19
Active contributors today

## How do you do? Solids of revolution

Steve
Featured 3 months ago

$4 \sqrt{3} \pi {a}^{3}$

#### Explanation:

We basically are being asked to calculate the volume of a spherical bead, that is, a sphere with a hole drilled through it.

Consider a cross section through the sphere, which we have centred on the origin of a Cartesian coordinate system:

The red circle has radius $2 a$, hence its equation is:

${x}^{2} + {y}^{2} = {\left(2 a\right)}^{2} \implies {x}^{2} + {y}^{2} = 4 {a}^{2}$

Method 1 - Calculate core and subtract from Sphere

First let us consider the volume of the entire Sphere, which has radius $2 a$. I will use the standard volume formula $V = \frac{4}{3} \pi {r}^{3}$

$\therefore {V}_{\text{sphere}} = \frac{4}{3} \pi {\left(2 a\right)}^{3}$
$\text{ } = \frac{4}{3} \pi 8 {a}^{3}$
$\text{ } = \frac{32}{3} \pi {a}^{3}$

For the bore we can consider a solid of revolution. (Note this is not a cylinder as it has a curved top and bottom from the sphere). We will use the method of cylindrical shells and rotate about $O y$, the shell formula is:

$V = 2 \pi \setminus {\int}_{\alpha}^{\beta} \setminus x f \left(x\right) \setminus \mathrm{dx}$

Also note that we require twice the volume because we have a portion above and below the $x$-axis. The shell volume of revolution about $O y$ for the bore is given by:

${V}_{\text{bore}} = 2 \pi \setminus {\int}_{0}^{a} \setminus x \left(2 \sqrt{4 {a}^{2} - {x}^{2}}\right) \setminus \mathrm{dx}$
$\text{ } = 2 \pi \setminus {\int}_{0}^{a} \setminus 2 x \setminus \sqrt{4 {a}^{2} - {x}^{2}} \setminus \mathrm{dx}$

We can evaluate using a substitution:

Let $u = 4 {a}^{2} - {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$
When $\left\{\begin{matrix}x = 0 \\ x = a\end{matrix}\right. \implies \left\{\begin{matrix}u = 4 {a}^{2} \\ u = 3 {a}^{2}\end{matrix}\right.$

And so:

${V}_{\text{bore}} = - 2 \pi \setminus {\int}_{4 {a}^{2}}^{3 {a}^{2}} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\int}_{3 {a}^{2}}^{4 {a}^{2}} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{3 {a}^{2}}^{4 {a}^{2}}$

$\text{ } = \frac{4 \pi}{3} \setminus {\left[{u}^{\frac{3}{2}}\right]}_{3 {a}^{2}}^{4 {a}^{2}}$
$\text{ } = \frac{4 \pi}{3} \setminus \left({\left(4 {a}^{2}\right)}^{\frac{3}{2}} - {\left(3 {a}^{2}\right)}^{\frac{3}{2}}\right)$
$\text{ } = \frac{4}{3} \pi \setminus \left(8 {a}^{3} - 3 \sqrt{3} {a}^{3}\right)$
$\text{ } = \frac{32}{3} \pi {a}^{3} - 4 \sqrt{3} \pi {a}^{3}$

And so the total volume is given by:

# V_("total") = V_("sphere") - V_("bore" #
$\text{ } = \frac{32}{3} \pi {a}^{3} - \left(\frac{32}{3} \pi {a}^{3} - 4 \sqrt{3} \pi {a}^{3}\right)$
$\text{ } = 4 \sqrt{3} \pi {a}^{3}$

Method 2 - Calculate volume of bead directly

We can use the same method of a solid of revolution using the method of cylindrical shells and rotate about $O y$, but this time we will consider the bead itself rather than the core that is removed, again we need to twice the volume because we have a portion above and below the $x$-axis. The shell volume of revolution about $O y$ for the bead is given by:

${V}_{\text{total}} = 2 \pi \setminus {\int}_{a}^{2 a} \setminus x \left(2 \sqrt{4 {a}^{2} - {x}^{2}}\right) \setminus \mathrm{dx}$
$\text{ } = 2 \pi \setminus {\int}_{a}^{2 a} \setminus 2 x \setminus \sqrt{4 {a}^{2} - {x}^{2}} \setminus \mathrm{dx}$

