Area Under Curves

The area of the region above the $x$-axis bounded by the curve $y=\tan x$, $0\le x\le\dfrac\pi2$ and the tangent to the curve at $x=\dfrac\pi4$ is :

The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ over the interval $\left[0,\frac\pi2\right]$ is

Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2$, $y=0$, and $x=0$ into two parts $R_1$ $(0\le x\le b)$ and $R_2$ $(b\le x\le1)$ such that $R_1-R_2=\frac14$. Then $b$ equals

Let $f:[-1,2]\to[0,\infty)$ be a continuous function such that $f(x)=f(1-x)$ for all $x\in[-1,2]$. Let $R_1=\displaystyle\int_{-1}^2xf(x)\,dx$, and $R_2$ be the area of the region bounded by $y=f(x)$, $x=-1$, $x=2$, and the $x$-axis. Then

The area bounded by the curve $y=f(x)$ and the lines $x=0$, $y=0$ and $x=t$, lies in the interval

Area of the region bounded by the curve $y=e^x$ and lines $x=0$ and $y=e$ is

The area of the region bounded by the curve $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$, where $-\infty<a<b<-2$, is

The area of the region between the curves $y=\sqrt{\dfrac{1+\sin x}{\cos x}}$ and $y=\sqrt{\dfrac{1-\sin x}{\cos x}}$ bounded by the lines $x=0$ and $x=\frac\pi4$ is