The area (in sq. units) in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1$,$x=0$,and $x=2$ is:

The area under the curve $y = x^2 - 4x$ within the $x$-axis and the line $x = 2$ is:

Let $A$ denote the area bounded by the curve $y=\frac{1}{x}$ and the lines $y=0, x=1, x=10$. Let $B=1+\frac{1}{2}+\ldots+\frac{1}{9}$ and $C=\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{10}$. Then,

The area between the parabola $y = x^2$ and the line $y = x$ is

Let $A$ be the area bounded by the curve $y = \cos^{-1}\sqrt{1 - x^2}$,the tangent to the curve $y = \sin^{-1}x$ at $x = 0$,and the line $x = 1$. Then the value of $2(\{A\} + \text{sgn}(A))$ is (where $\{.\}$ is the fractional part function and $\text{sgn}(x)$ is the signum function).

The area of the region lying in the first quadrant and bounded by the curve $y = 4x^2$, the $Y$-axis, and the lines $y = 2$ and $y = 4$ is: