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1

How do you findint x/sqrt(x^2 + 1) dx  using trigonometric substitution?

Cem Sentin
Featured 1 month ago

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

Explanation:

$\int \frac{x}{\sqrt{{x}^{2} + 1}} \cdot \mathrm{dx}$

After using $x = \tan y$ and $\mathrm{dx} = {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$ transforms, this integral became

$\int \tan \frac{y}{\sqrt{{\left(\tan y\right)}^{2} + 1}} \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$

=$\int \tan \frac{y}{\sqrt{{\left(\sec y\right)}^{2}}} \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$

=$\int \frac{{\left(\sec y\right)}^{2} \cdot \tan y}{\sec} y \cdot \mathrm{dy}$

=$\int \sec y \cdot \tan y \cdot \mathrm{dy}$

=$\sec y + C$

=$\sqrt{{\left(\tan y\right)}^{2} + 1} + C$

=$\sqrt{{x}^{2} + 1} + C$

1

How to determine the equation of the tangent line to y^3+xy=22 at (7,2)?

Shwetank Mauria
Featured 1 month ago

Equation of tangent at $\left(7 , 2\right)$ is $2 x + 19 y = 137$

Explanation:

Observe that $\left(7 , 2\right)$ lies on the curve ${y}^{3} + x y = 22$, as putting these values in the equation ${y}^{3} + x y = 22$ the equality is established as ${2}^{3} + 7 \times 2 = 8 + 14 = 22$.

Now, as slope of tangent at any point is given by the value of first derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ at that point. Hence, we need to first find $\frac{\mathrm{dy}}{\mathrm{dx}}$ using implicit differentiation.

As ${y}^{3} + x y = 22$

$3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + y + x \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

and putting $x = 7$ and $y = 2$,

$3 \cdot {2}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + 2 + 7 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

or $\frac{\mathrm{dy}}{\mathrm{dx}} \left(12 + 7\right) = - 2$

i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{19}$

As slope of tangent is $- \frac{2}{19}$ and it passes through $\left(7 , 2\right)$, its equation using point-slope form is

$y - 7 = - \frac{2}{19} \left(x - 2\right)$ i.e. $2 x + 19 y = 137$

1

Evaluate the definite integral int_2^4(x-3)^3dx ?

Shwetank Mauria
Featured 1 month ago

Integral is $0$.

Explanation:

Let $x - 3 = t$ then $\mathrm{dx} = \mathrm{dt}$ and

${\int}_{2}^{4} {\left(x - 3\right)}^{3} \mathrm{dx} = {\int}_{-} {1}^{1} {t}^{3} \mathrm{dt}$

= ${\left[{t}^{4} / 4\right]}_{-} {1}^{1}$

= $\left[\frac{1}{4} - \frac{1}{4}\right] = 0$

Note that when $x = 2$, $t = - 1$ and when $x = 4$, $t = 1$.

Now observe the graph of ${\left(x - 3\right)}^{3}$, as given below

graph{(x-3)^3 [0.542, 5.542, -1.2, 1.3]}

Note that area of curve in the interval $\left[2 , 3\right]$ is symmetric w.r.t. area of curve in the interval $\left[3 , 4\right]$, but they are in opposite direction, Hence the integral is $0$.

3

Can some one help on how to do this please? "shell method about y axis

Somebody N.
Featured 1 month ago

Volume $= \textcolor{b l u e}{\frac{31 \pi}{5} \text{cubic units.}}$

Explanation:

I am assuming that you already know about area under a curve.

In this problem we have the area under $y = {x}^{2}$ rotated about $y = - 9$. The area under this curve is made up of rectangles whose height is $y = {x}^{2}$ and whose width is $\delta x$. When we find the volume of this rotated about a horizontal line, we say that the height of these rectangles above the line of rotation is the radius of a disc or shell, that has a thickness of $\delta x$. The volume of each of these disc is then:

$\pi {r}^{2} \cdot \delta x$

Or $\textcolor{w h i t e}{88} \pi {\left(f \left(x\right)\right)}^{2} \cdot \delta x$

And the summation of these gives us the required volume.

In this problem we are rotating about the line $y = - 9$ instead of $y = 0$. This means that we have to find the required radius.

If you look at the diagram, you can see that we are 9 units below the line $y = 0$. If we add this 9 units to the function $y = {x}^{2}$ we would have the radius ${x}^{2} + 9$ marked $\boldsymbol{a}$. Our area is between the curve and the line $y = 0$, so if we then subtract the area between $y = - 9$ and $y = 0$ which has a radius $- 9$ marked $\boldsymbol{b}$

Our required integral will be:

$\pi {\int}_{0}^{1} \left({\left({x}^{2} + 9\right)}^{2} - {\left(- 9\right)}^{2}\right) \mathrm{dx}$

piint_(0)^(1)((x^4+18x^2)dx

Volume=V

V$= \pi \left\{{\left[\frac{1}{5} {x}^{5} + 6 {x}^{3}\right]}_{0}^{1}\right\}$

V$= \pi \left\{{\left[\frac{1}{5} {x}^{5} + 6 {x}^{3}\right]}^{1} - {\left[\frac{1}{5} {x}^{5} + 6 {x}^{3}\right]}_{0}\right\}$

Plugging in upper and lower bounds:

V$= \pi \left\{{\left[\frac{1}{5} {\left(1\right)}^{5} + 6 {\left(1\right)}^{3}\right]}^{1} - {\left[\frac{1}{5} {\left(0\right)}^{5} + 6 {\left(0\right)}^{3}\right]}_{0}\right\}$

V$= \pi \left\{{\left[\frac{31}{5}\right]}^{1} - {\left[0\right]}_{0}\right\} = \textcolor{b l u e}{\frac{31 \pi}{5} \text{cubic units.}}$

Volume of revolution:

I hope you can follow this. It is really quite simple to do, but really difficult to explain it in a simple way.

2

Evaluate the Limit?

Noah G
Featured 4 weeks ago

The limit is $2$

Explanation:

We can rewrite as

$L = {\lim}_{x \to \infty} \tan \frac{\frac{2}{x}}{\frac{1}{x}}$

We see that $\tan \left(\frac{2}{x}\right)$ converges to $0$ as $x \to \infty$ because the larger the value of $x$ the smaller the expression within the tangent becomes and the closer $\tan \left(\frac{2}{x}\right)$ becomes to $0$. Also, the limit ${\lim}_{x \to \infty} \frac{1}{x} = 0$ is commonly used. Therefore, we may use l;Hosptial's rule.

$L = {\lim}_{x \to \infty} \frac{- \frac{2}{x} ^ 2 \cdot {\sec}^{2} \left(\frac{2}{x}\right)}{- \frac{1}{x} ^ 2}$

$L = {\lim}_{x \to \infty} 2 {\sec}^{2} \left(\frac{2}{x}\right)$

The same principal applies with $\frac{2}{x}$ as $\frac{1}{x}$: the limit as $x$ approaches infinity always remains $0$.

$L = 2 {\sec}^{2} \left(0\right)$

$L = 2 \left(1\right)$

$L = 2$

A graphical verification confirms.

In the above graph, the red curve is $x \tan \left(\frac{2}{x}\right)$ and the blue line is $y = 2$. As you can see, the curve converges onto the line.

Hopefully this helps!

3

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

Cesareo R.
Featured 4 weeks 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

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

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

2

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

Jim H
Featured 4 weeks 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 $6.266$

2

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

Somebody N.
Featured 4 weeks 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 2 weeks 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.

3

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

Hammer
Featured 3 weeks 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.