Answer
$\int_0^\frac{3}{2}(-4x+6)^{\frac{3}{2}}dx=\frac{18\sqrt 6}{5}$
Work Step by Step
$\int_0^\frac{3}{2}(-4x+6)^{\frac{3}{2}}dx$
Let $u=-4x+6$
$\frac{du}{dx}=-4$
$dx=-\frac{1}{4}du$
$=-\frac{1}{4}\int_0^\frac{3}{2}u^{\frac{3}{2}}dx$
$=(-\frac{1}{4})(\frac{2}{5})(u^{\frac{5}{2}}|_0^\frac{3}{2})$
$=-\frac{1}{10}(-4x+6)^\frac{5}{2}|_0^\frac{3}{2}$
$=-\frac{1}{10}[(-4(\frac{3}{2})+6)^\frac{5}{2}-(-4(0)+6)^\frac{5}{2})]$
$=-\frac{1}{10}[(-6+6)^\frac{5}{2}-(0+6)^\frac{5}{2}]$
$=-\frac{1}{10}(0-36\sqrt 6)$
$=\frac{36\sqrt 6}{10}$
$=\frac{18\sqrt 6}{5}$
