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## Definite integral #int_0^(pi/8)cos^2(2x)dx# ?

Ratnaker Mehta
Featured 5 months ago

$\frac{1}{16} \left(\pi + 2\right)$.

#### Explanation:

We have, $1 + \cos 2 \theta = 2 {\cos}^{2} \theta$.

$\therefore {\int}_{0}^{\frac{\pi}{8}} {\cos}^{2} 2 x \mathrm{dx}$,

$= {\int}_{0}^{\frac{\pi}{8}} \frac{1 + \cos 4 x}{2} \mathrm{dx}$,

$= \frac{1}{2} {\left[x + \frac{1}{4} \sin 4 x\right]}_{0}^{\frac{\pi}{8}}$,

$= \frac{1}{2} \left[\frac{\pi}{8} + \frac{1}{4} \sin \left(4 \cdot \frac{\pi}{8}\right) - 0\right]$,

$= \frac{1}{2} \left(\frac{\pi}{8} + \frac{1}{4} \cdot 1\right)$,

$= \frac{1}{16} \left(\pi + 2\right)$.

## Find the coordinates of the points on the graph of #y=3x^2-2x# at which tangent line is parallel to the line #y=10x#?

Parabola
Featured 5 months ago

$\left(2 , 8\right)$

#### Explanation:

Two lines are parallel if they share the same slope.

Any line parallel to $y = 10 x$ will have the slope 10.

Now, we have: $y = 3 {x}^{2} - 2 x$

We try to find $\frac{\mathrm{dy}}{\mathrm{dx}}$

=>$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 2 x\right)$

We use the power rule:

$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ where $n$ is a constant.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 \cdot 2 {x}^{2 - 1} - 2 \cdot 1 {x}^{1 - 1}$

=>$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 x - 2$

Now, we set $\frac{\mathrm{dy}}{\mathrm{dx}} = 10$

=>$10 = 6 x - 2$

=>$12 = 6 x$

=>$2 = x$

We plug this value in to get our $y$, or $f \left(x\right)$.

$3 {\left(2\right)}^{2} - 2 \left(2\right) = y$

=>$12 - 4 = y$

=>$8 = y$

Therefore, the point is at $\left(2 , 8\right)$

## Find the mass of an object which is in the form of a cuboid#[0,1]×[2,4]×[1,3]#.The density at any point #(x,y,z)# on the cuboid is given by #delta(x,y,z)=x^2+y^2+z^2# ?

Cesareo R.
Featured 5 months ago

$56$

#### Explanation:

The mass is obtained by calculating

${\int}_{1}^{3} \left({\int}_{2}^{4} \left({\int}_{0}^{1} \delta \left(x , y , z\right) \mathrm{dx}\right) \mathrm{dy}\right) \mathrm{dz} =$
$= {\int}_{1}^{3} \left({\int}_{2}^{4} \left({\int}_{0}^{1} \left({x}^{2} + {y}^{2} + {z}^{2}\right) \mathrm{dx}\right) \mathrm{dy}\right) \mathrm{dz} =$
$= {\int}_{1}^{3} \left({\int}_{2}^{4} {\left(\frac{1}{3} {x}^{3} + {y}^{2} x + {z}^{2} x\right)}_{x = 0}^{x = 1} \mathrm{dy}\right) \mathrm{dz} =$
$= {\int}_{1}^{3} \left({\int}_{2}^{4} \left(\frac{1}{3} + {y}^{2} + {z}^{2}\right) \mathrm{dy}\right) \mathrm{dz} =$
$= {\int}_{1}^{3} {\left(\frac{1}{3} y + \frac{1}{3} {y}^{3} + {z}^{2} y\right)}_{y = 2}^{y = 4} \mathrm{dz} =$
$= {\int}_{1}^{3} \left(\frac{4 - 2}{3} + \frac{1}{3} \left({4}^{3} - {2}^{3}\right) + \left(4 - 2\right) {z}^{2}\right) \mathrm{dz} = 56$

## How do you find the second derivative of #x^2+y^2=1#?

Shwetank Mauria
Featured 4 months ago

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{1}{y} - {x}^{2} / {y}^{3}$

#### Explanation:

We use implicit differentiation as follows:

differentiating ${x}^{2} + {y}^{2} = 1$, we get

$2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

and differentiating it further

$2 + 2 \left[\frac{\mathrm{dy}}{\mathrm{dx}} \frac{\mathrm{dy}}{\mathrm{dx}} + y \cdot \frac{{d}^{2} y}{{\mathrm{dx}}^{2}}\right\} = 0$

or $2 + 2 {\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}^{2} + 2 y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

or $2 + 2 {x}^{2} / {y}^{2} + 2 y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

or $2 y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 2 - 2 {x}^{2} / {y}^{2}$

or $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{2}{2 y} - 2 {x}^{2} / {y}^{2} \cdot \frac{1}{2 y}$

= $- \frac{1}{y} - {x}^{2} / {y}^{3}$

## 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

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

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

## 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.

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