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3

## Evaluate the following term int_0^(3pi/2) 5|sinx|dx .How would i do this using FTC2(F(b)-F(a))?

Jim H
Featured 5 days ago

We need to split up the integral.

#### Explanation:

Recall that $\left\mid u \right\mid = \left\{\begin{matrix}u & \text{if" & u >= 0 \\ -u & "if} & u < 0\end{matrix}\right.$

so $\left\mid \sin x \right\mid = \left\{\begin{matrix}\sin x & \text{if" & sinx >= 0 \\ -sinx & "if} & \sin x < 0\end{matrix}\right.$.

We are integrating on $\left[0 , \frac{3 \pi}{2}\right]$ and we know that

$\left\{\begin{matrix}\sin x \ge 0 & \text{if" & 0 <=x <= pi \\ sinx < 0 & "if} & \pi < x \le \frac{3 \pi}{2}\end{matrix}\right.$

Therefore,

$\left\mid \sin x \right\mid = \left\{\begin{matrix}\sin x & \text{if" & 0 <= x <= pi \\ -sinx & "if} & \pi < x < \frac{3 \pi}{2}\end{matrix}\right.$.

${\int}_{0}^{\frac{3 \pi}{2}} 5 \left\mid \sin x \right\mid \mathrm{dx} = 5 {\int}_{0}^{\frac{3 \pi}{2}} \left\mid \sin x \right\mid \mathrm{dx}$

$= 5 \left[{\int}_{0}^{\pi} \sin x \mathrm{dx} + {\int}_{\pi}^{\frac{3 \pi}{2}} - \sin x \mathrm{dx}\right]$

$= 5 \left[{\int}_{0}^{\pi} \sin x \mathrm{dx} - {\int}_{\pi}^{\frac{3 \pi}{2}} \sin x \mathrm{dx}\right]$

Now use the fact that $\int \sin x \mathrm{dx} = - \cos x + C$ to find each of the integrals.

$= 5 \left[{\left.- \cos x\right]}_{0}^{\pi} + {\left.\cos x\right]}_{\pi}^{\frac{3 \pi}{2}}\right]$

$= 5 \left[\left(- \cos \pi + \cos 0\right) + \left(\cos \left(\frac{3 \pi}{2}\right) - \cos \pi\right)\right]$

 = 5[(-(-1)+1+0-(-1)]

$= 5 \left[3\right] = 15$

Bonus method

Some people prefer to integrate $\left\mid f \left(x\right) \right\mid$ by simply integrating from one zero to the next without first adjusting the sign. Any integral that comes out negative, we make positive.

The notation for this technique is

${\int}_{0}^{\frac{3 \pi}{2}} 5 \left\mid \sin x \right\mid \mathrm{dx} = \left\mid {\int}_{0}^{\pi} 5 \sin x \mathrm{dx} \right\mid + \left\mid {\int}_{\pi}^{\frac{3 \pi}{2}} 5 \sin x \mathrm{dx} \right\mid$

The first of these two integrals will be positive and the second will be negative. (That's why the first method changed the sign for the second integral before integrating.)

2

## How do you integrate (x^4+x^3+x^2+1)/(x^2+x-2) using partial fractions?

Douglas K.
Featured 5 days ago

Because the order of the numerator is greater than the denominator some reduction must be done before expanding the partial fractions. Please see the explanation.

#### Explanation:

Given: $\frac{{x}^{4} + {x}^{3} + {x}^{2} + 1}{{x}^{2} + x - 2}$

Begin reducing by adding 0 to the numerator in the form of $- 2 {x}^{2} + 2 {x}^{2}$

$\frac{{x}^{4} + {x}^{3} - 2 {x}^{2} + 2 {x}^{2} + {x}^{2} + 1}{{x}^{2} + x - 2}$

Break into two fractions:

$\frac{{x}^{4} + {x}^{3} - 2 {x}^{2}}{{x}^{2} + x - 2} + \frac{2 {x}^{2} + {x}^{2} + 1}{{x}^{2} + x - 2}$

Factor ${x}^{2}$ from the numerator of the first term:

