UNIT 01Integration Techniques
Core Concept
Key Integration Methods
Four essential techniques for evaluating integrals that cannot be done directly.
\( \int u\,dv = uv - \int v\,du \) (Integration by Parts)
\( \int \frac{1}{(x-a)(x-b)}\,dx \) — decompose via Partial Fractions
Trig sub: \( \sqrt{a^2-x^2} \Rightarrow x=a\sin\theta \)
IBP: choose \(u\) using LIATE — Logarithm, Inverse trig, Algebraic, Trig, Exponential.
\( \int x\,e^x\,dx = (x-1)e^x + C \) — classic IBP result.
Q 01
Evaluate \( \displaystyle\int x\,e^x\,dx \).
Solution
Use Integration by Parts with \(u=x\), \(dv=e^x\,dx\), so \(du=dx\), \(v=e^x\).\(\int x\,e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = (x-1)e^x + C\).
Q 02
Evaluate \( \displaystyle\int \frac{x}{x^2+1}\,dx \).
Solution
Let \(u = x^2+1\), so \(du = 2x\,dx\), meaning \(x\,dx = \tfrac{1}{2}du\).\(\int \frac{x}{x^2+1}\,dx = \tfrac{1}{2}\int\frac{du}{u} = \tfrac{1}{2}\ln|u|+C = \tfrac{1}{2}\ln(x^2+1)+C\).
Q 03
Evaluate \( \displaystyle\int_0^{1/2} \frac{1}{\sqrt{1-x^2}}\,dx \).
Solution
Recall \(\int \frac{dx}{\sqrt{1-x^2}} = \arcsin(x) + C\).Evaluating: \(\Big[\arcsin(x)\Big]_0^{1/2} = \arcsin\!\left(\tfrac{1}{2}\right) - \arcsin(0) = \dfrac{\pi}{6} - 0 = \dfrac{\pi}{6}\).
Q 04
Evaluate \( \displaystyle\int \frac{1}{(x+1)(x+2)}\,dx \) using partial fractions.
Solution
Decompose: \(\dfrac{1}{(x+1)(x+2)} = \dfrac{A}{x+1}+\dfrac{B}{x+2}\).Solving: \(A=1\), \(B=-1\).
\(\int\!\left(\dfrac{1}{x+1}-\dfrac{1}{x+2}\right)dx = \ln|x+1|-\ln|x+2|+C\).
UNIT 02Applications of Integration
Core Concept
Area, Volume & Arc Length
Area between curves: \( A = \int_a^b [f(x)-g(x)]\,dx \)
Disk method: \( V = \pi\int_a^b [f(x)]^2\,dx \)
Arc length: \( L = \int_a^b \sqrt{1+[f'(x)]^2}\,dx \)
Washer method subtracts inner radius: \(V = \pi\int(R^2-r^2)\,dx\).
Arc length integrand always has a \(\sqrt{\phantom{x}}\) — never forget it.
Q 05
Find the area of the region enclosed by \(y = x\) and \(y = x^2\) on \([0,1]\).
Solution
On \([0,1]\), \(x \ge x^2\), so: \(\displaystyle\int_0^1(x-x^2)\,dx = \left[\tfrac{x^2}{2}-\tfrac{x^3}{3}\right]_0^1 = \tfrac{1}{2}-\tfrac{1}{3} = \dfrac{1}{6}\).
Q 06
The region bounded by \(y=\sqrt{x}\), \(y=0\), and \(x=4\) is revolved about the \(x\)-axis. Find the volume using the disk method.
Solution
Disk method: \(V = \pi\displaystyle\int_0^4(\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\!\left[\dfrac{x^2}{2}\right]_0^4 = \pi\cdot 8 = 8\pi\).
Q 07
Find the arc length of \(y = x^{3/2}\) from \(x=0\) to \(x=1\).
Solution
\(y'=\tfrac{3}{2}\sqrt{x}\), so \(1+(y')^2 = 1+\tfrac{9}{4}x\).\(L=\displaystyle\int_0^1\sqrt{1+\tfrac{9}{4}x}\,dx\). Let \(u=1+\tfrac{9}{4}x\), \(du=\tfrac{9}{4}dx\).
\(L=\tfrac{4}{9}\cdot\tfrac{2}{3}\Big[u^{3/2}\Big]_1^{13/4} = \dfrac{8}{27}\!\left[\left(\tfrac{13}{4}\right)^{3/2}-1\right] = \dfrac{13\sqrt{13}-8}{27}\approx 1.44\).
Q 08
Evaluate the improper integral \( \displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx \).
Solution
\(\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx = \lim_{b\to\infty}\!\left[-\frac{1}{x}\right]_1^b = \lim_{b\to\infty}\!\left(-\frac{1}{b}+1\right) = 0+1 = 1\).
UNIT 03Sequences & Series
Core Concept
Convergence Tests & Power Series
Geometric series: \( \sum_{n=0}^{\infty} ar^n = \dfrac{a}{1-r} \), \(|r|<1\)
p-series: \(\sum \dfrac{1}{n^p}\) converges iff \(p > 1\)
Ratio test: \(L=\lim\dfrac{|a_{n+1}|}{|a_n|}\); converges if \(L<1\), diverges if \(L>1\)
The harmonic series \(\sum 1/n\) diverges (p=1 fails).
Radius of convergence R from ratio test: \(|x-c| < R\).
Q 09
Find the sum of the geometric series \( \displaystyle\sum_{n=0}^{\infty} 2\!\left(\frac{1}{3}\right)^n \).
Solution
\(a=2\), \(r=\tfrac{1}{3}\), \(|r|<1\), so the series converges.\(S = \dfrac{a}{1-r} = \dfrac{2}{1-\tfrac{1}{3}} = \dfrac{2}{\tfrac{2}{3}} = 3\).
Q 10
Which of the following series diverges?
Solution
The harmonic series \(\sum 1/n\) is the classic divergent p-series (p=1). By the p-series test, convergence requires \(p>1\). The other choices have \(p=2\), \(p=3\), or are convergent geometric series.
Q 11
Apply the Ratio Test to \( \displaystyle\sum_{n=1}^{\infty}\frac{n!}{2^n} \). The series:
Solution
\(\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)!/2^{n+1}}{n!/2^n} = \dfrac{n+1}{2}\).As \(n\to\infty\), this ratio \(\to\infty > 1\), so the series diverges by the Ratio Test.
Q 12
Find the radius of convergence of the power series \( \displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!} \).
Solution
Ratio test: \(\left|\dfrac{x^{n+1}/(n+1)!}{x^n/n!}\right| = \dfrac{|x|}{n+1} \to 0\) for all \(x\).Since the limit is 0 < 1 for every real \(x\), the series converges everywhere, so \(R = \infty\).
(This is in fact the Maclaurin series for \(e^x\).)
Q 13
The Maclaurin series for \(\cos x\) begins with which of the following?
Solution
\(\cos x = \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!} = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\)The series contains only even powers with alternating signs. Option A is \(e^x\), option B is \(\sin x\), option D has no alternating signs.
UNIT 04L'Hôpital's Rule & Limits
Core Concept
Indeterminate Forms
L'Hôpital's Rule: if \(\tfrac{f}{g}\to\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\), then \(\lim\dfrac{f(x)}{g(x)}=\lim\dfrac{f'(x)}{g'(x)}\)
Common forms: \(0/0,\;\infty/\infty,\;0\cdot\infty,\;1^\infty,\;0^0,\;\infty^0\).
\(\lim_{x\to 0}\dfrac{\sin x}{x}=1\) — memorize directly, no L'Hôpital needed.
Q 14
Evaluate \( \displaystyle\lim_{x\to 0}\frac{\sin x}{x} \).
Solution
The form is \(0/0\). Applying L'Hôpital's Rule: \(\displaystyle\lim_{x\to 0}\frac{\cos x}{1} = \cos 0 = 1\).This is one of the most fundamental limits in calculus and should be memorized directly.
