Tabansi
2021-10-14
Answered

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

Answered 2021-10-15
Author has **93** answers

Complete the square: $(8x-{x}^{2})=-({x}^{2}-8x+16-16)$

$\Rightarrow (8x-{x}^{2})=-{(x-4)}^{2}+16$

Substitute this integral, we get:

${\int}_{0}^{4}\sqrt{8x-{x}^{2}}dx={\int}_{0}^{4}\sqrt{-{(x-4)}^{2}+16}dx$

Apply u-substitution

u=x-4

$\Rightarrow du=dx$

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

$u=4\mathrm{sin}\left(v\right)$

$\Rightarrow du=4\mathrm{cos}\left(v\right)$

for$u=-4,v=\frac{-\pi}{2};u=0,v=0$

The integral in terms of v is:

${\int}_{-\frac{\pi}{2}}^{0}\sqrt{-16{\mathrm{sin}}^{2}v+16}\left(4\mathrm{cos}v\right)dv$

$\Rightarrow 4{\int}_{-\frac{\pi}{2}}^{0}\sqrt{16(1-{\mathrm{sin}}^{2}v)}\mathrm{cos}vdv$

$\Rightarrow 4{\int}_{-\frac{\pi}{2}}^{0}\sqrt{16{\mathrm{cos}}^{2}v}\mathrm{cos}vdv$

$\Rightarrow 4{\int}_{-\frac{\pi}{2}}^{0}4\mathrm{cos}v\mathrm{cos}vdv$

$\Rightarrow 16{\int}_{-\frac{\pi}{2}}^{0}{\mathrm{cos}}^{2}vdv$

$\Rightarrow 16{\int}_{-\frac{\pi}{2}}^{0}\frac{1+\mathrm{cos}\left(2v\right)}{2}dv$

$\Rightarrow 8{[v+\frac{1}{2}\mathrm{sin}\left(2v\right)]}_{-\frac{\pi}{2}}^{0}$

$\Rightarrow 8[(0+0)-(-\frac{\pi}{2}+\frac{1}{2}\mathrm{sin}(-\pi ))]$

$\Rightarrow 4\pi$

Answer:$4\pi$

Substitute this integral, we get:

Apply u-substitution

u=x-4

for x=0,u=-4;x=4,u=0

Thus the integral in terms of u is given by:

Use trigonometric substitution

for

The integral in terms of v is:

Answer:

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what is the integration of $\int \mathrm{cot}\left({e}^{x}\right)\cdot {e}^{x}dx$

This is my answer is it right

$u={e}^{x}$

$du={e}^{x}dx$

$dx=\frac{1}{{e}^{x}}du$

Then

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This is my answer is it right

Then

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We know that the solution of

${y}^{\prime}=y$

with $y(0)=1$ is

$y={e}^{x}$

As far as I see the Euler's method is explicitly used only to find the numerical approximation of e.g. $y(3)$. Can we use the Euler's method to solve this differential equation and find the exact solution?

Edit:

I thought of the following:

$y({t}_{1}+\delta t)=y({t}_{0})+{y}^{\prime}({t}_{1})\delta t$

$y({t}_{1}+2\delta t)=y({t}_{1}+\delta t)+{y}^{\prime}({t}_{1}+\delta t)\delta t$

$...$

and from that we could find a recursive solution for $y({t}_{0})$. Then by taking the limit $\underset{\delta t\to 0}{lim}$ of it, it could be possible to find the general solution of $y$ for every ${t}_{0}$ in the domain.

Is it impossible?

${y}^{\prime}=y$

with $y(0)=1$ is

$y={e}^{x}$

As far as I see the Euler's method is explicitly used only to find the numerical approximation of e.g. $y(3)$. Can we use the Euler's method to solve this differential equation and find the exact solution?

Edit:

I thought of the following:

$y({t}_{1}+\delta t)=y({t}_{0})+{y}^{\prime}({t}_{1})\delta t$

$y({t}_{1}+2\delta t)=y({t}_{1}+\delta t)+{y}^{\prime}({t}_{1}+\delta t)\delta t$

$...$

and from that we could find a recursive solution for $y({t}_{0})$. Then by taking the limit $\underset{\delta t\to 0}{lim}$ of it, it could be possible to find the general solution of $y$ for every ${t}_{0}$ in the domain.

Is it impossible?

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Whether the statement “ Let A be an $m\text{}\times \text{}n$ matrix.

The system$Ax=b\text{}\text{is consistent for all b in}\text{}{R}^{m}\text{}\text{if and only if the columns of A form a generating set for}\text{}{R}^{m}$ " is true or false.

The system