Integration by parts 2

Find this content useful?
Submitted by roamingfree on Fri, 02/05/2021 - 11:06
skills
You need to be able to apply the by parts formula
Question

Evaluate the integral

$$ \int_1^5 x\sqrt{x-1} dx $$
answer --- press button to toggle display
$\frac{272}{15}$
solution --- press button to toggle display
To integrate this function, we need to apply the by-parts formula once. Let $u = x$ then $u' = 1$, Let $v' = \sqrt{x-1}$ then $v = \frac{2}{3}(x-1)^{\frac{3}{2}}$. Applying the by parts formula ($\int uv' = uv - \int u'v$) gives $$ \begin{array}{rcl} \int x\sqrt{x-1}dx &=& \frac{2}{3}x(x-1)^{\frac{3}{2}} -\int \frac{2}{3}(x-1)^{\frac{3}{2}}dx \\ &=&(x-1)^{\frac{3}{2}}\left(\frac{2}{3}x - \frac{4}{25}(x-1)\right) \end{array} $$ Applying the limits gives $$ \int_1^5 x\sqrt{x-1}dx = (5-1)^\frac{3}{2}\left(\frac{2}{3}5 - \frac{4}{15}(5-1)\right) - (0) = \frac{272}{15} $$