We can evaluate using a substitution:

Let $u = 4 {a}^{2} - {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$
When $\left\{\begin{matrix}x = a \\ x = 2 a\end{matrix}\right. \implies \left\{\begin{matrix}u = 3 {a}^{2} \\ u = 0\end{matrix}\right.$

And so:

${V}_{\text{total}} = - 2 \pi \setminus {\int}_{3 {a}^{2}}^{0} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\int}_{0}^{3 {a}^{2}} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{0}^{3 {a}^{2}}$

$\text{ } = \frac{4 \pi}{3} \setminus {\left[{u}^{\frac{3}{2}}\right]}_{0}^{3 {a}^{2}}$
$\text{ } = \frac{4 \pi}{3} \setminus \left({\left(3 {a}^{2}\right)}^{\frac{3}{2}} - 0\right)$
$\text{ } = \frac{4 \pi}{3} \setminus \left(3 \sqrt{3} {a}^{3}\right)$
$\text{ } = 4 \sqrt{3} \pi {a}^{3}$, as above

## How do you evaluate #\int \frac { ( x - 2) } { x ( x ^ { 2} - 4x + 5) ^ { 2} } d x#?

Steve
Featured 2 months ago

$\int \setminus \frac{x - 2}{x {\left({x}^{2} - 4 x + 5\right)}^{2}} \setminus \mathrm{dx} = - \frac{2}{25} \ln | x | + \frac{1}{25} \ln | {x}^{2} - 4 x + 5 | - \frac{3}{50} \arctan \left(x - 2\right) + \frac{x - 4}{10 \left({x}^{2} - 4 x + 5\right)} + c$

#### Explanation:

Let us denote the required integral by $I$

$I = \int \setminus \frac{x - 2}{x {\left({x}^{2} - 4 x + 5\right)}^{2}} \setminus \mathrm{dx}$

Partial Fraction Decomposition

We can decompose the integrand using partial fractions, the partial fraction decomposition will be of the form:

$\frac{x - 2}{x {\left({x}^{2} - 4 x + 5\right)}^{2}} = \frac{A}{x} + \frac{B x + C}{{x}^{2} - 4 x + 5} + \frac{D x + E}{{x}^{2} - 4 x + 5} ^ 2$
$\text{ } = \frac{A {\left({x}^{2} - 4 x + 5\right)}^{2} + \left(B x + C\right) x \left({x}^{2} - 4 x + 5\right) + \left(D x + E\right) x}{x {\left({x}^{2} - 4 x + 5\right)}^{2}}$

$x - 2 = A {\left({x}^{2} - 4 x + 5\right)}^{2} + \left(B x + C\right) x \left({x}^{2} - 4 x + 5\right) + \left(D x + E\right) x$

$\therefore x - 2 = A \left({x}^{4} - 8 {x}^{3} + 26 {x}^{2} - 40 x + 25\right) + \left(B x + C\right) \left({x}^{3} - 4 {x}^{2} + 5 x\right) + \left(D x + E\right) x$

We can use various methods to find our unknown constants:

$\text{Put } x = 0 \implies - 2 = A \cdot {5}^{2}$
$\therefore A = - \frac{2}{25}$

Compare coefficients:

$\text{Coeff} \left({x}^{4}\right) \implies 0 = A + B$
$\therefore B = \frac{2}{25}$

$\text{Coeff} \left({x}^{3}\right) \implies 0 = - 8 A + C - 4 B$
$\therefore 0 = \frac{16}{25} + C - \frac{8}{25} \implies C = - \frac{8}{25}$

$\text{Coeff} \left({x}^{2}\right) \implies 0 = 26 A + 5 B - 4 C + D$
$\therefore 0 = - \frac{52}{25} + \frac{10}{25} + \frac{32}{25} + D \implies D = \frac{10}{25} = \frac{2}{5}$

