{-2t} \to 0\) while the trig factor stays bounded between \(\pm\sqrt{13}\), so the product \(\to 0\).
At \(t = 0\): \(e^0\big(-2(1) + 3(0)\big)/13 = -2/13\).
Convergence argument: \(|v(t)| \le e^{-2t}\), and \(\int_0^\infty e^{-2t}\,dt = 1/2\) converges. By comparison, so does ours. The oscillation is irrelevant — the exponential envelope is what decides it.
EE reading: this is the ringing of an underdamped RLC circuit, and the integral is the net charge delivered (if \(v\) were a current) as the transient dies out. The finite answer is exactly the statement that the transient carries finite total effect.
A circuit’s response has Laplace transform \(F(s) = \dfrac{2s + 1}{s(s + 1)(s + 3)}\). Decompose it into partial fractions, then write \(f(t)\) using \(1/(s + a) \rightarrow e^{-at}\) and \(1/s \rightarrow 1\). Verify your answer at \(t = 0\).
Solution
Three distinct linear factors, so cover-up on each root:
- \(s = 0\): \(\big(2(0) + 1\big) / \big[(0 + 1)(0 + 3)\big] = 1/3\)
- \(s = -1\): \((-2 + 1) / \big[(-1)(-1 + 3)\big] = (-1)/(-2) = 1/2\)
- \(s = -3\): \((-6 + 1) / \big[(-3)(-3 + 1)\big] = (-5)/(6) = -5/6\)
Reading off term by term:
Verification at \(t = 0\):
That is the right answer, and you can confirm it without inverting anything: the initial value theorem says \(f(0) = \lim_{s\to\infty} sF(s)\), and \(sF(s) \approx 2s^2/s^3 \to 0\).
Physically: the response starts at zero, transients \(e^{-t}\) and \(e^{-3t}\) decay away, and the circuit settles to the steady-state value 1/3 — which is also \(\lim_{s\to 0} sF(s) = 1/3\), the final value theorem. Two independent checks on one decomposition.
Scoring
| Score | Where you are |
|---|---|
| 90–100 | Solid. Move to Unit 2. |
| 75–89 | Technique is there; the losses are algebra and lost constants. Redo the misses without looking. |
| 60–74 | You know the methods but are picking the wrong one. Re-read Choosing a technique and drill recognition, not computation. |
| < 60 | Work back through the unit page section by section before retaking. Do not move on — every later unit integrates. |
In a real sitting of this exam, the most common losses are: forgetting “\(+\,C\)”, dropping the \(1/a\) in the arctangent form (Q6), not converting back from \(\theta\) to \(x\) (Q3), and writing a constant instead of \(Bx + C\) over an irreducible quadratic (Q5). None of those are conceptual. All of them are worth points.