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5

## Solve: Lim x->1 (sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2) ?

Cesareo R.
Featured 3 months ago

$\frac{3}{2}$

#### Explanation:

lim_(x->1)(sqrt(x)-sqrt(2-x^2))/(2x-sqrt(2+2x^2)

(sqrt[x] - sqrt[2 - x^2])/(2 x - sqrt[2 + 2 x^2]) *(sqrt[x] + sqrt[ 2 - x^2])/(2 x + sqrt[2 + 2 x^2])

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

= $\frac{1}{2} \left(\frac{x + 2}{x + 1}\right)$ and then

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

and then

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

= $\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{4}{2} = \frac{3}{2}$

4

## What is the shortest distance from the point (-3,3) to the curve y=(x-3)^3?

Jim H
Featured 3 months ago

#### Explanation:

Every point on the curve has coordinates $\left(x , {\left(x - 3\right)}^{3}\right)$ and the distance between such a point and the point $\left(- 3 , 3\right)$ is:

$\sqrt{{\left(x + 3\right)}^{2} + {\left({\left(x - 3\right)}^{3} - 3\right)}^{2}}$.

We can minimize the distance by minimizing the radicand:

$f \left(x\right) = {\left(x + 3\right)}^{2} + {\left({\left(x - 3\right)}^{3} - 3\right)}^{2}$.

Differentiate:

$f ' \left(x\right) = 2 \left(x + 3\right) + 2 \left({\left(x - 3\right)}^{3} - 3\right) \left(3 {\left(x - 3\right)}^{2}\right)$

Use some technology or approximation method to get

$x \approx 2.278$. (The other 4 solutions are imaginary.)

There cannot be a maximum. There is a minimum at this $x$.

Using the distance above, we find a minimum distance of approximately $6.266$

3

## Using the first derivative, what are the critical points of the curve and the intervals where it's increasing and decreasing?

Somebody N.
Featured 3 months ago

See below.

#### Explanation:

The critical points are the points where the first derivative equals zero, or the point doesn’t exist.

This is a polynomial so it is continuous for all $x \in \mathbb{R}$

First derivative of $f \left(x\right) = {x}^{4} - 8 {x}^{3} + 18 {x}^{2} - 5$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(f \left(x\right)\right) = 4 {x}^{3} - 24 {x}^{2} + 36 x$

Equating this to zero:

$4 {x}^{3} - 24 {x}^{2} + 36 x = 0$

$4 x \left({x}^{2} - 6 x + 9\right) = 0$

$4 x = 0 \implies x = 0$

$\left({x}^{2} - 6 x + 9\right) = 0$

Factor:

${\left(x - 3\right)}^{2} = 0 \implies x = 3$

Putting these values in $f \left(x\right)$

$f \left(0\right) = - 5$

$f \left(3\right) = 22$

Critical Points:

$\textcolor{b l u e}{x = 0}$ and $\textcolor{b l u e}{x = 3}$

$\left(0 , - 5\right)$ and $\left(3 , 22\right)$

$f \left(x\right)$ is increasing where $f ' \left(x\right) > 0$

$f \left(x\right)$ is decreasing where $f ' \left(x\right) < 0$

$4 {x}^{3} - 24 {x}^{2} + 36 x > 0$

${x}^{3} - 6 {x}^{2} + 9 x > 0$

$x \left({x}^{2} - 6 x + 9\right) > 0$

$x {\left(x - 3\right)}^{2} > 0$

$0 < x < 3$ , $3 < x < \infty$

Increasing in $\left(0 , 3\right)$ and $\left(3 , \infty\right)$

$4 {x}^{3} - 24 {x}^{2} + 36 x < 0$

${x}^{3} - 6 {x}^{2} + 9 x < 0$

$x \left({x}^{2} - 6 x + 9\right) < 0$

$x {\left(x - 3\right)}^{2} < 0$

$- \infty < x < 0$

Decreasing in $\left(- \infty , 0\right)$

GRAPH:

3

## Can anyone give me the idea of undefined Integral =?

George C.
Featured 3 months ago

A few thoughts...

#### Explanation:

Definite vs indefinite integral

A definite integral includes a specification of the set of values over which the integral should be calculated. As a result, it has a definite value, e.g. the area under a curve in a given interval.

