Question

If $g\left(x\right)=\sin x\underset{0}{\overset{x}{\int }}\cos tdt+{\cos }^{2}x-\frac{x^2}{2}+\underset{0}{\overset{x}{\int }}tdt+4,$ then the area bounded by the curve $y=3g\left(x\right)(x^2-3x+2)$ and the x-axis is

Answer 1

$g\left(x\right)=\sin x\underset{0}{\overset{x}{\int }}\cos tdt+{\cos }^{2}x-\frac{x^2}{2}+\underset{0}{\overset{x}{\int }}tdt+4$
$={\sin }^{2}x+{\cos }^{2}x-\frac{x^2}{2}+\frac{x^2}{2}+4=5$
$y=15(x^2-3x+2)$
$=15(x-1)(x-2)$


Required area $=\left|15\underset{1}{\overset{2}{\int }}\right(x^2-3x+2\left)dx\right|$
$=\left|15{[\frac{x^3}{3}-\frac{3x^2}{2}+2x]}_1^2\right|$
$=\frac{5}{2}=2.50$