Cumulative assessment: Calculus II

Apply integration by parts twice

Take $u=x^2$, $\dd v=\sin x\dd x$, giving $v=-\cos x$. Then

For the remaining integral, take $u=x$, $\dd v=\cos x\dd x$, so it equals $x\sin x+\cos x$. Thus

Differentiation gives $-2x\cos x+x^2\sin x+2\sin x+2x\cos x-2\sin x=x^2\sin x$. This check confirms both correction terms and their signs.

Factor the denominator and solve for constants

The domain excludes $x=\pm3$. Write

Multiplying through gives $1=A(x+3)+B(x-3)$. Evaluating at $3$ and $-3$ gives $A=1/6$, $B=-1/6$. Therefore

is an antiderivative on each interval avoiding the singularities. Constants may differ between disconnected intervals. Differentiating recovers $[(x+3)-(x-3)]/[6(x^2-9)]=1/(x^2-9)$.

A trigonometric substitution removes a quadratic denominator

Set $x=\tan\theta$ on $0\le\theta\le\pi/4$. Then $\dd x=\sec^2\theta\dd\theta$ and $(1+x^2)^2=\sec^4\theta$, so

The original integrand lies between $1/4$ and $1$ on an interval of length one, and the answer lies between those bounds. No sign ambiguity arises because the chosen tangent branch is one-to-one here.

Integration by parts leaves a vanishing tail

For a finite upper endpoint $b$, take $u=\ln x$, $\dd v=x^{-2}\dd x$, giving $v=-1/x$. Then

Since $\ln b/b\to0$ and $1/b\to0$, the improper integral converges to $\boxed1$. The boundary expression is evaluated through its limit, not by treating infinity as an ordinary argument.

Report a computable estimate and a theorem-based guarantee

With $h=1/4$,

The fourth derivative of cosine is cosine, so $|f^{(4)}|\le1$ on $[0,1]$. The error bound is

The exact integral is $\sin1$, whose comparison confirms an actual error of about $1.84\times10^{-5}$. The numerical comparison is a check; the derivative bound is what supplies the guarantee without already knowing the exact integral.

First convergence, then evaluation

For $a_n=n^2/3^n$, the ratio is $a_{n+1}/a_n=(n+1)^2/(3n^2)\to1/3<1$, so the series converges. For $|r|<1$, the geometric-series identity and legitimate termwise differentiation give

Apply $r\,\dd/\dd r$ again:

At $r=1/3$, the sum is $\boxed{3/2}$. The ratio test alone does not produce that value; the power-series identity is a separate argument.

A valid bound may demand many terms

The positive terms $1/\sqrt{n+1}$ decrease to zero, so the alternating series converges. The absolute series diverges by comparison with the $p$-series of exponent $1/2$ (or by an index shift). Thus convergence is conditional. After $N$ terms, $|R_N|\le1/\sqrt{N+2}$. Requiring this to be at most $0.02=1/50$ gives $N+2\ge2500$, or $\boxed{N\ge2498}$. This sufficient count shows how slowly this bound improves. A rapidly convergent reformulation, when available, may be preferable for computation.

The radius does not decide the endpoints

The absolute ratio tends to $|x-2|/4$, giving radius $\boxed{R=4}$. At $x=-2$, the series is $\sum(-1)^n/n$, which converges conditionally. At $x=6$, it is harmonic and diverges. Thus the interval is $\boxed{[-2,6)}$. Inside $|x-2|<4$, termwise differentiation gives

At either endpoint, the derivative-series terms have magnitude $1/4$ and do not tend to zero. Therefore its interval is $(-2,6)$, even though the original series converges at the left endpoint. An interior derivative identity does not automatically extend termwise to an endpoint.

Integrate an approximation together with a remainder bound

The polynomial is $P_5(x)=2x-(2x)^3/3!+(2x)^5/5!=2x-4x^3/3+4x^5/15$. The original quotient tends to $2$ at zero, so its continuous extension poses no improper singularity. Dividing the polynomial by $x$ and integrating gives

On $0\le x\le1/5$, the sine-series term magnitudes decrease; the next omitted term bounds the error by $(2x)^7/7!$. After division by $x>0$, the quotient error is at most $8x^6/315$, and the bound extends continuously as zero at $x=0$. Therefore the integral error is at most

The omitted sine term is negative, so this particular polynomial integral is an overestimate. The bound and direction have been justified on the whole integration interval, not only at its endpoint.

Use parameter speed for length and horizontal motion for area

Differentiate: $x'=1-\cos t$, $y'=\sin t$. Speed is

On $[0,2\pi]$, $\sin(t/2)\ge0$, so length is $L=\int_0^{2\pi}2\sin(t/2)\dd t=[-4\cos(t/2)]_0^{2\pi}=\boxed8$. For area, $x'\ge0$ and $y\ge0$, so

The endpoints have zero parameter velocity, but the curve is smooth inside and the speed integral remains valid. Length and area require different differential factors.

Subtract radial sector areas on the stated angle interval

In the first quadrant $0\le\theta\le\pi/2$, the cardioid radius $1+\cos\theta$ lies between $1$ and $2$, so the region has $1+\cos\theta\le r\le2$. Hence

The answer is positive and less than the first-quadrant disk area $\pi$. The area is not one-half the integral of a plain difference of radii.

Verify both the equation and the initial value

An integrating factor is $e^{2t}$. Multiplication gives $(e^{2t}y)'=4e^{2t}$, so $e^{2t}y=2e^{2t}+C$ and $y=2+Ce^{-2t}$. The initial condition gives $C=3$: $\boxed{y=2+3e^{-2t}}$. Then $y'=-6e^{-2t}$ and $y'+2y=-6e^{-2t}+4+6e^{-2t}=4$, while $y(0)=5$. The equilibrium is $y=2$ and this solution tends toward it from above as $t\to\infty$. The differential equation determines a family until an initial condition selects one member.