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2

## 1. Build a rectangular pen with three parallel partitions (meaning 4 total sections) using 500 feet of fencing. What dimensions will maximize the total area of the pen?

Steve
Featured 5 days ago

The total length of the partition should be $125$ feet, and the height of each partition should be $50$ feet (making each section $50$ feet x $31.25$ feet). This results in a maximum enclosed area of $6250 {\text{ feet}}^{2}$

#### Explanation:

Without loss of generality let us assume that the pen is divided as shown:

Let us set up the following variables:

$\left\{\begin{matrix}x & \text{Total height of the partition (feet)" \\ y & "Total length of the partition (feet)" \\ A & "Total Area enclosed by the partition (sq feet)}\end{matrix}\right.$

Our aim is to find $A \left(x\right)$, and to maximize the total area, A, wrt x (equally we could the same with $y$ and we would get the same result). ie we want a critical point of $\frac{\mathrm{dA}}{\mathrm{dx}}$.

Now, the total perimeter is given as $500$ (constant) and so:

$5 x + 2 y = 500$
$\therefore 2 y = 500 - 5 x$
$\therefore y = 250 - \frac{5}{2} x$ ..... [1]

And the total Area enclosed by the pen is given by:

$A = x y$

And substitution of the first result [1] gives us:

$A = x \left(250 - \frac{5}{2} x\right)$
$\setminus \setminus \setminus = 250 x - \frac{5}{2} {x}^{2}$

We no have the Area, A, as a function of a single variable, so Differentiating wrt $x$ we get:

$\frac{\mathrm{dA}}{\mathrm{dx}} = 250 - 5 x$ ..... [2]

At a critical point we have $\frac{\mathrm{dA}}{\mathrm{dx}} = 0 \implies$

$250 - 5 x = 0$
$\therefore \setminus \setminus \setminus \setminus \setminus 5 x = 250$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus x = 50$

And substituting $x = 50$ into [1] we get;

$y = 250 - \frac{5}{2} \left(50\right)$
$\setminus \setminus = 250 - 125$
$\setminus \setminus = 125$

We should check that $x = 50$ results in a maximum area. Differentiating [2] wrt x we get:

$\frac{{d}^{2} A}{\mathrm{dx}} ^ 2 = - 5 < 0$ when $x = 50$

Confirming that we have a maximum area, given by:

$A = \left(50\right) \left(125\right) = 6250 {\text{ feet}}^{2}$

We can visually verify that this corresponds to a maximum by looking at the graph of $y = A \left(x\right)$:
graph{250x - 5/2x^2 [-100, 200, -100, 7000]}

2

## What is the slope of the tangent line of  (xy-y/x)(xy-x/y) =C , where C is an arbitrary constant, at (-2,1)?

Steve
Featured 5 days ago

The slope of the tangent is $0$ at $\left(- 2 , 1\right)$, But $\left(- 2 , 1\right)$ only lies on the curve in the specific case of $C = 0$

#### Explanation:

We have:

$\left(x y - \frac{y}{x}\right) \left(x y - \frac{x}{y}\right) = C$

At $\left(- 2 , 1\right)$ we have:

$\left(- 2 + \frac{1}{2}\right) \left(- 2 + 2\right) = C \implies C = 0$

Multiplying out gives:

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

Differentiating wrt $x$ (Implicitly and using the product rule):

$\left({x}^{2}\right) \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(2 x\right) \left({y}^{2}\right) - 2 x - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} + x {y}^{2} - x - y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore \left({x}^{2} y - y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + x {y}^{2} - x = 0$
$\therefore \left({x}^{2} y - y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = x - x {y}^{2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - x {y}^{2}}{{x}^{2} y - y}$

So at $\left(- 2 , 1\right)$ we have:

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 - \left(- 2\right)}{4 - 1} = 0$

So the slope of the tangent is $0$ at $\left(- 2 , 1\right)$, But $\left(- 2 , 1\right)$ only lies on the curve in the specific case of $C = 0$

The curve of this family are quote interesting:

C=3

C=1

C=0.5

C=0.25

C=0 - The specific curve that (-2,1) lies on

C=-1

C=-2

3

## How do you find the positive values of p for which Sigma lnn/n^p from [2,oo) converges?

Andrea S.
Featured 5 days ago

The series:

${\sum}_{n = 2}^{\infty} \ln \frac{n}{n} ^ p$

is convergent for $p > 1$.

#### Explanation:

A necessary condition for a series to converge is that its general term is infinitesimal, that is:

(1) ${\lim}_{n \to \infty} \ln \frac{n}{n} ^ p = 0$

Consider the function $f \left(x\right) = \ln \frac{x}{x} ^ p$ and the limit:

(2) ${\lim}_{x \to \infty} \ln \frac{x}{x} ^ p$

Clearly if such limit exists it must coincide with (1).

Now, for $p = 0$ the limit (2) is $\infty$, while for $p \ne 0$ it is in the indeterminate form $\frac{\infty}{\infty}$ so we can solve it using l'Hospital's rule:

${\lim}_{x \to \infty} \ln \frac{x}{x} ^ p = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} {x}^{p}} = {\lim}_{x \to \infty} \left(\frac{1}{x}\right) \left(\frac{1}{p {x}^{p - 1}}\right) = {\lim}_{x \to \infty} \frac{1}{p {x}^{p}} = \left\{\begin{matrix}0 \text{ for " p > 0 \\ oo " for } p < 0\end{matrix}\right.$

So we know that for $p \le 0$ the series is not convergent.

To have a sufficient condition we can apply the integral test, using as test function: $f \left(x\right) = \ln \frac{x}{x} ^ p$.

For $p > 0$ such function is positive, decreasing and, as we have just seen, infinitesimal, so it satisfies the hypotheses of the integral test theorem, and the series is proven to be convergent if the improper integral:

${\int}_{2}^{\infty} \ln \frac{x}{x} ^ p$

also converges.

Solving the indefinite integral by parts:

$\int \ln \frac{x}{x} ^ p \mathrm{dx} = \int \ln x d \left({x}^{1 - p} / \left(1 - p\right)\right) = \ln x {x}^{1 - p} / \left(1 - p\right) - \int \frac{1}{x} {x}^{1 - p} / \left(1 - p\right) \mathrm{dx} = \frac{1}{1 - p} \ln \frac{x}{x} ^ \left(p - 1\right) - \int \frac{1}{x} {x}^{1 - p} / \left(1 - p\right) \mathrm{dx}$

$\int \frac{1}{x} {x}^{1 - p} / \left(1 - p\right) \mathrm{dx} = \frac{1}{p - 1} \int {x}^{- p} \mathrm{dx} = \frac{1}{1 - p} ^ 2 {x}^{1 - p} + C = \frac{1}{1 - p} ^ 2 \frac{1}{x} ^ \left(p - 1\right) + C$

So that:

${\int}_{2}^{\infty} \ln \frac{x}{x} ^ p = {\left[\frac{1}{1 - p} \ln \frac{x}{x} ^ \left(p - 1\right) - \frac{1}{1 - p} ^ 2 \frac{1}{x} ^ \left(p - 1\right)\right]}_{2}^{\infty}$

which is convergent for $p - 1 > 0 \implies p > 1$

In conclusion the series is convergent for $p > 1$

3

## Differentiation of cosec root x by first principle???

Eric Sia
Featured 4 days ago

Use the series expansion of the sine function.

#### Explanation:

$f \left(x\right) = \csc \left(\sqrt{x}\right)$

$f \left(x + \Delta x\right) = \csc \left(\sqrt{x + \Delta x}\right)$

By definition:

$f ' \left(x\right) = {\lim}_{\Delta x \to 0} \frac{f \left(x + \Delta x\right) - f \left(x\right)}{\Delta x}$

$= {\lim}_{\Delta x \to 0} \frac{1}{\Delta x} \left(\csc \left(\sqrt{x + \Delta x}\right) - \csc \left(\sqrt{x}\right)\right)$

$= {\lim}_{\Delta x \to 0} \frac{1}{\Delta x} \left(\frac{1}{\sin} \left(\sqrt{x + \Delta x}\right) - \frac{1}{\sin} \left(\sqrt{x}\right)\right)$

$= {\lim}_{\Delta x \to 0} \frac{- 1}{\Delta x} \left(\frac{\sin \left(\sqrt{x + \Delta x}\right) - \sin \left(\sqrt{x}\right)}{\sin \left(\sqrt{x + \Delta x}\right) \sin \left(\sqrt{x}\right)}\right)$

