Question

Evaluate the integral

\(\int\limits_0^1 {{x^3}} \sqrt {1 - {x^2}} .dx\)

Short answer

The integration \(\int\limits_0^1 {{x^3}\sqrt {1 - {x^2}} } .dx\)after solving will be equal to\(\frac{3}{{10.}}\)

Step-by-step solution

Step 1:- (substituting 1- x2 as u and solving it)

Given:- I= \(\int\limits_0^1 {{x^3}\sqrt {1 - {x^2}} } .dx\)

Let 1-x²=u

Diff.w.r.t.x

\(\frac{d}{{dx}}\left( {1 - {x^2}} \right) = \frac{d}{{dx}}\left( u \right)\)

\(\begin{aligned}{l} - 2x &= \frac{{du}}{{dx}}\\xdx &= \frac{{ - 1}}{2}du\end{aligned}\)

:- for x=0 , u=1 and x=1,

So ,I =\(\int\limits_1^0 {\left( {1 - u} \right)} \sqrt {u.du} \)

I =\(\frac{{ - 1}}{2}\left( {\int\limits_1^0 {\sqrt {u.du} - \int\limits_1^0 {u\sqrt {u.du} } } } \right)\)

I=\(\frac{{ - 1}}{2}\left( {\frac{2}{3}|{u^{3/2}}|_1^0 - \int\limits_1^0 {{u^{3/2}}} .du} \right)\)

Step 2:( continuing g the   soln  )1

I=\(\frac{{ - 1}}{2}\left( {\frac{2}{3}|{u^{3/2}}|_1^0 - \int\limits_1^0 {{u^{3/2}}} .du} \right)\)

I=\(\frac{{ - 1}}{2}\left( {\frac{{ - 2}}{3}x1 - \frac{2}{5}|0 - 1|} \right)\)

I=\(\frac{{ - 1}}{2}\left( {\frac{{ - 2}}{3} + \frac{2}{5}} \right)\)

I=\(\frac{{ - 1}}{2}\left( {\frac{{ - 4}}{{15}}} \right)\)

I=\(\frac{2}{{15}}\)

HENCE AFTER CALCULATING THE INTEGRAL

\(\int\limits_0^1 {{x^3}\sqrt {1 - {x^2}} } .dx\) is equal to \(\frac{2}{{15}}\)