Evaluate the integral.

${\int}_{2}^{3}[3{x}^{2}+2x+\frac{1}{{x}^{2}}]dx$

druczekq4
2021-11-20
Answered

Evaluate the integral.

${\int}_{2}^{3}[3{x}^{2}+2x+\frac{1}{{x}^{2}}]dx$

You can still ask an expert for help

breisgaoyz

Answered 2021-11-21
Author has **14** answers

Step 1

The given integral is${\int}_{2}^{3}[3{x}^{2}+2x+\frac{1}{{x}^{2}}]dx$ .

Step 2

Evaluate the given integral as shown below:

${\int}_{2}^{3}[3{x}^{2}+2x+\frac{1}{{x}^{2}}]dx={\int}_{2}^{3}\left[3{x}^{2}\right]dx+{\int}_{2}^{3}\left[2x\right]dx+{\int}_{2}^{3}\left[\frac{1}{{x}^{2}}\right]dx$

$=3{\int}_{2}^{3}{x}^{2}dx+2{\int}_{2}^{3}xdx+{\int}_{2}^{3}{x}^{-2}dx$

$=3{\left[\frac{{x}^{3}}{3}\right]}_{2}^{3}+2{\left[\frac{{x}^{2}}{2}\right]}_{2}^{3}+{\left[\frac{{x}^{-1}}{-1}\right]}_{2}^{3}$

$={\left[{x}^{3}\right]}_{2}^{3}+{\left[{x}^{2}\right]}_{2}^{3}-{\left[\frac{1}{x}\right]}_{2}^{3}$

$=[27-8]+[9-4]-[\frac{1}{3}-\frac{1}{2}]$

$=\left[19\right]+\left[5\right]+\left[\frac{1}{6}\right]$

$=\frac{145}{6}$

The given integral is

Step 2

Evaluate the given integral as shown below:

Rosemary McBride

Answered 2021-11-22
Author has **10** answers

Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:

${\int}_{a}^{b}f\left(x\right)dx=F\left(x\right){\mid}_{a}^{b}=F\left(b\right)-F\left(a\right)$

Step 2: In this case,$f\left(x\right)=(3{x}^{2}+2x+\frac{1}{{x}^{2}})$ . Find its integral.

$x}^{3}+{x}^{2}-\frac{1}{x}{\mid}_{2}^{3$

Step 3: Since$F\left(x\right){\mid}_{a}^{b}=F\left(b\right)-F\left(a\right)$ , expand the above into F(3)-F(2):

$({3}^{3}+{3}^{2}-\frac{1}{3})-({2}^{3}+{2}^{2}-\frac{1}{2})$

Step 4: Simplify.

$\frac{145}{6}$

Step 2: In this case,

Step 3: Since

Step 4: Simplify.

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Use a symbolic integration utility to find the indefinite integral.

$\int {(4t-3)}^{2}dt$

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Using the Euler's method (with $h={10}^{-n}$) to find $y(1)$

${y}^{\prime}=\frac{\mathrm{sin}(x)}{x}$

Since

${y}^{\prime}(x)=\frac{\mathrm{sin}(x)}{x}$

then

$f(x,y)=\frac{\mathrm{sin}(x)}{x}$

I know

${y}_{n+1}={y}_{n}+h\cdot f({x}_{n},{y}_{n})$

and given $y(0)=0$, so

${x}_{0}={y}_{0}=0$

Therefore,

${x}_{1}={x}_{0}+h=0+{10}^{-0}=1\Rightarrow {y}_{1}=y({x}_{1})=y(1)$

Then,

${y}_{1}=0+{10}^{-0}\cdot f({x}_{0},{y}_{0})$

${y}_{1}=f(0,0)$

But

$f(0,0)=\frac{\mathrm{sin}(0)}{0}=\frac{0}{0}$

How am I supposed to do it?

${y}^{\prime}=\frac{\mathrm{sin}(x)}{x}$

Since

${y}^{\prime}(x)=\frac{\mathrm{sin}(x)}{x}$

then

$f(x,y)=\frac{\mathrm{sin}(x)}{x}$

I know

${y}_{n+1}={y}_{n}+h\cdot f({x}_{n},{y}_{n})$

and given $y(0)=0$, so

${x}_{0}={y}_{0}=0$

Therefore,

${x}_{1}={x}_{0}+h=0+{10}^{-0}=1\Rightarrow {y}_{1}=y({x}_{1})=y(1)$

Then,

${y}_{1}=0+{10}^{-0}\cdot f({x}_{0},{y}_{0})$

${y}_{1}=f(0,0)$

But

$f(0,0)=\frac{\mathrm{sin}(0)}{0}=\frac{0}{0}$

How am I supposed to do it?

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Find the standard form of the equation of the circle having the following properties.

Center at the point (8,-6) Tangent to the x-axis

type the standard form of the equation of this circle

Center at the point (8,-6) Tangent to the x-axis

type the standard form of the equation of this circle

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Use a change of variables to evaluate the following integral.

$\int -({\mathrm{cos}}^{7}x-5{\mathrm{cos}}^{5}x-\mathrm{cos}x)\mathrm{sin}xdx$

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How can we use Euler's method to approximate the solutions for the following IVP below:

${y}^{\prime}=-y+t{y}^{1/2},\text{with}1\le t\le 2,\text{}y(1)=2,$

and with $h=0.5$

The main concern is the organization, i.e., set up of it for this particular example.

And, if the actual solution to the IVP above is:

$y(t)=(t-2+\sqrt{2}\mathrm{e}\cdot {\mathrm{e}}^{-t/2}{)}^{2}$

then, how to compare the actual error and compare the error bound?

${y}^{\prime}=-y+t{y}^{1/2},\text{with}1\le t\le 2,\text{}y(1)=2,$

and with $h=0.5$

The main concern is the organization, i.e., set up of it for this particular example.

And, if the actual solution to the IVP above is:

$y(t)=(t-2+\sqrt{2}\mathrm{e}\cdot {\mathrm{e}}^{-t/2}{)}^{2}$

then, how to compare the actual error and compare the error bound?