$= {\lim}_{\Delta x \to 0} \frac{- 1}{\Delta x \sin \left(\sqrt{x + \Delta x}\right) \sin \left(\sqrt{x}\right)} \left(\sin \left(\sqrt{x + \Delta x}\right) - \sin \left(\sqrt{x}\right)\right)$

Apply the series expansion of the sine function:

f'(x)=lim_(Deltax->0)(-1)/(Deltaxsin(sqrt(x+Deltax))sin(sqrtx))sum_(n=0)^oo(-1)^n/((2n+1)!)((sqrt(x+Deltax))^(2n+1)-(sqrtx)^(2n+1))

=lim_(Deltax->0)(-1)/(Deltaxsin(sqrt(x+Deltax))sin(sqrtx))sum_(n=0)^oo((-1)^n(sqrtx)^(2n+1))/((2n+1)!)((1+(Deltax)/x)^(n+1/2)-1)

Apply another series expansion:

f'(x)=lim_(Deltax->0)(-1)/(Deltaxsin(sqrt(x+Deltax))sin(sqrtx))sum_(n=0)^oo((-1)^n(sqrtx)^(2n+1))/((2n+1)!)(sum_(m=0)^oo((n+1/2),(m))((Deltax)/x)^m-1)

Isolate the $m = 0$ and $m = 1$ terms:

f'(x)=lim_(Deltax->0)(-1)/(Deltaxsin(sqrt(x+Deltax))sin(sqrtx))sum_(n=0)^oo((-1)^n(sqrtx)^(2n+1))/((2n+1)!)(1+(n+1/2)((Deltax)/x)+sum_(m=2)^oo((n+1/2),(m))((Deltax)/x)^m-1)

=lim_(Deltax->0)(-1)/(sin(sqrt(x+Deltax))sin(sqrtx))(1/(2sqrtx)sum_(n=0)^oo(-1)^n/((2n)!)(sqrtx)^(2n)+sum_(n=0)^oosum_(m=2)^oo((n+1/2),(m))((-1)^n(sqrtx)^(2n+1))/((2n+1)!)(Deltax)^(m-1)/x^m)

Recognize the series expansion of the cosine function:

f'(x)=lim_(Deltax->0)(-1)/(sin(sqrt(x+Deltax))sin(sqrtx))(1/(2sqrtx)cos(sqrtx)+sum_(n=0)^oosum_(m=2)^oo((n+1/2),(m))((-1)^n(sqrtx)^(2n+1))/((2n+1)!)(Deltax)^(m-1)/x^m)

Take the limit $\Delta x \to 0$:

$f ' \left(x\right) = \frac{- 1}{{\sin}^{2} \left(\sqrt{x}\right)} \left(\frac{1}{2 \sqrt{x}} \cos \left(\sqrt{x}\right)\right)$

$= - \frac{\csc \left(\sqrt{x}\right) \cot \left(\sqrt{x}\right)}{2 \sqrt{x}}$

2

## How do you use the integral test to determine if Sigma n^ke^-n from [1,oo) where k is an integer is convergent or divergent?

Andrea S.
Featured 4 days ago

The series

${\sum}_{= 1}^{\infty} {n}^{k} {e}^{- n}$

is convergent for any integer $k$

#### Explanation:

Choose as test function:

$f \left(x\right) = {x}^{k} {e}^{- x}$

This function is:

(i) non negative, as $f \left(x\right) > 0$ for $x > 0$

(ii) strictly decreasing in $\left(k , + \infty\right)$ as:

$f ' \left(x\right) = k {x}^{k - 1} {e}^{- x} - {x}^{k} {e}^{- x} = {x}^{k - 1} {e}^{- x} \left(k - x\right) < 0$ for $x > k$

(iii) infinitesimal, as:

${\lim}_{x \to \infty} {x}^{k} {e}^{- x} = 0$

(iv) $f \left(n\right) = {n}^{k} {e}^{- n}$

so the convergence of the series is equivalent to the convergence of the integral:

${\int}_{1}^{\infty} {x}^{k} {e}^{- x} \mathrm{dx}$

We have to find a general formula for this integral. Start from $k = 1$, integrating by parts:

${I}_{1} = {\int}_{1}^{\infty} x {e}^{- x} \mathrm{dx} = - {\int}_{1}^{\infty} x d \left({e}^{- x}\right) = {\left[- x {e}^{- x}\right]}_{1}^{\infty} + {\int}_{1}^{\infty} {e}^{- x} \mathrm{dx} = \frac{1}{e} - {\left[{e}^{- x}\right]}_{1}^{\infty} = \frac{2}{e}$

For $k = 2$

${I}_{2} = {\int}_{1}^{\infty} {x}^{2} {e}^{- x} \mathrm{dx} = - {\int}_{1}^{\infty} {x}^{2} d \left({e}^{- x}\right) = {\left[- {x}^{2} {e}^{- x}\right]}_{1}^{\infty} + 2 {\int}_{1}^{\infty} x {e}^{- x} \mathrm{dx} = \frac{1}{e} + 2 {I}_{1} = \frac{5}{e}$

and in general:

${I}_{k} = {\int}_{1}^{\infty} {x}^{k} {e}^{- x} \mathrm{dx} = - {\int}_{1}^{\infty} {x}^{k} d \left({e}^{- x}\right) = {\left[- {x}^{k} {e}^{- x}\right]}_{1}^{\infty} + k {\int}_{1}^{\infty} {x}^{k - 1} {e}^{- x} \mathrm{dx} = \frac{1}{e} + k {I}_{k - 1}$

I_k= 1/e + kI_(k-1) = 1/e +k(1/e+(k-1))I_(k-2) = ... = (1+k+k(k-1)+...+k!)/e = 1/esum_(j=1)^k (k!)/(j!)

So we can assert that:

int_1^oo x^ke^(-x)dx = 1/esum_(j=1)^k (k!)/(j!)

is convergent for every $k$ and this proves that also:

${\sum}_{= 1}^{\infty} {n}^{k} {e}^{- n}$

is convergent.

2

## How do you find the derivative of -ln(x-(x^2+1)^(1/2))?

Rory H.
Featured 4 days ago

The derivative of $- \ln \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)$ with respect to x is $\frac{1}{\sqrt{{x}^{2} + 1}} .$ To solve this use the chain rule carefully. See the explanation for details.

#### Explanation:

Starting with:

$\frac{d}{\mathrm{dx}} \left(- \ln \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)\right)$

use the chain rule to get:

$\frac{d}{\mathrm{dx}} \left(- \ln \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)\right)$

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

use the chain rule once again on the remaining derivative:

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

Simplify:

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

Note that $1 = {\left({x}^{2} + 1\right)}^{\frac{1}{2}} / {\left({x}^{2} + 1\right)}^{\frac{1}{2}}$, then substitute this for 1:

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

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

$= \left(\frac{x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}}{{x}^{2} + 1} ^ \left(\frac{1}{2}\right)\right) \div i \mathrm{de} \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)$

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

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

$= 1 \cdot \frac{1}{{\left({x}^{2} + 1\right)}^{\frac{1}{2}}}$

$= \frac{1}{{\left({x}^{2} + 1\right)}^{\frac{1}{2}}} = \frac{1}{\sqrt{\left({x}^{2} + 1\right)}} .$

Finally:

$\frac{d}{\mathrm{dx}} \left(- \ln \left(x - {\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)\right) = \frac{1}{\sqrt{\left({x}^{2} + 1\right)}} .$

If you have any questions about the use of the chain rule or any other part of this solution, then please ask.

Rory.

2

## What is the area of the largest rectangle that can be inscribed under the graph of y=2 cos x for -π /2 ≤x ≤π /2?

Steve
Featured 4 days ago

Maximum area is $2.244 {\text{ unit}}^{2}$ (3dp)

#### Explanation:

I assume that you man bounded by the x-axis also, otherwise the largest rectangle would be unbounded and therefore infinite.

This is a diagram depicting the problem:

Where $P \left(\alpha , \beta\right)$ is the point in Quadrant 1 where the rectangle intersects the curve $y = 2 \cos x$, and $P ' \left(- \alpha , \beta\right)$ is the corresponding point in quadrant 2.

Let us set up the following variables:

$\left\{\begin{matrix}\alpha & x \text{-coordinate of point "P \\ beta & y"-coordinate of point "P \\ A & "Total Area of inscribed rectangle}\end{matrix}\right.$

Our aim is to find $A$, as a function of a single variable and to maximize the total area, $A$ (It won't matter which variable we do this with as we will get the same result). ie we want a critical point of $A$ wrt the variable.

As $P$ lies on the curve $y = 2 \cos x$, we have:

$\beta = 2 \cos \alpha \setminus \setminus \setminus \setminus \setminus \ldots . . \left[1\right]$

And the total Area is that of a rectagle of widt $2 \alpha$ and height $\beta$, so:

$A = 2 \alpha \beta$
$\setminus \setminus \setminus = 2 \alpha \left(2 \cos \alpha\right) \setminus \setminus \setminus \setminus \setminus$ (from [1] )
$\setminus \setminus \setminus = 4 \alpha \cos \alpha$

We now have the Area, $A$, as a function of a single variable $\alpha$, so differentiating wrt $\alpha$ (using the product rule) we get:

$\frac{\mathrm{dA}}{d \alpha} = \left(4 \alpha\right) \left(- \sin \alpha\right) + \left(4\right) \left(\cos \alpha\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = 4 \left(\cos \alpha - \alpha \sin \alpha\right)$

At a critical point we have $\frac{\mathrm{dA}}{d \alpha} = 0 \implies$

$4 \left(\cos \alpha - \alpha \sin \alpha\right) = 0$
$\therefore \cos \alpha - \alpha \sin \alpha = 0$

In order to solve this equation we use Newton-Rhapson which gives $\alpha = 0.860333589 \ldots$

With this value of $\alpha$ we have:

$\beta = 1.30436924 \ldots$
$A = 2.24438535 \ldots$

We can visually verify that this corresponds to a maximum by looking at the graph of $y = A \left(\alpha\right)$:

graph{4xcosx [-4, 4, -5.5, 5.5]}

So maximum area is $2.244 {\text{ unit}}^{2}$ (3dp)

3

## What is lim_(xtooo)(5x^3+2x-3)/(4-x^2-2x^3)? I'm completely blank on this one please help

Jim H
Featured 14 hours ago

$- \frac{5}{2}$

#### Explanation:

Factor out of the numerator and denominator the greatest power of $x$ in the denominator, and then reduce. (Or, divide numerator and denominator by the greatest power of $x$ in the denominator.)

Use ${\lim}_{x \rightarrow \infty} \frac{c}{x} ^ n = 0$ for all $c$ and all positive $n$.

${\lim}_{x \rightarrow \infty} \frac{5 {x}^{3} + 2 x - 1}{4 - {x}^{2} - 2 {x}^{3}} = {\lim}_{x \rightarrow \infty} \frac{{x}^{3} \left(5 + \frac{2}{x} ^ 2 - \frac{1}{x} ^ 3\right)}{{x}^{3} \left(\frac{4}{x} ^ 3 - \frac{1}{x} - 2\right)}$

$= {\lim}_{x \rightarrow \infty} \frac{5 + \frac{2}{x} ^ 2 - \frac{1}{x} ^ 3}{\frac{4}{x} ^ 3 - \frac{1}{x} - 2}$

$= \frac{5 + 0 - 0}{0 - 0 - 2} = - \frac{5}{2}$

Note

The above is the long explanation. The short one is look at just the dominating terms -- the greatest powers of $x$-- in the numerator and denominator. At infinity, the quotient behaves like the ratio of those greatest powers.

Examples

${\lim}_{x \rightarrow \infty} \frac{3 {x}^{5} + 7 {x}^{2} - 19}{4 {x}^{5} - 2 {x}^{3} + 3 x - 2} = {\lim}_{x \rightarrow \infty} \frac{3 {x}^{5}}{4 {x}^{5}} = {\lim}_{x \rightarrow \infty} \frac{3}{4} = \frac{3}{4}$

${\lim}_{x \rightarrow \infty} \frac{7 {x}^{2} - 5 x + 7}{2 {x}^{3} + 5 x + 8} = {\lim}_{x \rightarrow \infty} \frac{7 {x}^{2}}{2 {x}^{3}} = {\lim}_{x \rightarrow \infty} \frac{7}{2 x} = 0$

${\lim}_{x \rightarrow \infty} \frac{{x}^{4} + 3 x - 9}{6 x + 2} = {\lim}_{x \rightarrow \infty} \frac{{x}^{4}}{6 x} = {\lim}_{x \rightarrow \infty} {x}^{3} / 6 = \infty$

2

## How do you find the arc length of the curve f(x)=x^3/6+1/(2x) over the interval [1,3]?

HSBC244
Featured 3 days ago

The arc length is $\frac{14}{3}$ units.

#### Explanation:

The arc length of a curve on the interval $\left[a , b\right]$ is given by evaluating ${\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$.

The derivative of $f ' \left(x\right)$, given by the power rule, is

$f ' \left(x\right) = \frac{1}{2} {x}^{2} - \frac{1}{2 {x}^{2}} = \frac{{x}^{4} - 1}{2 {x}^{2}}$

Substitute this into the above formula.

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

Expand.

${\int}_{1}^{3} \sqrt{1 + \frac{{x}^{8} - 2 {x}^{4} + 1}{4 {x}^{4}}} \mathrm{dx}$

Put on a common denominator.

int_1^3sqrt((x^8 + 2x^4 + 1)/(4x^4)dx

Factor the numerator as the perfect square trinomial, and recognize the denominator can be written of the form ${\left(a x\right)}^{2}$.

${\int}_{1}^{3} \sqrt{{\left({x}^{4} + 1\right)}^{2} / {\left(2 {x}^{2}\right)}^{2}} \mathrm{dx}$

Eliminate the square root using ${\left({a}^{2}\right)}^{\frac{1}{2}} = a$

${\int}_{1}^{3} \frac{{x}^{4} + 1}{2 {x}^{2}} \mathrm{dx}$

Factor out a $\frac{1}{2}$ and put it in front of the integral.

$\frac{1}{2} {\int}_{1}^{3} \frac{{x}^{4} + 1}{x} ^ 2 \mathrm{dx}$

Separate into different fractions.

$\frac{1}{2} {\int}_{1}^{3} {x}^{4} / {x}^{2} + \frac{1}{x} ^ 2 \mathrm{dx}$

Simplify using ${a}^{n} / {a}^{m} = {a}^{n - m}$ and $\frac{1}{a} ^ n = {a}^{-} n$.

$\frac{1}{2} {\int}_{1}^{3} {x}^{2} + {x}^{-} 2 \mathrm{dx}$

Integrate using $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$, with $n \in \mathbb{R} , n \ne - 1$.

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

Evaluate using the second fundamental theorem of calculus, which states that for ${\int}_{a}^{b} F \left(x\right) = f \left(b\right) - f \left(a\right)$, if $f \left(x\right)$ is continuous on $\left[a , b\right]$ and where $f ' \left(x\right) = F \left(x\right)$.

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

Combine fractions and simplify.

$\frac{1}{2} \left(9 - \frac{1}{3} - \frac{1}{3} + 1\right)$

$\frac{1}{2} \left(10 - \frac{2}{3}\right)$

$5 - \frac{1}{3}$

$\frac{14}{3}$

Hopefully this helps!

3

## Let f_n (x) = (Sigma_(n=1))^n (sin^2x)/(cos^2(x/2) - cos^2((2n+1)/2)x) and g_n (x) = (Pi_(n=1))^n f_n (x) Let I_n = int_0^pi (f_n (x))/(g_n (x)) dx and if (Sigma_(n=1))^n I_n = K pi . Then K is??

Steve
Featured yesterday

$K = 1$

#### Explanation:

Firstly poor notation for the summation and product. The standard notation is to use a "dummy" variable, usually $i$ or $r$ as the loop counter, as in

${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$

So using the correct notation we have:

${f}_{n} \left(x\right) = {\sum}_{r = 1}^{n} \setminus {\sin}^{2} \frac{x}{{\cos}^{2} \left(\frac{x}{2}\right) - {\cos}^{2} \left(\frac{\left(2 r + 1\right) x}{2}\right)}$

If we focus in the denominator for a moment, which desperately needs simplification, we see that it is the difference of two squares, so we can factorise prior to simplifying, and we can use the identities:

cos(A)+cos(B) = \ \ \ \ 2cos((​A+B)/2​​ )cos((​A−B)/2​​ )
cos(A) - cos(B) = -2 sin((​A+B)/2​​ )sin((​A−B)/2​​ )
$\sin 2 A = 2 \sin A \cos A$

to get;

${\cos}^{2} \left(\frac{x}{2}\right) - {\cos}^{2} \left(\frac{\left(2 r + 1\right) x}{2}\right)$
$\setminus \setminus \setminus = {\cos}^{2} \left(\frac{x}{2}\right) - {\cos}^{2} \left(r x + \frac{x}{2}\right)$
$\setminus \setminus \setminus = \left(\cos \left(\frac{x}{2}\right) - \cos \left(r x + \frac{x}{2}\right)\right) \left(\cos \left(\frac{x}{2}\right) - \cos \left(r x + \frac{x}{2}\right)\right)$
$\setminus \setminus \setminus = 2 \cos \left(\frac{r x + x}{2}\right) \cos \left(\frac{- r x}{2}\right) \left(- 2\right) \sin \left(\frac{r x + x}{2}\right) \sin \left(\frac{- r x}{2}\right)$
$\setminus \setminus \setminus = 2 \cos \left(\frac{r x + x}{2}\right) \cos \left(\frac{r x}{2}\right) \left(- 2\right) \sin \left(\frac{r x + x}{2}\right) \left(- \sin \left(\frac{r x}{2}\right)\right)$
$\setminus \setminus \setminus = \left\{2 \sin \left(\frac{r x}{2}\right) \cos \left(\frac{r x}{2}\right)\right\} \left\{2 \sin \left(\frac{r x + x}{2}\right) \cos \left(\frac{r x + x}{2}\right)\right\}$
$\setminus \setminus \setminus = \sin \left(\frac{2 r x}{2}\right) \sin \left(\frac{2 \left(r x + x\right)}{2}\right)$
$\setminus \setminus \setminus = \sin \left(r x\right) \sin \left(\left(r + 1\right) x\right)$

So we can therefore write ${f}_{n} \left(x\right)$ as:

${f}_{n} \left(x\right) = {\sum}_{r = 1}^{n} \setminus {\sin}^{2} \frac{x}{\sin \left(r x\right) \sin \left(\left(r + 1\right) x\right)}$

Let us examine the first few expansions of $f$ and $g$

${f}_{1} \left(x\right) = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}$
${f}_{2} \left(x\right) = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}$
${f}_{2} \left(x\right) = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)} + {\sin}^{2} \frac{x}{\sin \left(3 x\right) \sin \left(4 x\right)}$

And we have:

${g}_{n} \left(x\right) = {\prod}_{r = 1}^{n} \setminus {f}_{n} \left(x\right)$

${g}_{1} \left(x\right) = {f}_{1} \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}$

${g}_{2} \left(x\right) = {f}_{1} \left(x\right) \cdot {f}_{2} \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}\right)$

${g}_{3} \left(x\right) = {f}_{1} \left(x\right) \cdot {f}_{2} \left(x\right) \cdot {f}_{3} \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)} + {\sin}^{2} \frac{x}{\sin \left(3 x\right) \sin \left(4 x\right)}\right)$

So using the definition of ${I}_{n}$:

${I}_{n} = {\int}_{0}^{\pi} \setminus {f}_{n} \frac{x}{g} _ n \left(x\right) \setminus \mathrm{dx}$

We have:

${I}_{1} = {\int}_{0}^{\pi} \setminus {f}_{1} \frac{x}{g} _ 1 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \setminus \frac{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}}{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\left[x\right]}_{0}^{\pi}$
$\setminus \setminus \setminus = \pi$

So if ${\sum}_{r = 1}^{n} \setminus {I}_{r} = K \pi$ then;

$n = 1 \implies {I}_{1} = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies \pi = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies K = 1$

Let's see if this holds with $n = 2$, if so then perhaps we can prove the proposition by Induction.

${I}_{2} = {\int}_{0}^{\pi} \setminus {f}_{2} \frac{x}{g} _ 2 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \setminus \frac{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}}{\left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}\right)} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \frac{1}{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \frac{\sin \left(x\right) \sin \left(2 x\right)}{\sin} ^ 2 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \frac{\sin \left(x\right) 2 \sin x \cos x}{\sin} ^ 2 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} 2 \cos x \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 2 \setminus {\left[\sin x\right]}_{0}^{\pi} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 2 \setminus \left(\sin \pi - \sin 0\right)$
$\setminus \setminus \setminus = 0$

So if ${\sum}_{r = 1}^{n} \setminus {I}_{r} = K \pi$ then;

$n = 2 \implies {I}_{1} + {I}_{2} = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies \pi + 0 = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies K = 1$, consistent with the above case $n = 1$

I think it's fairly easy to see that ${I}_{k} = 0 \forall k \ge 2$, and if I have a bit more time later I will attempt to prove that by Induction.