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ANTONII [103]
3 years ago
15

What are four possible energy sources for a circuit

Advanced Placement (AP)
2 answers:
Contact [7]3 years ago
5 0
Electric, solar, wind and geothermal
asambeis [7]3 years ago
3 0
Hi
<span> 

They are : electric, solar, wind, and geothermal.</span>
 

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Researchers have found that the more World Cup soccer a student watches the better they do on their exams. Students reported the
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Robin is to recieve the personal Appearance Ribbon today. What type of award is this
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AP CALC QUESTION!! WILL MARK BRAINLIEST (TWO QUESTIONS)
kodGreya [7K]

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}

8 0
3 years ago
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