By way of contrast, an indefinite integral does not specify the set of values over which the integral should be calculated. It basically identifies what the antiderivative function looks like, including some constant of integration to be determined. For example:

$\int {x}^{2} \mathrm{dx} = \frac{1}{3} {x}^{3} + C$

Non-elementary integrals of elementary functions

Unlike derivatives, the integral of an elementary function is not necessarily elementary. The term "elementary function" denotes functions constructed using basic arithmetic operations, $n$th roots, trigonometric, hyperbolic, exponential and logarithms.

There are some very useful non-elementary functions expressible as integrals of elementary functions. For example, the Gamma function:

$\Gamma \left(x\right) = {\int}_{0}^{\infty} {t}^{x - 1} {e}^{- t} \mathrm{dt}$

The Gamma function extends the definition of factorial to values apart from non-negative integers.

Poles and Cauchy principal value

If a function has a singularity such as a simple pole, then its definite integral over a range including that pole is not automatically well defined. A workaround for such cases is provided by the Cauchy principal value.

For example:

${\int}_{- 1}^{1} \frac{\mathrm{dt}}{t} = {\lim}_{\epsilon \to 0 +} \left({\int}_{- 1}^{-} \epsilon \frac{\mathrm{dt}}{t} + {\int}_{\epsilon}^{1} \frac{\mathrm{dt}}{t}\right) = 0$

Non measurable sets

If the set over which you are trying to integrate is non-measurable, then the integral is usually not defined. An exception would be if the value of the function on that set was zero.

To 'construct' a non-measurable set you would typically use the axiom of choice.

For example, you could define an equivalence relation on $\mathbb{R}$ by:

a ~ b <=> (a-b) " is rational"

This equivalence relation partitions $\mathbb{R}$ into an uncountable infinity of countable sets.

Use the axiom of choice to choose exactly one element of each equivalence class to create a subset $S \subset \mathbb{R}$.

For any rational number $x$, we can define ${S}_{x}$ to consist of the elements of $S$ offset by $x$. Then the sets ${S}_{x} : x \in \mathbb{Q}$ form a partition of $\mathbb{R}$ into a countable infinity of subsets.

We can define a non-integrable function by:

$f \left(t\right) = \left\{\begin{matrix}1 \text{ if " t in S_x " where " x = p/q " in lowest terms and " q " is even" \\ 0 " otherwise}\end{matrix}\right.$

This function is not integrable over any interval.

5

## lim_(n->oo)(sum_(k=1)^n(sqrt(2n^2+k)/(2n^2+k)))?

Hammer
Featured 3 months ago

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{\sqrt{2 {n}^{2} + k}}{2 {n}^{2} + k} = \frac{1}{\sqrt{2}}$

#### Explanation:

We have

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{\sqrt{2 {n}^{2} + k}}{2 {n}^{2} + k} = {\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{\sqrt{2 {n}^{2} + k}} =$

= lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2)

Now, let's try to remember the definition of an integral. This might seem strange, but it will come of great use.

The definite integral ${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$ represents the $\textcolor{red}{\text{net area}}$ defined by the graph of the function $f \left(x\right)$ and the x-axis. One way to find the value of an integral is by aproximating it with infinitely many rectangles ("boxes") of equal or not width.

This concept is called a $\textcolor{red}{\text{Riemann Sum}}$.

There are infinitely many rectangles, therefore, if there are $n$ rectangles, we have to take the limit as $n \to \infty$.

Now, let's take the case where the width is constant between all boxes. If we define ${x}_{1} , {x}_{2} , \ldots , {x}_{n}$ to be some "marks" on the x-axis such that the width of the $i$-th rectangle is ${x}_{i + 1} - {x}_{i} = \Delta {x}_{i}$, then its lenght is $f \left({x}_{i}\right)$. Since the width is equal, $\Delta {x}_{i}$ is the same for all $i$'s. We can more easily call it color(red)(Delta.

The Riemann Sum aproximates an integral by being the sum of all the rectangles. In order to simplify this further, let's take the particular case where color(red)(a = 0 and color(red)(b=1. We have:

${\int}_{0}^{1} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \Delta \cdot f \left({x}_{i}\right)$

This is only true if ${x}_{n}$ is equal to $1$.

