Use geometry and properties of integrals to evaluate the following definite inte

Use geometry and properties of integrals to evaluate the following definite integrals.
${\int }_{0}^{4}\sqrt{8x-{x}^{2}}dx$
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Asma Vang
Complete the square: $\left(8x-{x}^{2}\right)=-\left({x}^{2}-8x+16-16\right)$
$⇒\left(8x-{x}^{2}\right)=-{\left(x-4\right)}^{2}+16$
Substitute this integral, we get:
${\int }_{0}^{4}\sqrt{8x-{x}^{2}}dx={\int }_{0}^{4}\sqrt{-{\left(x-4\right)}^{2}+16}dx$
Apply u-substitution
u=x-4
$⇒du=dx$
for x=0,u=-4;x=4,u=0
Thus the integral in terms of u is given by:
${\int }_{0}^{4}\sqrt{8x-{x}^{2}}dx={\int }_{-4}^{0}\sqrt{-{u}^{2}+16}du$
Use trigonometric substitution
$u=4\mathrm{sin}\left(v\right)$
$⇒du=4\mathrm{cos}\left(v\right)$
for $u=-4,v=\frac{-\pi }{2};u=0,v=0$
The integral in terms of v is:
${\int }_{-\frac{\pi }{2}}^{0}\sqrt{-16{\mathrm{sin}}^{2}v+16}\left(4\mathrm{cos}v\right)dv$
$⇒4{\int }_{-\frac{\pi }{2}}^{0}\sqrt{16\left(1-{\mathrm{sin}}^{2}v\right)}\mathrm{cos}vdv$
$⇒4{\int }_{-\frac{\pi }{2}}^{0}\sqrt{16{\mathrm{cos}}^{2}v}\mathrm{cos}vdv$
$⇒4{\int }_{-\frac{\pi }{2}}^{0}4\mathrm{cos}v\mathrm{cos}vdv$
$⇒16{\int }_{-\frac{\pi }{2}}^{0}{\mathrm{cos}}^{2}vdv$
$⇒16{\int }_{-\frac{\pi }{2}}^{0}\frac{1+\mathrm{cos}\left(2v\right)}{2}dv$
$⇒8{\left[v+\frac{1}{2}\mathrm{sin}\left(2v\right)\right]}_{-\frac{\pi }{2}}^{0}$
$⇒8\left[\left(0+0\right)-\left(-\frac{\pi }{2}+\frac{1}{2}\mathrm{sin}\left(-\pi \right)\right)\right]$
$⇒4\pi$
Answer: $4\pi$