Question

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

Answer 1

Hint: 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.