One way to make $\Delta$ constant is to define ${x}_{k} = \frac{k}{n}$, for some $k$, $1 \le k \le n$. This way, ${x}_{n} = 1$ and

$\textcolor{red}{\Delta} = {x}_{i + 1} - {x}_{i} = \frac{i + 1}{n} - \frac{i}{n} = \textcolor{red}{\frac{1}{n}}$.

${\int}_{0}^{1} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \frac{1}{n} \cdot f \left(\frac{i}{n}\right)$.

This looks familiar to our original sum, ${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}}$.

All we are left to do is to assume they're equal and then solve the integral.

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \frac{1}{n} \cdot f \left(\frac{i}{n}\right)$

$i$ and $k$ are just the indexes, the name of which doesn't matter. Both sums could be counting $i$ or $k$, it's irrelevant.

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}} = {\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot f \left(\frac{k}{n}\right)$

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

We have to find the integral of this function:

${\lim}_{n \to \infty} {\int}_{0}^{1} \frac{\mathrm{dx}}{\sqrt{2 + \frac{x}{n}}}$

Since $n \to \infty$, this means that $\frac{x}{n} = 0$, as $x$ is finite.

${\lim}_{n \to \infty} {\int}_{0}^{1} \frac{\mathrm{dx}}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.

This is because it forms a rectangle with the x-axis, the width and lenght of which is $1$ and $\frac{1}{\sqrt{2}}$, respectively.

As a conclusion,

color(blue)(lim_(n->oo) sum_(k=1)^(n) sqrt(2n^2+k)/(2n^2+k) = 1/sqrt2.

4

## Find the Limit?

Hammer
Featured 2 months ago

${\lim}_{n \to \infty} {P}_{n} = \frac{2}{3}$

#### Explanation:

We see that

${P}_{n} = {\prod}_{k = 2}^{n} \frac{{k}^{3} - 1}{{k}^{3} + 1} = {\prod}_{k = 2}^{n} \frac{{k}^{3} - {1}^{3}}{{k}^{3} + {1}^{3}}$

Both the numerator and denominator are a sum or difference of cubes, for which we have a formula:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

Thus, we have

${P}_{n} = {\prod}_{k = 2}^{n} \frac{\left(k - 1\right) \left({k}^{2} + k + 1\right)}{\left(k + 1\right) \left({k}^{2} - k + 1\right)}$

${P}_{n} = \frac{\left(2 - 1\right) \left({2}^{2} + 2 + 1\right)}{\left(2 + 1\right) \left({2}^{2} - 2 + 1\right)} \cdot \frac{\left(3 - 1\right) \left({3}^{2} + 3 + 1\right)}{\left(3 + 1\right) \left({3}^{2} - 3 + 1\right)} \ldots \frac{\left(n - 2\right) \left({n}^{2} - n + 1\right)}{\left(n\right) \left({n}^{2} - 3 n + 3\right)} \frac{\left(n - 1\right) \left({n}^{2} + n + 1\right)}{\left(n + 1\right) \left({n}^{2} - n + 1\right)}$

We can group ${P}_{n}$ into the following products:

${P}_{n} = \textcolor{red}{{\prod}_{k = 2}^{n} \frac{k - 1}{k + 1}} \cdot \left(\frac{{2}^{2} + 2 + 1}{{2}^{2} - 2 + 1}\right) \left(\frac{{3}^{2} + 3 + 1}{{3}^{2} - 3 + 1}\right) \ldots \left(\frac{{n}^{2} - n + 1}{{n}^{2} - 3 n + 3}\right) \left(\frac{{n}^{2} + n + 1}{{n}^{2} - n + 1}\right)$

Let the product in $\textcolor{red}{\text{red}}$ be ${p}_{n}$.

Notice how some things cancel each other:

${P}_{n} = \textcolor{red}{{p}_{n}} \cdot \left(\frac{\cancel{{2}^{2} + 2 + 1}}{{2}^{2} - 2 + 1}\right) \left(\frac{\cancel{{3}^{2} + 3 + 1}}{\cancel{{3}^{2} - 3 + 1}}\right) \ldots \left(\frac{\cancel{{n}^{2} - n + 1}}{\cancel{{n}^{2} - 3 n + 3}}\right) \left(\frac{{n}^{2} + n + 1}{\cancel{{n}^{2} - n + 1}}\right)$

It might seem confusing, but the numerator of the $k$-th term is being cancelled by the denominator of the $k + 1$-th term.

