Practice Exam 1

Techniques of Integration

Questions10

Time90 min

Points100

CalculatorQ7 only

Use this properly

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.

1By parts8 pts

Evaluate ∫ x ln x dx.

Solution

LIATE: L beats A, so u = ln x and dv = x dx.

Then du = dx/x and v = x²/2.

∫ x ln x dx = x²2 ln x − ∫ x²2 · 1x dx = x²2 ln x − 12∫ x dx

= x²2 ln x − x²4 + C

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.

2Trig integrals8 pts

Evaluate ∫ sin²x cos³x dx.

Solution

Cosine carries the odd power, so peel one off and let u = sin x, du = cos x dx.

∫ sin²x cos²x · cos x dx = ∫ sin²x (1 − sin²x) cos x dx

= ∫ u²(1 − u²) du = ∫ (u² − u⁴) du = u³3 − u⁵5 + C

= sin³x3 − sin⁵x5 + C

3Trig substitution12 pts

Evaluate ∫ dxx² √(9 − x²).

Solution

Form √(a²−x²) with a = 3, so x = 3 sin θ, dx = 3 cos θ dθ, √(9−x²) = 3 cos θ.

∫ 3 cos θ dθ(9 sin²θ)(3 cos θ) = 19 ∫ dθsin²θ = 19 ∫ csc²θ dθ = −19 cot θ + C

From the sine triangle: opposite = x, hypotenuse = 3, adjacent = √(9−x²), so cot θ = √(9−x²)/x.

= −√(9 − x²)9x + C

Common loss of points: stopping at −¹⁄₉cot θ. The answer must be in terms of x.

4Partial fractions10 pts

Evaluate ∫ 5x − 3x² − 2x − 3 dx.

Solution

Degree 1 over degree 2 — proper, no long division. Factor: x²−2x−3 = (x−3)(x+1). Distinct linear factors, so use cover-up.

A: cover (x−3), evaluate (5x−3)/(x+1) at x = 3 → 12/4 = 3
B: cover (x+1), evaluate (5x−3)/(x−3) at x = −1 → (−8)/(−4) = 2

∫ [3x−3 + 2x+1] dx = 3 ln|x−3| + 2 ln|x+1| + C

Check: recombining 3/(x−3) + 2/(x+1) gives (3x+3+2x−6)/((x−3)(x+1)) = (5x−3)/(x²−2x−3). ✓

5Partial fractions · quadratic factor10 pts

Evaluate ∫ x² + x + 2(x+1)(x²+1) dx.

Solution

x²+1 is irreducible, so it gets a linear numerator:

x²+x+2(x+1)(x²+1) = Ax+1 + Bx+Cx²+1

Clearing denominators: x²+x+2 = A(x²+1) + (Bx+C)(x+1).

x = −1: 1−1+2 = 2 = 2A ⇒ A = 1
x² coefficient: 1 = A + B ⇒ B = 0
constant term: 2 = A + C ⇒ C = 1

∫ [1x+1 + 1x²+1] dx = ln|x+1| + arctan x + C

The lesson: irreducible quadratics produce arctangents (and sometimes logarithms, when the numerator supplies the derivative of the denominator). Writing a bare constant over x²+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.

6Complete the square8 pts

Evaluate ∫ dxx² + 4x + 13.

Solution

The quadratic is irreducible (discriminant 16 − 52 < 0), so partial fractions does not apply. Complete the square:

x² + 4x + 13 = (x+2)² + 9

Now it matches ∫ du/(u²+a²) with u = x+2 and a = 3:

= 13 arctan(x+23) + C

Do not lose the 1/a. Dropping the 1/3 is the most common error on this exact problem type.

7Numerical integration10 pts

Use Simpson’s rule with n = 4 to estimate ∫₀¹ e^−x² dx, to five decimal places. Why can this integral not be done exactly?

Solution

Δx = 1/4 = 0.25. Evaluate at the five nodes:

`x`	0.00	0.25	0.50	0.75	1.00
`e`^−x²	1.00000	0.93941	0.77880	0.56978	0.36788
weight	1	4	2	4	1

S₄ = 0.253[1.00000 + 4(0.93941) + 2(0.77880) + 4(0.56978) + 0.36788]

= 0.253[1.00000 + 3.75764 + 1.55760 + 2.27912 + 0.36788] = 0.253(8.96224) = 0.74685

The true value is 0.74682, so four intervals already gives five-digit accuracy — that is the 1/n⁴ convergence earning its keep.

Why no exact answer: e^−x² 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, erf(x) — which you will meet again in noise and probability analysis.

8Improper integral10 pts

Determine whether ∫₁^∞ ln xx² dx converges, and evaluate it if so.

Solution

Type I. Write the limit first, then integrate by parts with u = ln x, dv = x⁻²dx, so v = −1/x:

∫₁^T ln xx² dx = [−ln xx]₁^T + ∫₁^T dxx²

= −ln TT + 0 + [−1x]₁^T = −ln TT − 1T + 1

Now take T → ∞. By L’Hôpital, ln T/T → 0, and 1/T → 0:

∫₁^∞ ln xx² dx = 1 (converges)

The idea worth keeping: ln x grows, but any positive power of x beats it. The 1/x² decay wins comfortably, so the logarithm in the numerator does not save the integrand from convergence.

9Cyclic parts · EE12 pts

A damped signal has instantaneous value v(t) = e^−2tcos(3t). Evaluate ∫₀^∞ v(t) dt and explain why the integral converges.

Solution

Use the cyclic-parts result with a = −2, b = 3:

∫ e^at cos(bt) dt = e^at(a cos bt + b sin bt)a² + b²

Here a² + b² = 4 + 9 = 13, so the antiderivative is

e^−2t(−2 cos 3t + 3 sin 3t)13

At the upper limit, e^−2t → 0 while the trig factor stays bounded between ±√13, so the product → 0.

At t = 0: e⁰(−2(1) + 3(0))/13 = −2/13.

∫₀^∞ v dt = 0 − (−213) = 213 ≈ 0.154

Convergence argument: |v(t)| ≤ e^−2t, and ∫₀^∞e^−2tdt = 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.

10Partial fractions · inverse Laplace12 pts

A circuit’s response has Laplace transform F(s) = 2s + 1s(s+1)(s+3). Decompose it into partial fractions, then write f(t) using 1/(s+a) → e^−at and 1/s → 1. Verify your answer at t = 0.

Solution

Three distinct linear factors, so cover-up on each root:

s = 0: (2(0)+1) / [(0+1)(0+3)] = 1/3
s = −1: (−2+1) / [(−1)(−1+3)] = (−1)/(−2) = 1/2
s = −3: (−6+1) / [(−3)(−3+1)] = (−5)/(6) = −5/6

F(s) = 1/3s + 1/2s+1 − 5/6s+3

Reading off term by term:

f(t) = 13 + 12e^−t − 56e^−3t

Verification at t = 0:

f(0) = 13 + 12 − 56 = 2 + 3 − 56 = 0 ✓

That is the right answer, and you can confirm it without inverting anything: the initial value theorem says f(0) = lim_s→∞ sF(s), and sF(s) ≈ 2s²/s³ → 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→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.

Where the points usually go

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 θ 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.