$\text{Coeff} \left({x}^{1}\right) \implies 1 = - 40 A + 5 C + E$
$\therefore 1 = \frac{80}{25} - \frac{40}{25} + E \implies E = - \frac{3}{5}$

Hence we can the integral as:

$I = \int \setminus \frac{- \frac{2}{25}}{x} + \frac{\frac{2}{25} x - \frac{8}{25}}{{x}^{2} - 4 x + 5} + \frac{\frac{2}{5} x - \frac{3}{5}}{{x}^{2} - 4 x + 5} ^ 2 \setminus \mathrm{dx}$

$\setminus \setminus = - \frac{2}{35} \setminus \int \setminus \frac{1}{x} \mathrm{dx} + \frac{2}{25} \int \frac{x - 4}{{x}^{2} - 4 x + 5} \setminus \mathrm{dx} + \frac{1}{5} \int \frac{2 x - 3}{{x}^{2} - 4 x + 5} ^ 2 \setminus \mathrm{dx}$

$\setminus \setminus = - \frac{2}{25} \setminus {I}_{1} + \frac{2}{25} \setminus {I}_{2} + \frac{1}{5} \setminus {I}_{3}$

Where:

${I}_{1} = \int \setminus \frac{1}{x} \mathrm{dx} + \frac{2}{25} \setminus \mathrm{dx}$
${I}_{2} = \int \frac{x - 4}{{x}^{2} - 4 x + 5} \setminus \mathrm{dx}$
${I}_{3} = \int \frac{2 x - 3}{{x}^{2} - 4 x + 5} ^ 2 \setminus \mathrm{dx}$

Now let us take each of these three separate integrals in turn:

Integral 1: ${I}_{1}$

The first integral, ${I}_{1}$, we can just evaluate directly:

${I}_{1} = \int \setminus \frac{1}{x} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \ln | x |$

Integral 2: ${I}_{2}$

The second integral, ${I}_{2}$, we manipulate by completing the square:

${I}_{2} = \int \setminus \frac{x - 4}{{x}^{2} - 4 x + 5} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \setminus \frac{x - 4}{{\left(x - 2\right)}^{2} - {2}^{2} + 5} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \setminus \frac{x - 4}{{\left(x - 2\right)}^{2} + 1} \setminus \mathrm{dx}$

Let $u = x - 2 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 1$, and $x - 4 = u - 2$, then substituting gives

${I}_{2} = \int \setminus \frac{u - 2}{{u}^{2} + 1} \setminus \mathrm{du}$
$\setminus \setminus \setminus = \int \setminus \frac{u}{{u}^{2} + 1} - \frac{2}{{u}^{2} + 1} \setminus \mathrm{du}$
$\setminus \setminus \setminus = \frac{1}{2} \int \setminus \frac{2 u}{{u}^{2} + 1} \setminus \mathrm{du} - 2 \int \setminus \frac{1}{{u}^{2} + 1} \setminus \mathrm{du}$
$\setminus \setminus \setminus = \frac{1}{2} \ln | {u}^{2} + 1 | - 2 \arctan u$

Restoring the substitution we get:

${I}_{2} = \frac{1}{2} \ln | {x}^{2} - 4 x + 5 | - 2 \arctan \left(x - 2\right)$

Integral 3: ${I}_{3}$

The third integral, ${I}_{3}$ we continue as earlier by completing the square to get:

${I}_{3} = \int \setminus \frac{2 x - 3}{{x}^{2} - 4 x + 5} ^ 2 \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \int \frac{2 x - 3}{{\left(x - 2\right)}^{2} + 1} ^ 2 \setminus \mathrm{dx}$

Let $u = x - 2 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 1$, and $2 x - 3 = 2 u + 1$, then substituting gives

${I}_{3} = \int \frac{2 u + 1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du}$
$\setminus \setminus \setminus = \int \frac{2 u}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} + \int \frac{1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du}$

Consider the integral:

$\int \frac{2 u}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du}$

Let $w = {u}^{2} + 1 \implies \frac{\mathrm{dw}}{\mathrm{du}} = 2 u$, so we can substitute to get:

$\int \frac{2 u}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} = \int \setminus \frac{1}{w} ^ 2 \setminus \mathrm{dw}$
$\text{ } = - \frac{1}{w}$

Restoring the substitutions we get:

$\int \frac{2 u}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} = - \frac{1}{{u}^{2} + 1}$
$\text{ } = - \frac{1}{{\left(x - 2\right)}^{2} + 1}$
$\text{ } = - \frac{1}{{x}^{2} - 4 x + 5}$

And now we consider:

$\int \frac{1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du}$

Let $u = \tan \theta \implies \frac{\mathrm{du}}{d \theta} = {\sec}^{2} \theta$, then substituting gives

$\int \frac{1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} = \int \setminus \frac{1}{{\tan}^{2} \theta + 1} ^ 2 \setminus {\sec}^{2} \theta \setminus d \theta$
$\text{ } = \int \setminus \frac{1}{{\sec}^{2} \theta} ^ 2 \setminus {\sec}^{2} \theta \setminus d \theta$
$\text{ } = \int \setminus \frac{1}{{\sec}^{2} \theta} \setminus d \theta$
$\text{ } = \int \setminus {\cos}^{2} \theta \setminus d \theta$
$\text{ } = \int \setminus \frac{1}{2} \left(1 + \cos 2 \theta\right) \setminus d \theta$
$\text{ } = \frac{1}{2} \setminus \int \setminus 1 + \cos 2 \theta \setminus d \theta$
$\text{ } = \frac{1}{2} \left(\theta + \frac{1}{2} \sin 2 \theta\right)$
$\text{ } = \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta$

Now, $u = \tan \theta \implies u = \sin \frac{\theta}{\cos} \theta$

$\therefore \sin \theta \cos \theta = {\cos}^{2} \theta \cdot u$
$\text{ } = \frac{1}{\sec} ^ 2 \theta \cdot u$
$\text{ } = \frac{1}{1 + {\tan}^{2} \theta} \cdot u$
$\text{ } = \frac{u}{{u}^{2} + 1}$

Restoring the substitutions we get:

$\int \frac{1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} = \frac{1}{2} \arctan u + \frac{1}{2} \frac{u}{{u}^{2} + 1}$
$\text{ } = \frac{1}{2} \arctan \left(x - 2\right) + \frac{1}{2} \frac{x - 2}{{\left(x - 2\right)}^{2} + 1}$
$\text{ } = \frac{1}{2} \arctan \left(x - 2\right) + \frac{1}{2} \frac{x - 2}{{x}^{2} - 4 x + 5}$

Hence, we can write the third integral as:

${I}_{3} = \int \frac{2 u}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du} + \int \frac{1}{{u}^{2} + 1} ^ 2 \setminus \mathrm{du}$
$\setminus \setminus \setminus = - \frac{1}{{x}^{2} - 4 x + 5} + \frac{1}{2} \arctan \left(x - 2\right) + \frac{1}{2} \frac{x - 2}{{x}^{2} - 4 x + 5}$

Combine Results

Now we return back to our original partial fraction result from earlier, and replace the three integrals with the above results:

$I = - \frac{2}{25} \setminus {I}_{1} + \frac{2}{25} \setminus {I}_{2} + \frac{1}{5} \setminus {I}_{3}$

$\setminus \setminus = - \frac{2}{25} \left\{\ln | x |\right\} + \frac{2}{25} \left\{\frac{1}{2} \ln | {x}^{2} - 4 x + 5 | - 2 \arctan \left(x - 2\right)\right\} + \frac{1}{5} \left\{- \frac{1}{{x}^{2} - 4 x + 5} + \frac{1}{2} \arctan \left(x - 2\right) + \frac{1}{2} \frac{x - 2}{{x}^{2} - 4 x + 5}\right\} + c$

$\setminus \setminus = - \frac{2}{25} \ln | x | + \frac{1}{25} \ln | {x}^{2} - 4 x + 5 | - \frac{4}{25} \arctan \left(x - 2\right) - \frac{1}{5} \frac{1}{{x}^{2} - 4 x + 5} + \frac{1}{10} \arctan \left(x - 2\right) + \frac{1}{10} \frac{x - 2}{{x}^{2} - 4 x + 5} + c$