$\therefore$

${P}_{n} = {p}_{n} \frac{{n}^{2} + n + 1}{3}$

Let's write out ${p}_{n}$:

${p}_{n} = \frac{2 - 1}{2 + 1} \cdot \frac{3 - 1}{3 + 1} \ldots \frac{n - 2}{n} \cdot \frac{n - 1}{n + 1}$

${p}_{n} = \frac{1 \cdot 2 \cdot 3 \cdot \ldots \cdot \left(n - 2\right) \left(n - 1\right)}{3 \cdot 4 \cdot 5 \cdot \ldots \cdot n \left(n + 1\right)}$

Again, lots of things cancel out.

${p}_{n} = \frac{1 \cdot 2 \cdot \cancel{3 \cdot 4 \cdot 5 \cdot \ldots \cdot \left(n - 2\right) \left(n - 1\right)}}{\cancel{3 \cdot 4 \cdot 5 \cdot \ldots \cdot \left(n - 2\right) \left(n - 1\right)} n \left(n + 1\right)} = \frac{2}{n \left(n + 1\right)}$

Therefore, we have:

${P}_{n} = \textcolor{red}{\frac{2}{n \left(n + 1\right)}} \cdot \left[\frac{n \left(n + 1\right)}{3} + \frac{1}{3}\right]$

${P}_{n} = \frac{2}{3} \left[1 + \frac{1}{n \left(n + 1\right)}\right]$

Now, let's take the $\textcolor{red}{\text{limit}}$ as $n \to \infty$ of both sides.

${\lim}_{n \to \infty} {P}_{n} = {\lim}_{n \to \infty} \frac{2}{3} \left[1 + \frac{1}{n \left(n + 1\right)}\right]$

As $n$ grows boundless, $\frac{1}{n \left(n + 1\right)}$ becomes arbitrarily small.

In other words, it approaches $0$.

$\implies {\lim}_{n \to \infty} {P}_{n} = \frac{2}{3} \left[1 + 0\right] = \frac{2}{3}$

$\textcolor{red}{{\lim}_{n \to \infty} {P}_{n} = \frac{2}{3}}$

3

## Find the area of the region enclosed by y=ln(x) ,the x-axis,the y-axis and y=1 ? (a) dx select (b) dy select

Parabola
Featured 2 months ago

$A = e - 1$

#### Explanation:

I have the graphs of the functions and lines here:

We want the area of the green region.

We can think of the situation like this:

Now, let's find the intercept and the intersection.

The intercept:

$\ln \left(x\right) = 0$

$\implies x = {e}^{0}$

$\implies x = 1$

The intersection:

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

${e}^{1} = x$

$e = x$

We can now form a rectangle like the following:

The area of the green region is the area of the rectangle minus the area under the curve of $\ln \left(x\right)$ from $1$ to $e$

$A = e \cdot 1 - {\int}_{1}^{e} \ln \left(x\right) \mathrm{dx}$

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

Remember that $\int \ln \left(x\right) \mathrm{dx} = x \left(\ln \left(x\right) - 1\right) + C$

$\implies A = e - {\int}_{1}^{e} \ln \left(x\right) \mathrm{dx}$

$\implies A = e - \left(e \left(\ln \left(e\right) - 1\right) - 1 \left(\ln \left(1\right) - 1\right)\right)$

$\implies A = e - \left(e \left(1 - 1\right) - 1 \left(0 - 1\right)\right)$

=>A=e-(e(0)-1(-1)

$\implies A = e - \left(0 + 1\right)$

$\implies A = e - 1$

$\implies A \approx 1.71828182846$

In case you are wondering how I solved $\int \ln \left(x\right) \mathrm{dx}$...

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

$u = \ln \left(x\right)$

$\mathrm{dv} = 1$

$\implies \mathrm{du} = \frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right)$

$\implies \mathrm{du} = \frac{1}{x}$

$\implies v = \int 1 \mathrm{dx}$

$\implies v = x$

$\implies x \ln \left(x\right) - \int x \cdot \frac{1}{x} \mathrm{dx}$

$\implies x \ln \left(x\right) - \int 1 \mathrm{dx}$

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

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

2

## How do I determine If int_0^1 3/(4x-1) dx converges or diverges?

Bryson
Featured 2 months ago

Diverges

#### Explanation:

Many students fall into a trap when solving problems similar to this coming up with an answer of $3 \ln \left(3\right)$ do not make this mistake.

