Answer
$\dfrac{92}{3}$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{1}^{4}[\dfrac{1}{4} x^2+x \sqrt y]_0^4dy=\int_{1}^{4}[4+4y^{1/2}] dy$
or, $ \int_{1}^{4}[\dfrac{16}{4}+4 \sqrt y]dy =\int_{1}^{4}[4+4y^{1/2}] dy$
This implies that
$[4y+\dfrac{8}{3} y^{3/2}]_{1}^{4}=\dfrac{92}{3}$