$\frac{{x}^{2} \left(\cancel{{x}^{2} + x - 2}\right)}{\cancel{{x}^{2} + x - 2}} + \frac{2 {x}^{2} + {x}^{2} + 1}{{x}^{2} + x - 2}$

The first term becomes ${x}^{2}$ and we combine like terms in the numerator of the second fraction:

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

Add 0 to the numerator of the second term in the form of $3 x - 6 - 3 x + 6$:

${x}^{2} + \frac{3 {x}^{2} + 3 x - 6 - 3 x + 6 + 1}{{x}^{2} + x - 2}$

Break the second term into two fractions:

${x}^{2} + \frac{3 {x}^{2} + 3 x - 6}{{x}^{2} + x - 2} + \frac{- 3 x + 6 + 1}{{x}^{2} + x - 2}$

Remove a factor of 3 from the first numerator and combine like terms in the last fraction:

${x}^{2} + \frac{3 \left(\cancel{{x}^{2} + x - 2}\right)}{\cancel{{x}^{2} + x - 2}} + \frac{7 - 3 x}{{x}^{2} + x - 2}$

The second term becomes 3:

${x}^{2} + 3 + \frac{7 - 3 x}{{x}^{2} + x - 2}$

Partial Fraction Expansion of the last term:

$\frac{7 - 3 x}{{x}^{2} + x - 2} = \frac{A}{x + 2} + \frac{B}{x - 1}$

$7 - 3 x = A \left(x - 1\right) + B \left(x + 2\right)$

Make B disappear by letting x = -2

$7 - 3 \left(- 2\right) = A \left(- 2 - 1\right) + B \left(- 2 + 2\right)$

$A$ = -13/3

Make A disappear by Letting x = 1:

$7 - 3 \left(1\right) = A \left(1 - 1\right) + B \left(1 + 2\right)$

$B = \frac{4}{3}$

Check

$\frac{- \frac{13}{3}}{x + 2} + \frac{\frac{4}{3}}{x - 1}$

$\frac{- \frac{13}{3}}{x + 2} \frac{x - 1}{x - 1} + \frac{\frac{4}{3}}{x - 1} \frac{x + 2}{x + 2}$

$\frac{\left(- \frac{13}{3}\right) \left(x - 1\right) + \left(\frac{4}{3}\right) \left(x + 2\right)}{\left(x + 2\right) \left(x - 1\right)}$

$\frac{7 - 3 x}{\left(x + 2\right) \left(x - 1\right)}$

This checks

Returning to the main problem:

$\int \frac{{x}^{4} + {x}^{3} + {x}^{2} + 1}{{x}^{2} + x - 2} \mathrm{dx} = \int {x}^{2} \mathrm{dx} + 3 \int \mathrm{dx} - \frac{13}{3} \int \frac{1}{x + 2} \mathrm{dx} + \frac{4}{3} \int \frac{1}{x - 1} \mathrm{dx}$

$\int \frac{{x}^{4} + {x}^{3} + {x}^{2} + 1}{{x}^{2} + x - 2} \mathrm{dx} = \frac{1}{3} {x}^{3} + 3 x - \frac{13}{3} \ln \left(x + 2\right) + \frac{4}{3} \ln \left(x - 1\right) + C$

1

## Evaluate the given sum?

HSBC244
Featured 5 days ago

The integral is $\ln | \frac{1 + \sqrt{1 - {x}^{2}}}{x} | - {\cos}^{-} 1 \frac{x}{x} + C$

#### Explanation:

This can be rewritten as $\int {\cos}^{-} 1 x {x}^{-} 2 \mathrm{dx}$. Whenever you see a product within an integral, you should consider using integration by parts.

The integration by parts formula states that an integral $\int u \mathrm{dv} = u v - \int \left(v \mathrm{du}\right)$. ${\cos}^{-} 1 x$ is not that simple to integrate, so let's say $u = {\cos}^{-} 1 x$ and $\mathrm{dv} = {x}^{-} 2$. Then $\mathrm{du} = - \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$ and $v = - \frac{1}{x}$.