Q 15
Evaluate \( \displaystyle\lim_{x\to\infty}\frac{x^2}{e^x} \).
Solution
Form is \(\infty/\infty\). Apply L'Hôpital twice:\(\dfrac{x^2}{e^x}\xrightarrow{L'H}\dfrac{2x}{e^x}\xrightarrow{L'H}\dfrac{2}{e^x}\to 0\) as \(x\to\infty\).
Exponential growth always dominates polynomial growth.
UNIT 05Parametric Curves & Polar Coordinates
Core Concept
Parametric & Polar Calculus
Parametric arc length: \( L = \int_\alpha^\beta\sqrt{\left(\tfrac{dx}{dt}\right)^2+\left(\tfrac{dy}{dt}\right)^2}\,dt \)
Polar area: \( A = \dfrac{1}{2}\int_\alpha^\beta r^2\,d\theta \)
For \(x=r\cos\theta\), \(y=r\sin\theta\): \(x^2+y^2=r^2\).
Q 16
Find the arc length of the parametric curve \(x=\cos t\), \(y=\sin t\) for \(t\in[0,\tfrac{\pi}{2}]\).
Solution
\(\tfrac{dx}{dt}=-\sin t\), \(\tfrac{dy}{dt}=\cos t\).\(L=\displaystyle\int_0^{\pi/2}\!\sqrt{\sin^2t+\cos^2t}\,dt = \int_0^{\pi/2}1\,dt = \dfrac{\pi}{2}\).
This traces a quarter of the unit circle, confirming \(L = \tfrac{1}{4}(2\pi\cdot 1) = \tfrac{\pi}{2}\).
Q 17
Find the area enclosed by the polar curve \(r=2\) from \(\theta=0\) to \(\theta=\tfrac{\pi}{2}\).
Solution
Polar area: \(A = \dfrac{1}{2}\displaystyle\int_0^{\pi/2}r^2\,d\theta = \dfrac{1}{2}\int_0^{\pi/2}4\,d\theta = 2\cdot\dfrac{\pi}{2} = \pi\).This is one quarter of a circle with radius 2, whose total area is \(\pi(2)^2=4\pi\), and \(\tfrac{1}{4}\cdot 4\pi=\pi\). ✓
Q 18
Evaluate \( \displaystyle\int_0^2\sqrt{4-x^2}\,dx \).
Solution
Geometric approach: \(\sqrt{4-x^2}\) is the upper semicircle of \(x^2+y^2=4\) (radius 2).Integrating from 0 to 2 gives a quarter-circle area: \(\dfrac{1}{4}\pi(2)^2 = \pi\).
Alternatively, trig sub \(x=2\sin\theta\) yields the same result.
UNIT 06Differential Equations
Core Concept
Separable & Linear First-Order ODEs
Separable: \( \dfrac{dy}{dx} = f(x)g(y) \Rightarrow \int\dfrac{dy}{g(y)} = \int f(x)\,dx \)
Linear 1st-order: \( y'+P(x)y = Q(x) \), integrating factor \(\mu = e^{\int P\,dx}\)
After finding \(\mu\): multiply both sides, left side becomes \((\mu y)'\), then integrate.
Q 19
Find the general solution of \( \dfrac{dy}{dx} = xy \).
Solution
Separate variables: \(\dfrac{dy}{y} = x\,dx\).Integrate both sides: \(\ln|y| = \dfrac{x^2}{2} + C_1\).
Exponentiate: \(y = Ce^{x^2/2}\), where \(C = \pm e^{C_1}\) is an arbitrary constant.
Q 20
Find the general solution of \( \dfrac{dy}{dx} + y = e^x \).
Solution
Linear ODE with \(P(x)=1\). Integrating factor: \(\mu = e^{\int 1\,dx} = e^x\).Multiply: \((ye^x)' = e^x\cdot e^x = e^{2x}\).
Integrate: \(ye^x = \dfrac{e^{2x}}{2} + C\).
Divide by \(e^x\): \(y = \dfrac{e^x}{2} + Ce^{-x}\).