The area of the region bounded by the line $y=2x+1$,the $X$-axis,and the ordinates $x=-1$ and $x=1$ is

The area of the region $\{(x, y): x^{2}+y^{2} \leq 1 \leq x+y\}$ is

Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$,and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then $b$ equals

Let $f(x) = x^3 - 3x^2 + 3x + 1$ and $g$ be the inverse of $f$. The area bounded by the curve $y = g(x)$ with the $x$-axis between $x = 1$ and $x = 2$ is (in square units):

The area bounded by the curve $y = x^2 + 2$, the $x$-axis, and the lines $x = 1$ and $x = 2$ is:

Area enclosed by the curve $y = f(x)$ that is defined parametrically as $x = \frac{1 - t^2}{1 + t^2}, y = \frac{2t}{1 + t^2}$ (where $t \in R$) is equal to