Use the formula now:

$\int {\cos}^{-} 1 x {x}^{-} 2 \mathrm{dx} = - {\cos}^{-} 1 \frac{x}{x} - \int \left(- \frac{1}{x} \cdot - \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}\right)$

$\int {\cos}^{-} 1 x {x}^{-} 2 \mathrm{dx} = - {\cos}^{-} 1 \frac{x}{x} - \int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx}$

The remaining integral will have to be integrated by trigonometric substitution. This is a technique of integration where you make use of a pythagorean identity to get rid of the √. When the square root within the integral is of the form $\sqrt{{a}^{2} - {x}^{2}}$ like the one above, use the substitution $\textcolor{red}{x = \sin \theta}$. Then $\mathrm{dx} = \cos \theta d \theta$.

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \frac{1}{\sin \theta \sqrt{1 - {\sin}^{2} \theta}} \cdot \cos \theta d \theta$

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \frac{1}{\sin \theta \sqrt{{\cos}^{2} \theta}} \cdot \cos \theta d \theta$

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \frac{1}{\sin} \theta d \theta$

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \csc \theta d \theta$

This is a known integral that is derived here.

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = - \ln | \csc \theta + \cot \theta | + C$

We now reverse the substitutions by drawing a triangle and finding the correct ratios. We know from our original substitution that $\frac{x}{1} = \sin \theta$.

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = - \ln | \frac{1}{x} + \frac{\sqrt{1 - {x}^{2}}}{x} | + C$

$\int \frac{1}{x \sqrt{1 - {x}^{2}}} \mathrm{dx} = - \ln | \frac{1 + \sqrt{1 - {x}^{2}}}{x} | + C$

Let's return our attention to the whole integral.

$\int {\cos}^{-} 1 x {x}^{-} 2 \mathrm{dx} = - {\cos}^{-} 1 \frac{x}{x} - \left(- \ln | \frac{1 + \sqrt{1 - {x}^{2}}}{x} |\right) + C$

$\int {\cos}^{-} 1 x {x}^{-} 2 \mathrm{dx} = \ln | \frac{1 + \sqrt{1 - {x}^{2}}}{x} | - {\cos}^{-} 1 \frac{x}{x} + C$

Hopefully this helps!

1

## How do you evaluate the definite integral int (x^2+2x)/(x+1)^2 from [0, 1]?

HSBC244
Featured 4 days ago

$\frac{1}{2}$

#### Explanation:

Before evaluating, we always have to find the integral. We look to simplify the integral as much as possible to see if anything can be cancelled.

${\int}_{0}^{1} \frac{x \left(x + 2\right)}{x + 1} ^ 2$

We can't cancel anything, but a u-substitution would be effective. Let $u = x + 1$. Then $\mathrm{du} = \mathrm{dx}$. Also note that $x = u - 1$ and $x + 2 = u + 1$. The integral therefore becomes:

${\int}_{0}^{1} \frac{\left(u - 1\right) \left(u + 1\right)}{u} ^ 2 \mathrm{du}$

Expand this:

${\int}_{0}^{1} \frac{{u}^{2} - u + u - 1}{u} ^ 2 \mathrm{du}$

${\int}_{0}^{1} \frac{{u}^{2} - 1}{u} ^ 2 \mathrm{du}$

Break into separate fractions.

${\int}_{0}^{1} {u}^{2} / {u}^{2} - \frac{1}{u} ^ 2 \mathrm{du}$

${\int}_{0}^{1} 1 - \frac{1}{u} ^ 2 \mathrm{du}$

${\int}_{0}^{1} 1 - {u}^{-} 2 \mathrm{du}$

You can now integrate this as $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$, where $n \ne - 1$.

${\left[u + \frac{1}{u}\right]}_{0}^{1}$

Reverse the substitution, since the initial variable wasn't $u$, it was $x$.

${\left[x + 1 + \frac{1}{x + 1}\right]}_{0}^{1}$

Evaluate using the second fundamental theorem of calculus, which states that ${\int}_{a}^{b} F \left(x\right) \mathrm{dx} = f \left(b\right) - f \left(a\right)$, where $f ' \left(x\right) = F \left(x\right)$ in all $\left[a , b\right]$.