However, let us begin by noting that this integral does not follow our formal definition by which the integral is continuous on the given interval. Note that $x = \frac{1}{4}$ results in a indefinite solution to the given equation.

If this original integral did not contain the value $\frac{1}{4}$ then we could compute the integral. In this case the only reason we cannot is that $\frac{1}{4}$ is on the interval $\left[0 , 1\right]$.

However in this case we are allow to set limits for the integral. From the limits we set it allows the user to determine if the integral approaches a value or diverges. In this case divergency meaning going off in two opposite directions towards $\infty$.

Let us start by computing the integral if we were on a different interval that did not include $\frac{1}{4}$

${\int}_{0}^{1} \frac{3}{4 x - 1}$

$3 \ln \left(\left\mid 4 x - 1 \right\mid\right)$

Now in this case let us define a new integral ${\int}_{0}^{b} \frac{3}{4 x - 1}$ and ${\int}_{b}^{1} \frac{3}{4 x - 1}$ by which we add the sums of these two integrals. We are allow to do this because in the case of these two integrals the sum is equal to the original integral.

Now since we experience a bad spot at $x = \frac{1}{4}$ let us approach $x = \frac{1}{4}$

In this case let us define $\lim b \to {1}^{+} / 4 \left[{\int}_{0}^{b} \frac{3}{4 x - 1}\right]$

$= \lim b \to {1}^{+} / 4 \left[3 \ln \left(\left\mid 4 b - 1 \right\mid\right) - 3 \ln \left(\left\mid 4 \left(0\right) - 1 \right\mid\right)\right]$

$= \lim b \to {1}^{+} / 4 \left[3 \ln \left(\left\mid 4 b - 1 \right\mid\right) - 3 \ln \left(1\right)\right]$

Let us then find $b$ from the opposite direction.

$\lim b \to {1}^{-} / 4 \left[{\int}_{b}^{1} \frac{3}{4 x - 1}\right]$

$= \lim b \to {1}^{-} / 4 \left[3 \ln \left(\left\mid 4 \left(1\right) - 1 \right\mid\right) - 3 \ln \left(\left\mid 4 b - 1 \right\mid\right)\right]$

$= \lim b \to {1}^{-} / 4 \left[3 \ln \left(3\right) - 3 \ln \left(\left\mid 4 b - 1 \right\mid\right)\right]$

let us now note that $\lim b \to {1}^{-} / 4 \left[3 \ln \left(\left\mid 4 b - 1 \right\mid\right)\right] = - \infty$

Remember that $\lim b \to 0 \ln \left(b\right) = - \infty$ in this case our interior function approaches zero. In cases like this we are allowed to substitute or shall I say formulate a new limit by which b approaches zero.

Now from summing and replacing our limits that:

$\left\{\lim b \to {1}^{+} / 4 \left[3 \ln \left(\left\mid 4 b - 1 \right\mid\right) - 3 \ln \left(1\right)\right]\right\} + \left\{\lim b \to {1}^{-} / 4 \left[3 \ln \left(3\right) - 3 \ln \left(\left\mid 4 b - 1 \right\mid\right)\right]\right\}$

$\left[3 \ln \left(3\right) + \infty\right] - \left[\infty\right]$
since $\ln \left(1\right) = 0$

We now have an indefinite form. In this case $\infty - \infty$ meaning our original integral approaches the area of $\infty$ minus $\infty$ $\therefore$ one point on the graph bends towards negative infinity to make an area beneath the curve of $\infty$ and the other end of our function bends in the positive direction such the make another area of $\infty$

2

## The region R is bounded by the graph of f(x)=2x(2-x) and the x-axis. Which is greater, the volume of the solid generated when R is revolved about the line y=2 or the volume of the solid generated when R is revolved about the line y=0?

Sean
Featured 2 months ago

The solid from revolving around the $x$-axis is greater

#### Explanation:

.

$R$ is the area bounded by:

$y = 2 x \left(2 - x\right)$ and $y = 0$ ($x$-axis).

$2 x \left(2 - x\right) = 0 , \therefore x = 0 \mathmr{and} 2$

This is a parabola that opens down and crosses the $x$-axis at $x = 0 \mathmr{and} 2$

If this area is revolved about the $x$-axis the resulting solid will look like the following figure:

If we cut a vertical slice through the solid with a thickness $= \mathrm{dx}$ we will have a circular disc whose radius is $y = 2 x \left(2 - x\right)$.

The area of this circular disc is:

$A = \pi {r}^{2} = \pi {\left[2 x \left(2 - x\right)\right]}^{2} = \pi \left[4 {x}^{2} \left({x}^{2} - 4 x + 4\right)\right]$

$A = \pi \left(4 {x}^{4} - 16 {x}^{3} + 16 {x}^{2}\right)$

Then the volume of this disc would br:

${V}_{\text{disc}} = \pi \left(4 {x}^{4} - 16 {x}^{3} + 16 {x}^{2}\right) \mathrm{dx}$

Now, imagine that the thickness $\mathrm{dx}$ is infinitely small and there are infinite number of discs with varying radii between $x = 0 \mathmr{and} 2$.

If we take the integral of the area of this disc and evaluate it from $0 \to 2$ we will get the volume of the solid.

${V}_{\text{Solid}} = {\int}_{0}^{2} \pi \left(4 {x}^{4} - 16 {x}^{3} + 16 {x}^{2}\right) \mathrm{dx}$

${V}_{\text{Solid}} = \pi {\left(\frac{4}{5} {x}^{5} - 4 {x}^{4} + \frac{16}{3} {x}^{3}\right)}_{0}^{2} = \pi \left(\frac{4}{5} \left(32\right) - 4 \left(16\right) + \frac{16}{3} \left(8\right)\right) = \pi \left(\frac{128}{5} - 64 + \frac{128}{3}\right) = \pi \frac{384 - 960 + 640}{15} = \frac{64}{15} \pi$

If the area between the parabola and the $x$-axis is revolved around the line $y = 2$ the solid generated will look like the following figure:

If we cut a vertical slice through this solid with a thickness of $\mathrm{dx}$ we will have a circular washer (a disc with a hole in the middle). The outer radius of this washer is ${r}_{2} = 2$ and the inner radius of the washer is ${r}_{1} = y = 2 x \left(2 - x\right)$.

We can calculate the area of the washer by subtracting the area of the inner circle from the area of the outer circle:

${A}_{\text{Washer}} = \pi {r}_{2}^{2} - \pi {r}_{1}^{2} = \pi \left({r}_{2}^{2} - {r}_{1}^{2}\right) = \pi \left[4 - {\left(2 x \left(2 - x\right)\right)}^{2}\right]$

${A}_{\text{Washer}} = \pi \left(- 4 {x}^{4} + 16 {x}^{3} - 16 {x}^{2} + 4\right)$

The volume of the washer is:

${V}_{\text{Washer}} = \pi \left(- 4 {x}^{4} + 16 {x}^{3} - 16 {x}^{2} + 4\right) \mathrm{dx}$

Now, imagine that the thickness of the washer $\mathrm{dx}$ is infinitely small and there are infinite number of washers between $x = 0 \mathmr{and} 2$. If we add the volumes of these washers together we will have the volume of the solid.

To do this, we will take the integral of the area function:

${V}_{\text{Solid}} = {\int}_{0}^{2} \pi \left(- 4 {x}^{4} + 16 {x}^{3} - 16 {x}^{2} + 4\right) \mathrm{dx}$

${V}_{\text{Solid}} = \pi {\left(- \frac{4}{5} {x}^{5} + 4 {x}^{4} - \frac{16}{3} {x}^{3} + 4 x\right)}_{0}^{2}$

${V}_{\text{Solid}} = \pi \left(- \frac{4}{5} \left(32\right) + 4 \left(16\right) - \frac{16}{3} \left(8\right) + 4 \left(2\right)\right)$

${V}_{\text{Solid}} = \pi \left(- \frac{128}{5} + 64 - \frac{128}{3} + 8\right)$

${V}_{\text{Solid}} = \pi \frac{- 384 + 960 - 640 + 120}{15}$

${V}_{\text{Solid}} = \frac{56}{15} \pi$

Therefore, the solid from revolving around the $x$-axis is greater

2

## How do you integrate int 1/sqrt(2x-12sqrtx-5)  using trigonometric substitution?

maganbhai P.
Featured 1 month ago

I=sqrt(2x-12sqrtx-5)+3sqrt2ln|sqrt(2x)-3sqrt2+sqrt(2x-12sqrtx- 5)|+C,
There is no any trig.function in the answer .So it is difficult to proceed with trig substitution.

