Answer
$$A = \frac{{6 - \sqrt 2 }}{{10}}$$
Work Step by Step
$$\eqalign{
& y = \sin 3x\cos 2x \cr
& {\text{The area of the shaded region is given by}} \cr
& A = \int_0^{\pi /4} {\sin 3x\cos 2x} dx \cr
& {\text{Use the sine - cosine product}} \cr
& \sin mx\cos nx = \frac{1}{2}\left( {\sin \left[ {\left( {m - n} \right)x} \right] + \sin \left[ {\left( {m + n} \right)x} \right]} \right) \cr
& \sin 3x\cos 2x = \frac{1}{2}\left( {\sin \left[ {\left( {3 - 2} \right)x} \right] + \sin \left[ {\left( {3 + 2} \right)x} \right]} \right) \cr
& \sin 3x\cos 2x = \frac{1}{2}\left( {\sin x + \sin 5x} \right) \cr
& A = \int_0^{\pi /4} {\frac{1}{2}\left( {\sin x + \sin 5x} \right)} dx \cr
& A = \frac{1}{2}\int_0^{\pi /4} {\left( {\sin x + \sin 5x} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \frac{1}{2}\left[ { - \cos x - \frac{1}{5}\cos 5x} \right]_0^{\pi /4} \cr
& A = \frac{1}{2}\left[ { - \cos \left( {\frac{\pi }{4}} \right) - \frac{1}{5}\cos 5\left( {\frac{\pi }{4}} \right)} \right] - \frac{1}{2}\left[ { - \cos \left( 0 \right) - \frac{1}{5}\cos 5\left( 0 \right)} \right] \cr
& A = \frac{1}{2}\left[ { - \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{{10}}} \right] - \frac{1}{2}\left[ { - 1 - \frac{1}{5}} \right] \cr
& A = \frac{1}{2}\left( { - \frac{2}{5}\frac{{\sqrt 2 }}{2}} \right) - \frac{1}{2}\left( { - \frac{6}{5}} \right) \cr
& A = - \frac{{\sqrt 2 }}{{10}} + \frac{3}{5} \cr
& A = \frac{{6 - \sqrt 2 }}{{10}} \cr} $$