$1 + 1 + \frac{1}{1 + 1} - \left(0 + 1 + \frac{1}{0 + 1}\right)$

$2 + \frac{1}{2} - 2$

$\frac{1}{2}$

Hopefully this helps!

1

## How do you evaluate the integral int lnx/(1+x)^2?

mason m
Featured yesterday

$- \ln \frac{x}{1 + x} + \ln x - \ln \left(1 + x\right) + C$

#### Explanation:

$I = \int \ln \frac{x}{1 + x} ^ 2 \mathrm{dx}$

Integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. Let:

$u = \ln x$
$\mathrm{dv} = \frac{1}{1 + x} ^ 2 \mathrm{dx}$

Differentiate $u$ and integrate $\mathrm{dv}$. The integration of $\mathrm{dv}$ is best performed with the substitution $t = 1 + x \implies \mathrm{dt} = \mathrm{dx}$. You should get:

$\mathrm{du} = \frac{1}{x} \mathrm{dx}$
$v = - \frac{1}{1 + x}$

Then:

$I = u v - \int v \mathrm{du}$

$I = - \ln \frac{x}{1 + x} - \int \frac{1}{x} \left(- \frac{1}{1 + x}\right) \mathrm{dx}$

$I = - \ln \frac{x}{1 + x} + \int \frac{1}{x \left(1 + x\right)} \mathrm{dx}$

Perform partial fraction decomposition on $\frac{1}{x \left(1 + x\right)}$:

$\frac{1}{x \left(1 + x\right)} = \frac{A}{x} + \frac{B}{1 + x}$

Then:

$1 = A \left(1 + x\right) + B x$

Letting $x = - 1$:

$1 = A \left(1 - 1\right) + B \left(- 1\right)$

$B = - 1$

Letting $x = 0$:

$1 = A \left(1 + 0\right) + B \left(0\right)$

$1 = A$

Then:

$\frac{1}{x \left(1 + x\right)} = \frac{1}{x} - \frac{1}{1 + x}$

So:

$I = - \ln \frac{x}{1 + x} + \int \frac{1}{x} \mathrm{dx} - \int \frac{1}{1 + x} \mathrm{dx}$

These are simple integrals. The second can be performed with the substitution $s = 1 + x \implies \mathrm{ds} = \mathrm{dx}$.

$I = - \ln \frac{x}{1 + x} + \ln x - \ln \left(1 + x\right) + C$

3

## What is the integrating factor of 3(x^2+y^2)dx + x(x^2+3y+6y)dy = 0?

Truong-Son N.
Featured 4 days ago

I got:

$\mu \left(y\right) = {e}^{y}$

The point of an integrating factor is to turn an inexact differential into an exact one. One physical application of this is to turn a path function into a state function in chemistry (such as dividing by $T$ to turn ${q}_{\text{rev}}$, a path function, into $S$, a state function, entropy).

I assume that the second terms include $3 {y}^{2}$, not $3 y$ (it would be odd to not simply write $9 y$).

Two options to find the special integrating factor, as defined by Nagle, are:

$\boldsymbol{\mu \left(x\right) = \text{exp} \left[\int \frac{{\left(\frac{\partial M}{\partial y}\right)}_{x} - {\left(\frac{\partial N}{\partial x}\right)}_{y}}{N \left(x , y\right)} \mathrm{dx}\right]}$,

$\boldsymbol{\mu \left(y\right) = \text{exp} \left[\int \frac{{\left(\frac{\partial N}{\partial x}\right)}_{y} - {\left(\frac{\partial M}{\partial y}\right)}_{x}}{M \left(x , y\right)} \mathrm{dy}\right]}$,

for the differential

$\boldsymbol{\mathrm{dF} \left(x , y\right) = {\left(\frac{\partial F}{\partial x}\right)}_{y} \mathrm{dx} + {\left(\frac{\partial F}{\partial y}\right)}_{x} \mathrm{dy}}$,

where $M = {\left(\frac{\partial F}{\partial x}\right)}_{y}$ and $N = {\left(\frac{\partial F}{\partial y}\right)}_{x}$.

