Question

Evaluate the definite integral \(\int\limits_{\rm{1}}^{\rm{2}} {{\rm{x}}\sqrt {{\rm{x - 1}}} } {\rm{ dx }}\).

Short answer

The value of the definite integral \(\int\limits_1^2 {x\sqrt {x - 1} } dx\)is \(\frac{{16}}{{15}}\).

Step-by-step solution

Given Information.

The given definite integral is \(\int\limits_1^2 {x\sqrt {x - 1} } dx\).

Make assumptions.

Let \(x - 1 = u\) then

\(\begin{aligned}{l}\frac{{du}}{{dx}} = \frac{d}{{dx}}\left( {x - 1} \right)\\ \Rightarrow \frac{{du}}{{dx}} = 1\\ \Rightarrow du = dx\end{aligned}\)

and if \(x - 1 = u\)then \(x = u + 1\)

Change the limits.

If \(x = 1\) then \(u = 0\) and if \(x = 2\) then \(u = 1\)

So, the limits will be \(0,1\).

Solve the given definite integral.

\(\int\limits_1^2 {x\sqrt {x - 1} } dx = \int\limits_0^1 {(u + 1)\sqrt u du} \)

\(\begin{aligned}{l} = \int\limits_1^2 {(u\sqrt u } + \sqrt u )dx\\ = \int\limits_1^2 {\left( {{u^{3/2}} + {u^{1/2}}} \right)} du\\ = \int\limits_1^2 {\left( {{u^{3/2}}} \right)} du + \int\limits_1^2 {\left( {{u^{1/2}}} \right)} du\\ = \left( {\frac{{{u^{5/2}}}}{{5/2}} + \frac{{{u^{3/2}}}}{{3/2}}} \right)_1^2\\ = \frac{2}{5} + \frac{2}{3}\\ = \frac{{6 + 10}}{{15}}\\ = \frac{{16}}{{15}}\end{aligned}\)

Therefore, \(\int\limits_1^2 {x\sqrt {x - 1} } dx = \frac{{16}}{{15}}\).