$\setminus \setminus = - \frac{2}{25} \ln | x | + \frac{1}{25} \ln | {x}^{2} - 4 x + 5 | - \frac{3}{50} \arctan \left(x - 2\right) - \frac{1}{5} \frac{1}{{x}^{2} - 4 x + 5} + \frac{1}{10} \frac{x - 2}{{x}^{2} - 4 x + 5} + c$

$\setminus \setminus = - \frac{2}{25} \ln | x | + \frac{1}{25} \ln | {x}^{2} - 4 x + 5 | - \frac{3}{50} \arctan \left(x - 2\right) + \frac{x - 2 - 2}{10 \left({x}^{2} - 4 x + 5\right)} + c$

$\setminus \setminus = - \frac{2}{25} \ln | x | + \frac{1}{25} \ln | {x}^{2} - 4 x + 5 | - \frac{3}{50} \arctan \left(x - 2\right) + \frac{x - 4}{10 \left({x}^{2} - 4 x + 5\right)} + c$

## How do you find the area of the region bounded by the given curves #y = 6x^2lnx # and #y = 24lnx#?

Anjali G
Featured 1 month ago

${\int}_{1}^{2} \left(24 \ln x - 6 {x}^{2} \ln x\right) \mathrm{dx}$
$= 32 \ln 2 - \frac{58}{3}$

#### Explanation:

First set the equations equal to each other, and find the intersection points, then find the area between the curves.

$\textcolor{red}{y = 6 {x}^{2} \ln x}$
$\textcolor{b l u e}{y = 24 \ln x}$

$6 {x}^{2} \ln x = 24 \ln x$
$0 = 24 \ln x - 6 {x}^{2} \ln x$
$0 = \left(- 6\right) \left(\ln x\right) \left({x}^{2} - 4\right)$
$x = 1 , 2$

By setting the equations equal to each other, I found that the functions intersect at $x = 1$ and $x = 2$:

$\text{Area} = {\int}_{1}^{2} \left(24 \ln x - 6 {x}^{2} \ln x\right) \mathrm{dx}$

Focus on the unbounded (indefinite) integral to solve:
$\int \left(24 \ln x - 6 {x}^{2} \ln x\right) \mathrm{dx}$
$= - 6 \int \left(\ln x\right) \left({x}^{2} - 4\right) \mathrm{dx}$

Use integration by parts: $\int \textcolor{b l u e}{u} \textcolor{g r e e n}{\mathrm{dv}} = v u - \int v \mathrm{du}$
$\left(\begin{matrix}u = \textcolor{b l u e}{\ln x} \\ \mathrm{du} = \frac{1}{x} \mathrm{dx}\end{matrix}\right) \left(\begin{matrix}v = \frac{1}{3} {x}^{3} - 4 x \\ \mathrm{dv} = \textcolor{g r e e n}{{x}^{2} - 4}\end{matrix}\right)$

$= - 6 \left[\left(\ln x\right) \left(\frac{1}{3} {x}^{3} - 4 x\right) - \int \left(\frac{1}{x}\right) \left(\frac{1}{3} {x}^{3} - 4 x\right) \mathrm{dx}\right]$

$= - 6 \left[\left(\ln x\right) \left(\frac{1}{3} {x}^{2} - 4 x\right) - \int \left(\frac{1}{3} {x}^{2} - 4\right) \mathrm{dx}\right]$

$= - 6 \left[\left(\ln x\right) \left(\frac{1}{3} {x}^{2} - 4 x\right) - \left(\frac{1}{9} {x}^{3} - 4 x\right)\right]$

$= - 6 \left(\ln x\right) \left(\frac{1}{3} {x}^{3} - 4 x\right) + \frac{2}{3} {x}^{3} - 24 x$

Now go back to the definite integral:
${\int}_{1}^{2} \left(24 \ln x - 6 {x}^{2} \ln x\right) \mathrm{dx}$
$= {\left[- 6 \left(\ln x\right) \left(\frac{1}{3} {x}^{3} - 4 x\right) + \frac{2}{3} {x}^{3} - 24 x\right]}_{1}^{2}$

$= - 6 \ln 2 \left(\frac{8}{3} - 8\right) + \frac{16}{3} - 48 - \frac{2}{3} + 24$

$= 32 \ln 2 - \frac{58}{3}$

## What happens to the convergence when you raise the sum of an equation to the power k?

yasinsale
Featured 3 weeks ago

Under some conditions, which are given below, letter series converges.