For now, let's find the partial derivatives. For your differential:

$\textcolor{g r e e n}{M \left(x , y\right) = 3 {x}^{2} + 3 {y}^{2}}$
$\textcolor{g r e e n}{N \left(x , y\right) = {x}^{3} + 3 x {y}^{2} + 6 x y}$

Therefore:

$\textcolor{g r e e n}{{\left(\frac{\partial M}{\partial y}\right)}_{x} = 6 y}$

$\textcolor{g r e e n}{{\left(\frac{\partial N}{\partial x}\right)}_{y} = 3 {x}^{2} + 3 {y}^{2} + 6 y}$

which are clearly not equal, so the current differential is inexact. So, let us divide by $N \left(x , y\right)$ and see if the integral with respect to $x$ is reasonable to do:

$\ln \mu \left(x\right) = \int \frac{6 y - 3 {x}^{2} - 3 {y}^{2} - 6 y}{{x}^{3} + 3 x {y}^{2} + 6 x y} \mathrm{dx}$

$= \int \frac{- 3 {x}^{2} - 3 {y}^{2}}{{x}^{3} + 3 x {y}^{2} + 6 x y} \mathrm{dx}$

This doesn't look all that nice (it cannot be readily factored to eliminate $y$ terms), so what if we try integrating with respect to $y$ instead, and dividing by $M \left(x , y\right)$ instead? Then:

$\mu \left(y\right) = \text{exp} \left[\int \frac{{\left(\frac{\partial N}{\partial x}\right)}_{y} - {\left(\frac{\partial M}{\partial y}\right)}_{x}}{M \left(x , y\right)} \mathrm{dy}\right]$

and:

$\ln \mu \left(y\right) = \int \frac{3 {x}^{2} + 3 {y}^{2} + 6 y - 6 y}{3 {x}^{2} + 3 {y}^{2}} \mathrm{dy}$

$= \int {\cancel{\frac{3 {x}^{2} + 3 {y}^{2}}{3 {x}^{2} + 3 {y}^{2}}}}^{1} \mathrm{dy}$

And this integral is easy. It's just $y$. Therefore, $\textcolor{b l u e}{\mu \left(y\right) = {e}^{y}}$ would be your integrating factor. Let's test it out:

$3 {e}^{y} \left({x}^{2} + {y}^{2}\right) \mathrm{dx} + x {e}^{y} \left({x}^{2} + 3 {y}^{2} + 6 y\right) \mathrm{dy} = 0$

$\left(3 {x}^{2} {e}^{y} + 3 {y}^{2} {e}^{y}\right) \mathrm{dx} + \left({x}^{3} {e}^{y} + 3 x {y}^{2} {e}^{y} + 6 x y {e}^{y}\right) \mathrm{dy} = 0$

Checking for exactness, we obtain:

((delM)/(dely))_x stackrel(?" ")(=) ((delN)/(delx))_y

3x^2e^y + 3(y^2e^y + 2ye^y) stackrel(?" ")(=) 3x^2e^y + 3y^2e^y + 6ye^y

$3 {x}^{2} {e}^{y} + 3 {y}^{2} {e}^{y} + 6 y {e}^{y} = 3 {x}^{2} {e}^{y} + 3 {y}^{2} {e}^{y} + 6 y {e}^{y}$ color(blue)(sqrt"")

so we know our integrating factor is correct!

2

## How do you find the equations for the tangent plane to the surface x=y(2z-3) through (4, 4, 2)?

Steve
Featured 2 days ago

$x - y - 8 z = - 16$

#### Explanation:

First we rearrange the equation of the surface into the form $f \left(x , y , z\right) = 0$

$x = y \left(2 z - 3\right)$
$\therefore x = 2 y z - 3 y$
$\therefore x + 3 y - 2 y z = 0$

And so we have our function:

$f \left(x , y , z\right) = x + 3 y - 2 y z$

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

$\nabla f \left(x , y , z\right) = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} + \frac{\partial f}{\partial z} \hat{k}$

