Question

AP CALC QUESTION!! WILL MARK BRAINLIEST (TWO QUESTIONS)

Answer 1

1. Using washers, the volume is given by the integral

$\displaystyle\pi\int_1^{e^2}((2+2)^2-(\ln x+2)^2)\,\mathrm dx$

$=\boxed{\displaystyle\pi\int_1^{e^2}(20+4\ln x-(\ln x)^2)\,\mathrm dx}$

We're using washers whose centers depend on the value of $x$, hence we integrate with respect to

2. The area of the given region is given by the integral

$\displaystyle\int_0^2\sin^{-1}\frac x2\,\mathrm dx$

To compute the integral, first consider the substitution $u=\frac x2$, or $2u=x$ so that $2\,\mathrm du=\mathrm dx$. Then $x\to0\implies u\to0$ and $x\to2\implies u\to1$, so the integral is equivalently

$\displaystyle2\int_0^1\sin^{-1}u\,\mathrm du$

Integrate by parts, taking

$f=\sin^{-1}u\implies\mathrm df=\dfrac{\mathrm du}{\sqrt{1-u^2}}$

$\mathrm dg=\mathrm du\implies g=u$

so that

$\displaystyle2\int_0^1\sin^{-1}u\,\mathrm du=2\left(u\sin^{-1}u\bigg|_0^1-\int_0^1\frac u{\sqrt{1-u^2}}\,\mathrm du\right)$

$\sin^{-1}0=0$ and $\sin^{-1}1=\frac\pi2$, so the area is

$\displaystyle\pi-2\int_0^1\frac u{\sqrt{1-u^2}}\,\mathrm du$

For the remaining integral, substitute $w=1-u^2$, so that $\mathrm dw=-2u\,\mathrm du$. Then $u\to0\implies w\to1$ and $u\to1\implies w\to0$:

$\displaystyle\pi-\int_0^1\frac{\mathrm dw}{\sqrt w}$

(notice that the integral is improper)

$\displaystyle\pi-\lim_{t\to0^+}2\sqrt w\bigg|_t^1$

$\displaystyle\pi-2\left(1-\lim_{t\to0^+}\sqrt t\right)=\boxed{\pi-2}$