#### Explanation:

It dependes on properties of ${a}_{n}$. There is a more general theorem about this problem:

Theorem: Let ${a}_{n}$ be a bounded sequnce and $\sum {b}_{n}$ be a convergent series with all ${b}_{n} > 0$. Then $\sum {a}_{n} {b}_{n}$ absolutely converges, means converges.

For our purpose, your hypothesis to be true, we need the followings to be true:

• For all ${a}_{n} > 0$,
• ${a}_{n}$ sequence is bounded (Let's say $| {a}_{n} | < \setminus \alpha$),
• $\setminus \sum {a}_{n}$ be convergent (Let's say its equal to $a$).

If these two items are true, then your statement would be true. Moreover we can find an upper limit for $\sum {\left({a}_{n}\right)}^{k}$, of coarse for positive finite integers $k$.

Let's start with $k = 2$:

$\sum {a}_{n}^{2} = \sum {a}_{n} {a}_{n} \le \sum | {a}_{n} {a}_{n} | \le \alpha a$

$\sum {a}_{n}^{2} \le \alpha a$,

which means $\sum {a}_{n}^{2}$ converges. Let's pass to $k = 3$:

$\sum {a}_{n}^{3} = \sum {a}_{n} {a}_{n}^{2} \le \sum | {a}_{n} {a}_{n}^{2} | \le \alpha \sum | {a}_{n}^{2} | \le \alpha \left(\alpha a\right) = {\alpha}^{2} a$

$\sum {a}_{n}^{3} \le {\alpha}^{2} a$

If we continue in this respect we can find by induction (!),

$\sum {\left({a}_{n}\right)}^{k} \le {\alpha}^{k - 1} a$.

## How to find formula for nth derivative of #f(x)=e^(2x)(x^2-3x+2)#?

Ratnaker Mehta
Featured 2 weeks ago

${2}^{n - 2} {e}^{2 x} \left\{4 {x}^{2} + 4 \left(n - 3\right) x + \left({n}^{2} - 7 n + 8\right)\right\} .$

#### Explanation:

We will solve this Problem, using the following Leibnitz Theorem

(LT) for ${n}^{t h}$ Derivative of Product of two functions $u \mathmr{and} v .$

Denoting the ${n}^{t h}$ der. of fun. $u$ by ${u}_{n} ,$ we have,

${\left(u v\right)}_{n} = {u}_{n} v + {\text{_nC_1u_(n-1)v_1+}}_{n} {C}_{2} {u}_{n - 2} {v}_{2} + \ldots + u {v}_{n} .$

We take, $u = {e}^{2 x} \Rightarrow {u}_{n} = {2}^{n} \cdot {e}^{2 x} , \mathmr{and} , v = {x}^{2} - 3 x + 2.$

#:. v_1=2x-3, v_2=2, &, v_3=v_4=...=v_n=0, n>=3.#

$\therefore \frac{{d}^{n}}{\mathrm{dx}} ^ n \left\{\left({x}^{2} - 3 x + 2\right) {e}^{2 x}\right\} = \left({2}^{n} \cdot {e}^{2 x}\right) \left({x}^{2} - 3 x + 2\right)$
$+ n \cdot \left({2}^{n - 1} {e}^{2 x}\right) \left(2 x - 3\right) + \frac{n}{2} \cdot \left(n - 1\right) \cdot \left({2}^{n - 2} {e}^{2 x}\right) \cdot 2 ,$

$= {2}^{n} \cdot {e}^{2 x} \left({x}^{2} - 3 x + 2\right) + {2}^{n - 1} {e}^{2 x} \left(2 n x - 3 n\right) + {2}^{n - 2} {e}^{2 x} \left({n}^{2} - n\right) ,$

$= {2}^{n - 2} {e}^{2 x} \left\{4 \left({x}^{2} - 3 x + 2\right) + 2 \left(2 n x - 3 n\right) + \left({n}^{2} - n\right)\right\} ,$

$= {2}^{n - 2} {e}^{2 x} \left\{4 {x}^{2} + 4 \left(n - 3\right) x + \left({n}^{2} - 7 n + 8\right)\right\} .$

Enjoy Maths.!