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

$\nabla f = \left(\frac{\partial}{\partial x} \left(x + 3 y - 2 y z\right)\right) \hat{i} +$
$\text{ } \left(\frac{\partial}{\partial y} \left(x + 3 y - 2 y z\right)\right) \hat{j} +$
$\text{ } \left(\frac{\partial}{\partial z} \left(x + 3 y - 2 y z\right)\right) \hat{k}$
$\text{ } = \left(1\right) \hat{i} + \left(3 - 2 z\right) \hat{j} + \left(- 2 y\right) \hat{k}$
$\text{ } = \hat{i} + \left(3 - 2 z\right) \hat{j} - \left(2 y\right) \hat{k}$

So for the particular point $\left(4 , 4 , 2\right)$ the normal vector to the surface is given by:

$\nabla f \left(4 , 4 , 2\right) = \hat{i} + \left(3 - 2 \cdot 2\right) \hat{j} - \left(2 \cdot 4\right) \hat{k}$
$\text{ } = \hat{i} - \hat{j} - 8 \hat{k}$
$\text{ } = \left(\begin{matrix}1 \\ - 1 \\ - 8\end{matrix}\right)$

So the tangent plane to the surface $x = y \left(2 z - 3\right)$ has this normal vector and it also passes though the point $\left(4 , 4 , 2\right)$. It will therefore have a vector equation of the form:

$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

Where $\vec{r} = \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)$; $\vec{n}$ is the normal, and $a$ is any point in the plane

Hence, the tangent plane equation is:

$\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) \cdot \left(\begin{matrix}1 \\ - 1 \\ - 8\end{matrix}\right) = \left(\begin{matrix}4 \\ 4 \\ 2\end{matrix}\right) \cdot \left(\begin{matrix}1 \\ - 1 \\ - 8\end{matrix}\right)$
$\therefore \left(x\right) \left(1\right) + \left(y\right) \left(- 1\right) + \left(z\right) \left(- 8\right) = \left(4\right) \left(1\right) + \left(4\right) \left(- 1\right) + \left(2\right) \left(- 8\right)$
$\therefore x - y - 8 z = 4 - 4 - 16$
$\therefore x - y - 8 z = - 16$

We can confirm this graphically: Here is the surface with the normal vector:

and here is the surface with the tangent plane and the normal vector:

1

## What is a solution to the differential equation dy/dx=1+2xy?

Steve
Featured yesterday

$y = \frac{\sqrt{\pi}}{2} {e}^{{x}^{2}} e r f \left(x\right) + A {e}^{{x}^{2}}$

Where $e r f \left(x\right)$ is the Error Function :

$e r f \left(x\right) = \frac{2}{\sqrt{\pi}} {\int}_{0}^{x} {e}^{- {t}^{2}} \setminus \mathrm{dt}$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 + 2 x y$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x y = 1$ ..... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$\frac{\mathrm{dy}}{\mathrm{dx}} + P \left(x\right) y = Q \left(x\right)$

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

$I = {e}^{\int P \left(x\right) \mathrm{dx}}$
$\setminus \setminus = {e}^{\int \setminus - 2 x \setminus \mathrm{dx}}$
$\setminus \setminus = {e}^{- {x}^{2}}$

And if we multiply the DE [1] by this Integrating Factor we will have a perfect product differential;

$\frac{\mathrm{dy}}{\mathrm{dx}} - 2 x y = 1$
$\therefore {e}^{- {x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x y {e}^{- {x}^{2}} = 1 \cdot {e}^{- {x}^{2}}$
$\therefore \frac{d}{\mathrm{dx}} \left(y {e}^{- {x}^{2}}\right) = {e}^{- {x}^{2}}$

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

$y {e}^{- {x}^{2}} = \int \setminus {e}^{- {x}^{2}} \setminus \mathrm{dx}$

The RHS integral does not have an elementary form, but we can use the definition of the Error Function :

$e r f \left(x\right) = \frac{2}{\sqrt{\pi}} {\int}_{0}^{x} {e}^{- {t}^{2}} \setminus \mathrm{dt}$

Which gives us:

$y {e}^{- {x}^{2}} = \frac{\sqrt{\pi}}{2} e r f \left(x\right) + A$
$y = \frac{\sqrt{\pi}}{2} {e}^{{x}^{2}} e r f \left(x\right) + A {e}^{{x}^{2}}$

2

## The parabola y=x^2 is paramatized by x(t)=t & y(t)=t^2. At the point A(t,t^2) a line segment AP 1 unit long is perpendicular to the parabola extending inward. What is the parametric equation of the curve traced by the point P as A moves along parabola?

