If the area bounded by $y = x^3 + ax$ (where $a > 0$), the $x$-axis and the lines $x = -2$ and $x = 1$ is $\frac{37}{4}$ square units, then:

$A$ farmer $F_1$ has a land in the shape of a triangle with vertices at $P(0,0)$,$Q(1,1)$,and $R(2,0)$. From this land,a neighbouring farmer $F_2$ takes away the region which lies between the side $PQ$ and a curve of the form $y = x^n$ $(n > 1)$. If the area of the region taken away by the farmer $F_2$ is exactly $30\%$ of the area of $\triangle PQR$,then the value of $n$ is:

The area of the region bounded by the curve $y=x^{2}+1$,the lines $x=1, x=2$ and the $X$-axis is

The slope of the tangent to a curve $y = f(x)$ at $(x, f(x))$ is $2x + 1$. If the curve passes through the point $(1, 2)$,then the area of the region bounded by the curve,the $x$-axis,and the line $x = 1$ is:

The area of the region $\{(x, y):|x-y| \leq y \leq 4 \sqrt{x}\}$ is

Find the area under the curve $y=x^{4}$ bounded by the lines $x=1$,$x=5$ and the $x$-axis.