## What is a x b?

Truong-Son N.
Featured 1 week ago

I'm guessing the question refers to the $x$ coordinates where $f$ changes direction, and that $a \times b = - \frac{1}{7}$:

$a = - \sqrt{\frac{1}{7}}$, local minimum

#b = sqrt((1/7)#, local maximum

$f \left(x\right) = \frac{6 x}{1 + 7 {x}^{2}}$:

graph{(6x)/(1 + 7x^2) [-3.464, 3.464, -1.732, 1.732]}

$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{6 \left(7 {x}^{2} - 1\right)}{1 + 7 {x}^{2}} ^ 2$:

graph{-(6(7x^2 - 1))/(1 + 7x^2)^2 [-3.464, 3.464, -1.732, 1.732]}

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{84 x \left(7 {x}^{2} - 3\right)}{1 + 7 {x}^{2}} ^ 3$:

graph{(84x (7x^2 - 3))/(1 + 7 x^2)^3 [-2.464, 2.464, -24, 24]}

Well, first of all, I think the question has a funky typo. It should say, "changes from decreasing to increasing at the point $a$ . . . "

Whenever a function changes from increasing to decreasing or vice versa, it exhibits either a local maximum or minimum.

For example:

• $y = {x}^{2}$ has one local minimum at $x = 0$, since the graph has only an upwards concavity.

graph{x^2 [-2.464, 2.464, -2, 2]}

• $y = \sin x$ from $0$ to $2 \pi$ changes from increasing to decreasing at $x = \frac{\pi}{2}$, and changes from decreasing to increasing at $x = \frac{3 \pi}{2}$.

graph{sinx [-0.05, 6.25, -1.2, 1.01]}

For your graph, the first derivative with respect to $x$, denoted $\frac{\mathrm{df}}{\mathrm{dx}}$, shows you whether the function increases, decreases, or neither at a certain $x$.

Here, we have (using the product rule):

$\frac{\mathrm{df}}{\mathrm{dx}} = 6 x \cdot - \frac{1}{1 + 7 {x}^{2}} ^ 2 \cdot 14 x + \frac{1}{1 + 7 {x}^{2}} \cdot 6$

$= \frac{- 6 \left(14 {x}^{2}\right)}{1 + 7 {x}^{2}} ^ 2 - \frac{- 6 \left(1 + 7 {x}^{2}\right)}{1 + 7 {x}^{2}} ^ 2$

$= \frac{- 6 \left(14 {x}^{2} - \left(1 + 7 {x}^{2}\right)\right)}{1 + 7 {x}^{2}} ^ 2$

$= - \frac{6 \left(7 {x}^{2} - 1\right)}{1 + 7 {x}^{2}} ^ 2$

So, you took the derivative correctly. Now, to find the points $a$ and $b$ where the function has a local maximum or minimum, find the zeroes of this derivative (i.e. find when $\frac{\mathrm{df}}{\mathrm{dx}} = 0$, the function changes direction).

$0 = - \frac{{\cancel{6}}^{\ne 0} \left(7 {x}^{2} - 1\right)}{\cancel{{\left(1 + 7 {x}^{2}\right)}^{2}}} ^ \left(\ne 0\right)$

The denominator can never equal $0$ because it's an upwards-shifted quadratic, so we don't have to worry about it.