Steve
Featured 21 hours ago

$P = P \left(t - 2 t \sqrt{\frac{1}{4 {t}^{2} + 1}} , {t}^{2} + \sqrt{\frac{1}{4 {t}^{2} + 1}}\right)$

So the parametric equations of P are:

${P}_{x} \left(t\right) = t - 2 t \sqrt{\frac{1}{4 {t}^{2} + 1}}$
${p}_{y} \left(t\right) = {t}^{2} + \sqrt{\frac{1}{4 {t}^{2} + 1}}$

#### Explanation:

Our parabola $y = {x}^{2}$ is parametrised by $x = t$ and $y = {t}^{2}$, and $A \left(t , {t}^{2}\right)$ is a general point on the parabola, and we give the point $P$ the coordinates (to be determined) $P \left(\alpha , \beta\right)$

Differentiating wrt $x$ we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x = 2 t$

So the gradient of the tangent at $A$ is given by $\frac{\mathrm{dy}}{\mathrm{dx}} = t$, We are told that $A P$ is perpendicular to to parabola, in other words, it is perpendicular to the tangent at A. Hence the gradient of AP is $- \frac{1}{2 t}$ as the product of their gradients are -1.

We can now form the equation of the normal $A P$ using the point-slope form of the straight line, $y - {y}_{1} = m \left(x - {x}_{1}\right)$:

$y - {t}^{2} = - \frac{1}{2 t} \left(x - t\right)$
$\therefore 2 t y - 2 {t}^{3} = - \left(x - t\right)$
$\therefore 2 t y - 2 {t}^{3} = + t - x$

$P \left(\alpha , \beta\right)$ lies on this line, giving:

$\therefore 2 t \beta - 2 {t}^{3} = t - \alpha$

And we are also told that $A P$ has length 1, so by Pythagoras:

${\left(t - \alpha\right)}^{2} + {\left({t}^{2} - \beta\right)}^{2} = {1}^{2}$

We can combine these equations to eliminate $t - \alpha$:

${\left(2 t \beta - 2 {t}^{3}\right)}^{2} + {\left({t}^{2} - \beta\right)}^{2} = 1$
$\therefore 4 {t}^{2} {\beta}^{2} - 8 {t}^{4} \beta + 4 {t}^{6} + {t}^{4} - 2 {t}^{2} \beta + {\beta}^{2} = 1$
$\therefore \left(4 {t}^{2} + 1\right) {\beta}^{2} - 2 {t}^{2} \left(4 {t}^{2} + 1\right) \beta + {t}^{4} \left(4 {t}^{2} + 1\right) = 1$
$\therefore {\beta}^{2} - 2 {t}^{2} \beta + {t}^{4} = \frac{1}{4 {t}^{2} + 1}$
$\therefore {\beta}^{2} - 2 {t}^{2} \beta + {t}^{4} - \frac{1}{4 {t}^{2} + 1} = 0$

We can solve this quadratic in $\beta$ by completing the square to get:

${\left(\beta - {t}^{2}\right)}^{2} - {t}^{4} + {t}^{4} - \frac{1}{4 {t}^{2} + 1} = 0$
$\therefore {\left(\beta - {t}^{2}\right)}^{2} = \frac{1}{4 {t}^{2} + 1}$
$\therefore \beta - {t}^{2} = \pm \sqrt{\frac{1}{4 {t}^{2} + 1}}$
$\therefore \beta = {t}^{2} \pm \sqrt{\frac{1}{4 {t}^{2} + 1}}$

We have found two solutions because there is one point $A P$ distance $1$ unit extending inwards,and one point extending outwards. Intuitively we choose $\beta = {t}^{2} + \sqrt{\frac{1}{4 {t}^{2} + 1}}$.