#### Explanation:

Here,

$I = \int \frac{1}{\sqrt{2 x - 12 \sqrt{x} - 5}} \mathrm{dx}$

Let, $\sqrt{x} = t \implies x = {t}^{2} \implies \mathrm{dx} = 2 t \mathrm{dt}$

$I = \int \frac{2 t}{\sqrt{2 {t}^{2} - 12 t - 5}} \mathrm{dt}$

$= \frac{1}{2} \int \frac{4 t - 12 + 12}{\sqrt{2 {t}^{2} - 12 t - 5}} \mathrm{dt}$

=color(red)(1/2int(4t-12)/sqrt(2t^2-12t-5)dt)+color(blue) (1/2int12/sqrt(2t^2-12t-5)dt

$I = \textcolor{red}{{I}_{1}} + \textcolor{b l u e}{{I}_{2}} \ldots \to \left(A\right)$

Now, ${I}_{1} = \frac{1}{2} \int \frac{4 t - 12}{\sqrt{2 {t}^{2} - 12 t - 5}} \mathrm{dt}$

Take, $\sqrt{2 {t}^{2} - 12 t - 5} = u \implies 2 {t}^{2} - 12 t - 5 = {u}^{2}$

$\implies 4 t - 12 = 2 u \mathrm{du}$

$\therefore {I}_{1} = \frac{1}{2} \int \frac{2 u}{u} \mathrm{du} = \int \mathrm{du} = u + {c}_{1} , \to u = \sqrt{2 {t}^{2} - 12 t - 5}$

$= \sqrt{2 {t}^{2} - 12 t - 5} + {c}_{1} , w h e r e , \sqrt{x} = t$

$\therefore {I}_{1} = \sqrt{2 x - 12 \sqrt{x} - 5} + {c}_{1}$

$A l s o , {I}_{2} = \frac{1}{2} \int \frac{12}{\sqrt{2 {t}^{2} - 12 t - 5}} \mathrm{dt}$

${I}_{2} = \frac{6}{\sqrt{2}} \int \frac{1}{\sqrt{{t}^{2} - 6 t - \frac{5}{2}}} \mathrm{dt}$

$= \frac{6}{\sqrt{2}} \int \frac{1}{\sqrt{{t}^{2} - 6 t + 9 - \frac{23}{2}}} \mathrm{dt} \to \left[- \frac{5}{2} = \frac{18 - 23}{2}\right]$

$= 3 \sqrt{2} \int \frac{1}{\sqrt{{\left(t - 3\right)}^{2} - {\left(\sqrt{\frac{23}{2}}\right)}^{2}}} \mathrm{dt}$

$= 3 \sqrt{2} \ln | t - 3 + \sqrt{{\left(t - 3\right)}^{2} - {\left(\sqrt{\frac{23}{2}}\right)}^{2}} | + c$

$= 3 \sqrt{2} \ln | t - 3 + \textcolor{v i o \le t}{\sqrt{{t}^{2} - 6 t - \frac{5}{2}}} | + c$

=3sqrt2ln|t-3+sqrt(2t^2-12t-5)/color(orange)(sqrt2)|+color(orange)(c

I_2=3sqrt2ln|sqrt2t-3sqrt2+sqrt(2t^2-12t-5)|+color(orange)(c_2

Subst .back $t = \sqrt{x}$

${I}_{2} = 3 \sqrt{2} \ln | \sqrt{2 x} - 3 \sqrt{2} + \sqrt{2 x - 12 \sqrt{x} - 5} | + {c}_{2}$

Hence, from $\left(A\right)$,

I=sqrt(2x-12sqrtx-5)+3sqrt2ln|sqrt(2x)-3sqrt2+sqrt(2x-12sqrtx- 5)|+C,

$w h e r e , C = {c}_{1} + {c}_{2}$