$\implies 7 {x}^{2} = 1$

So, your critical x values (for your local maxima and/or minima) are at

$\implies \textcolor{g r e e n}{x} = \pm \sqrt{\frac{1}{7}} \approx \textcolor{g r e e n}{\pm 0.3780}$

Now that you know the $x$ coordinates, you'll need the $y$ coordinates, so plug in your $x$ values back into the ORIGINAL function:

$\textcolor{g r e e n}{f \left(\pm \sqrt{\frac{1}{7}}\right)} = \frac{6 \left(\pm \sqrt{\frac{1}{7}}\right)}{1 + 7 {\left(\pm \sqrt{\frac{1}{7}}\right)}^{2}}$

$= \frac{\pm \frac{6}{\sqrt{7}}}{1 + 7 \cdot \frac{1}{7}} = \pm \frac{3}{\sqrt{7}}$

$\approx \textcolor{g r e e n}{\pm 1.134}$

Right now, all you know is where the points $a$ and $b$ are, but not which is which.

The second derivative tells you the concavity; up, down, or an inflection point. By the question wording, one is concave up and the other is concave down.

This derivative would be somewhat ugly

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = - \frac{d}{\mathrm{dx}} \left[\frac{42 {x}^{2} - 6}{1 + 7 {x}^{2}} ^ 2\right]$,

but you should get:

$= \frac{84 x \left(7 {x}^{2} - 3\right)}{1 + 7 {x}^{2}} ^ 3$

At the same critical $x$ values you got earlier, evaluate this second derivative.

• If it is positive, the function is concave up and is at a minimum.
• If it is negative, the function is concave down and is at a maximum.
• Otherwise, the function is at an inflection point.

At $\underline{x = - \sqrt{\frac{1}{7}}}$:

$\underline{\frac{{d}^{2} f}{{\mathrm{dx}}^{2}}} = \frac{84 \left(- \sqrt{\frac{1}{7}}\right) \left(7 {\left(- \sqrt{\frac{1}{7}}\right)}^{2} - 3\right)}{1 + 7 {\left(- \sqrt{\frac{1}{7}}\right)}^{2}} ^ 3$

$= \frac{\left(- \frac{84}{\sqrt{7}}\right) \left(- 2\right)}{2} ^ 3 \text{ } \underline{> 0}$

At $\underline{x = + \sqrt{\frac{1}{7}}}$, we similarly get:

$\underline{\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} < 0}$

So, the left-hand critical point is a minimum and the right-hand critical point is a maximum.

$\textcolor{b l u e}{\left(- \sqrt{\frac{1}{7}} , - \frac{3}{\sqrt{7}}\right) \approx \left(- 0.3780 , - 1.134\right)}$ (minimum)

$\textcolor{b l u e}{\left(\sqrt{\frac{1}{7}} , \frac{3}{\sqrt{7}}\right) \approx \left(0.3780 , 1.134\right)}$ (maximum)

And we can check by graphing. Here is the function, followed by its first derivative, and then its second derivative.

$f \left(x\right) = \frac{6 x}{1 + 7 {x}^{2}}$:

graph{(6x)/(1 + 7x^2) [-3.464, 3.464, -1.732, 1.732]}

$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{6 \left(7 {x}^{2} - 1\right)}{1 + 7 {x}^{2}} ^ 2$:

graph{-(6(7x^2 - 1))/(1 + 7x^2)^2 [-3.464, 3.464, -1.732, 1.732]}

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{84 x \left(7 {x}^{2} - 3\right)}{1 + 7 {x}^{2}} ^ 3$:

graph{(84x (7x^2 - 3))/(1 + 7 x^2)^3 [-2.464, 2.464, -24, 24]

##### Questions
• · 21 minutes ago
• 25 minutes ago · in Integration by Parts
• · 27 minutes ago
• · 31 minutes ago
• · 50 minutes ago
• · An hour ago
• · An hour ago
• · An hour ago
• · An hour ago
• · An hour ago
• · An hour ago
• · An hour ago
• · An hour ago
· An hour ago
• · 2 hours ago
• · 2 hours ago
• · 3 hours ago
• · 3 hours ago
• · 3 hours ago
• · 4 hours ago
• · 5 hours ago
• · 5 hours ago
• · 5 hours ago
• 5 hours ago · in Chain Rule
• · 6 hours ago · in Chain Rule
• · 6 hours ago
• · 6 hours ago
• · 6 hours ago
• · 10 hours ago