Now from before we have:

$2 t \beta - 2 {t}^{3} = t - \alpha$ :
$\therefore \alpha = t - 2 t \beta + 2 {t}^{3}$

Substituting our valu of $\beta$, from above, gives us:

$\alpha = t - 2 t \left({t}^{2} + \sqrt{\frac{1}{4 {t}^{2} + 1}}\right) + 2 {t}^{3}$
$\alpha = t - 2 {t}^{3} - 2 t \sqrt{\frac{1}{4 {t}^{2} + 1}} + 2 {t}^{3}$
$\alpha = t - 2 t \sqrt{\frac{1}{4 {t}^{2} + 1}}$

And so the general coordinates of $P \left(\alpha , \beta\right)$ in term of $t$ are:

$P = P \left(t - 2 t \sqrt{\frac{1}{4 {t}^{2} + 1}} , {t}^{2} + \sqrt{\frac{1}{4 {t}^{2} + 1}}\right)$

So the parametric equations of P are:

${P}_{x} \left(t\right) = t - 2 t \sqrt{\frac{1}{4 {t}^{2} + 1}}$
${p}_{y} \left(t\right) = {t}^{2} + \sqrt{\frac{1}{4 {t}^{2} + 1}}$

The loci traced out by $P$ as $A$ moves along the parabola is shown in the following graph in green:

2

## How do you find the power series for f(x)=int tln(1-t)dt from [0,x] and determine its radius of convergence?

Andrea S.
Featured 21 hours ago

$f \left(x\right) = - {\sum}_{n = 0}^{\infty} {x}^{n + 3} / \left(\left(n + 3\right) \left(n + 1\right)\right)$ with radius of convergence $R = 1$.

#### Explanation:

We have:

$f \left(x\right) = {\int}_{0}^{x} t \ln \left(1 - t\right) \mathrm{dt}$

Focus on the function:

$\ln \left(1 - x\right)$

We know that:

$\ln \left(1 - x\right) = - {\int}_{0}^{x} \frac{\mathrm{dt}}{1 - t}$

where the integrand function is the sum of the geometric series:

${\sum}_{n = 0}^{\infty} {t}^{n} = \frac{1}{1 - t}$

then we can integrate term by term, and we have:

$\ln \left(1 - x\right) = - {\int}_{0}^{x} \left({\sum}_{n = 0}^{\infty} {t}^{n}\right) \mathrm{dt} = - {\sum}_{n = 0}^{\infty} {\int}_{0}^{x} {t}^{n} \mathrm{dt} = - {\sum}_{n = 0}^{\infty} {x}^{n + 1} / \left(n + 1\right)$

and mutiplying by x term by term:

$x \ln \left(1 - x\right) = - {\sum}_{n = 0}^{\infty} {x}^{n + 2} / \left(n + 1\right)$

We can substitute this expression in the original integral and integrate again term by term:

$f \left(x\right) = {\int}_{0}^{x} t \ln \left(1 - t\right) \mathrm{dt} = - {\int}_{0}^{x} \left({\sum}_{n = 0}^{\infty} {t}^{n + 2} / \left(n + 1\right)\right) \mathrm{dt} = - {\sum}_{n = 0}^{\infty} {\int}_{0}^{x} {t}^{n + 2} / \left(n + 1\right) \mathrm{dt} = - {\sum}_{n = 0}^{\infty} {x}^{n + 3} / \left(\left(n + 3\right) \left(n + 1\right)\right)$

To determine the radius of convergence we can then use the ratio test:

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left\mid \frac{{x}^{n + 4} / \left(\left(n + 4\right) \left(n + 2\right)\right)}{{x}^{n + 3} / \left(\left(n + 3\right) \left(n + 1\right)\right)} \right\mid = \left\mid x \right\mid \frac{\left(n + 3\right) \left(n + 1\right)}{\left(n + 4\right) \left(n + 2\right)}$

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left\mid x \right\mid$

and the series is absolutely convergent for $\left\mid x \right\mid < 1$ which means the radius of convergence is $R = 1$