Question

How do you integrate \left\\{ {\dfrac{x}{{\sqrt {4 + 4{x^2}} }}} \right\\}dx from 0 to 2?

Answer 1

Here in this question given a definite integral, we have to find the integrated value of given function. This can be solved by substitution method means substitute denominator as t and later integrated by using the general power formula of integration. And further simplify by substituting the limit points we get the required solution.

Complete step by step solution:
Definite integral as integral expressed as the difference between the values of the integral at specified upper and lower limits of the independent variable.
Consider the given integral
\left\\{ {\dfrac{x}{{\sqrt {4 + 4{x^2}} }}} \right\\}dx--------(1)
The given limit point is [0,2].
Integrate equation (1) with respect to $x$ with limit point [0,2], then
$\Rightarrow \,\,\,\,\int_0^2 {\dfrac{x}{{\sqrt {4 + 4{x^2}} }}} \,dx$------(2)
This can be solve by using the general power formula of integration
i.e., $\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$, $\left( {n \ne - 1} \right)$
Put the denominator of equation 2 as t
i.e., $4 + 4{x^2} = t$
differentiate with respect to x, then
$0 + 4.2x\,dx = dt$
$8x\,dx = dt$
Divide both side by 8
$x\,dx = \dfrac{{dt}}{8}$
Substitute this in equation (2), then
$\Rightarrow \,\,\,\,\int_0^2 {\dfrac{{\dfrac{{dt}}{8}}}{{\sqrt t }}} \,$
$\Rightarrow \,\,\,\,\dfrac{1}{8}\int_0^2 {\dfrac{{dt}}{{\sqrt t }}} \,$
$\sqrt t$ can be written as ${t^{\dfrac{1}{2}}}$, then
$\Rightarrow \,\,\,\,\dfrac{1}{8}\int_0^2 {\dfrac{{dt}}{{{t^{\dfrac{1}{2}}}}}} \,$
Take denominator to the numerator
$\Rightarrow \,\,\,\,\dfrac{1}{8}\int_0^2 {{t^{ - \dfrac{1}{2}}}} \,dt$
Using the general power formula of integration then
$\Rightarrow \,\,\,\,\dfrac{1}{8}\left[ {\dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}}} \right]_0^2$
On simplification, we get
$\Rightarrow \,\,\,\,\dfrac{1}{8}\left[ {\dfrac{{{t^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right]_0^2$
$\Rightarrow \,\,\,\,\dfrac{1}{8}\left[ {2{t^{\dfrac{1}{2}}}} \right]_0^2$
Or it can be written as
$\Rightarrow \,\,\,\,\dfrac{1}{8}\left[ {2\sqrt t } \right]_0^2$
Where $t = 4 + 4{x^2}$, then
$\Rightarrow \,\,\,\,\dfrac{1}{8}\left[ {2\sqrt {4 + 4{x^2}} } \right]_0^2$
Take 4 as common inside the square root
$\Rightarrow \,\,\,\,\dfrac{1}{8}\left[ {2\sqrt {4\left( {1 + {x^2}} \right)} } \right]_0^2$
Take 4 outside the root
$\Rightarrow \,\,\,\,\dfrac{1}{8}\left[ {2 \cdot 2\sqrt {\left( {1 + {x^2}} \right)} } \right]_0^2$
$\Rightarrow \,\,\,\,\dfrac{1}{8}\left[ {4\sqrt {\left( {1 + {x^2}} \right)} } \right]_0^2$
On simplification, we get
$\Rightarrow \,\,\,\,\dfrac{1}{2}\left[ {\sqrt {\left( {1 + {x^2}} \right)} } \right]_0^2$
Now, apply the upper and limit to the independent variable $x$.
$\Rightarrow \,\,\,\,\dfrac{1}{2}\left[ {\sqrt {\left( {1 + {2^2}} \right)} - \sqrt {\left( {1 + {0^2}} \right)} } \right]$
$\Rightarrow \,\,\,\,\dfrac{1}{2}\left[ {\sqrt {\left( {1 + 4} \right)} - \sqrt {\left( {1 + 0} \right)} } \right]$
$\Rightarrow \,\,\,\,\dfrac{1}{2}\left[ {\sqrt 5 - \sqrt 1 } \right]$
$\Rightarrow \,\,\,\,\dfrac{1}{2}\left[ {\sqrt 5 - 1} \right]$

Hence the integrated value of $\int_0^2 {\dfrac{x}{{\sqrt {4 + 4{x^2}} }}} \,dx$ is $\dfrac{1}{2}\left[ {\sqrt 5 - 1} \right]$.

Note: In integration we have two different kinds of integration one is definite integral and the indefinite integral. In a definite integral we have a limits point, by substituting the limits points we simplify the given function. whereas in indefinite integral the limit points are not mentioned.