suppose

$f(x) \geqslant 0$

The area bounded by the curves

$y=f(x), x=a, x=b$

and the x-axis is given by

$\int_a^b f(x) d x$

The area between the curves $y=f(x)$,$y=g(x)$ and $x=a$ and $x=b$ is given by

$\int_a^b[f(x)-g(x)] d x \quad$ if $f(x) \geqslant g(x)$

in general,

area

= $\int_a^b|f(x)-g(x)| d x$

Find the area enclosed by the curves

$y=\sin x+\cos x \text { and } y=|\cos x-\sin x|$

over the interval $[0, \frac{\pi}{2}]$.

solution: Let $f(x)=\sin x+\cos x$

$g(x)=|\cos x-\sin x|$

for $x \in[0, \frac{\pi}{2}]$

Here, $g(x)=|\cos x-\sin x| \leq|\cos x|+|\sin x|$

=$\cos x+\sin x \quad$ for $x \in[0, \frac{\pi}{2}]$ = f(x)

$f(x) \geqslant g(x)$ for $x \in[0, \frac{\pi}{2}]$

$\therefore$ Required area $=\int_0^{\pi / 2}[f(x)-g(x)] d x$

$=\int_0^{\pi / 2}[\sin x+\cos x-(\cos x-\sin x)] d x$

Now, $|\cos x-\sin x|= \cos x-\sin x \qquad$ if $0 \leq x \leq \frac{\pi}{4}$

$|\cos x-\sin x|= \sin x-\cos x \qquad$ if $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$

Area $= \int_0^{\pi / 4}[(\sin x+\cos x)-(\cos x-\sin x)] d x$ $+\int_{\pi / 4}^{\pi / 4}[(\sin x+\cos x)-(\sin x-\cos x)] d x$

$= \int_0^{\pi / 4} 2 \sin x d x+ \int_{\pi / 4}^{\pi /2 }2\cos x d x$ $-2 \cos x\mid_0 ^{\pi / 4}+(2 \sin x)\mid_{\pi / 4} ^{\pi / 2}$ $=-2(\frac{1}{\sqrt{2}}-1)+2(1-\frac{1}{\sqrt{2}})$ $=4(1-\frac{1}{\sqrt{2}})=4-2 \sqrt{2}$

Find the area of the region in the first quadrant given by ${(x, y): x y \leq 8,1 \leq y \leq x^2}$.

Solve Upper curve,

$f(x)=x^2, 1 \leq x \leq 2$

$8 / x, 2 \leq x \leq 8$

Upper curve $f(x)$ = $\begin{cases} x^2, & 1 \leq x \leq 2 \\ \frac{8}{x}, & 2 \leq x \leq 8\end{cases}$

Lower curve, $g(x)=1,1 \leq x \leq 8$

Area $=\int_1^8[f(x)-g(x)] d x$

$=\int_1^2(x^2-1) d x+\int_2^8(\frac{8}{x}-1) d x$

$=(\frac{x^3}{3}-x)|_1 ^2+(8 \ln \mid x\mid-x)|_2 ^8$

$=(\frac{8}{3}-2)-(\frac{1}{3}-1)$ $+(8 \ln 8-8)-(8 \ln 2-2)$

$\frac{7}{3}-1+8 \ln 4-6=16 \ln 2-\frac{14}{3}$.

Another way:

Area $=\int_1^4(\frac{8}{y}-\sqrt{y}) d y$

$=8 \ln |y||_1 ^4-\frac{2}{3} y^{3 / 2}|_1 ^4$

$=8 \ln 4-\frac{2}{3}(4^{3 / 2}-1)$

$=16 \ln 2-\frac{2}{3}(8-1)=16 \ln 2-\frac{14}{3}.$

If the line $x=\alpha$ divides the area of the region

$R={(x, y) \in \mathbb{R}^2: x^3 \leq y \leq x, 0 \leq x \leq 1}$ into equal parts, then

(A) $0<\alpha \leq \frac{1}{2}$

(B) $\frac{1}{2}<\alpha<1$

(C) $2 \alpha^4-4 \alpha^2+1=0$

(D) $\alpha^4+4 \alpha^2-1=0$

Area of the region $R =\int_0^1(x-x^3) d x$

$=(\frac{x^2}{2}-\frac{x^4}{4})|_0 ^1$ $=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

$\int_0^\alpha(x-x^3) d x=\frac{1}{2} \times \frac{1}{4}$

$=\frac{1}{8}$

$\Rightarrow \frac{\alpha^2}{2}-\frac{\alpha^4}{4}$

$=\frac{1}{8}$

$\Rightarrow 4 \alpha^2-2 \alpha^4=1$

$\Rightarrow 2 \alpha^4-4 \alpha^2+1=0$

$\int_0^{1 / 2}(x-x^3) dx$

$=\frac{1}{8}-\frac{(\frac{1}{2})^4}{4}$ $=\frac{1}{8}-\frac{1}{64}$ $=\frac{7}{64}<\frac{1}{2}$ $\Rightarrow \frac{1}{2}<\alpha<1$

$\therefore$ Option (B) is correct.

Another way: $2 \alpha^4-4 \alpha^2+1=0$ $\Rightarrow \quad \alpha^2=\frac{4 \pm \sqrt{16-8}}{4}=1 \pm \frac{1}{\sqrt{2}}$

since $\quad \alpha<1 \Rightarrow \alpha^2=1-\frac{1}{\sqrt{2}}$

$\Rightarrow \alpha=\sqrt{1-\frac{1}{\sqrt{2}}}>\frac{1}{2}$

Let $f:[-1,2] \rightarrow(0, \infty)$

be a continuous function

such that : $f(x)=f(1-x)$ for all $x \in[-1,2]$.

Let, $R_1=\int_{-1}^2 x f(x) d x$ and, $R_2$ be the area of the region bounded by

$y=f(x), x=-1, x=2$ and the $x$-axis.

Then,

(A) $R_1=2 R_2 \qquad$ (B) $R_1=3 R_2$

(C) $2 R_1=R_2 \qquad$ (D) $3 R_1=R_2$

$R_1 =\int_{-1}^2 x f(x) d x$

$=\int_{-1}^2 x f(1-x) d x$

Putting, $1-x=y$,

$\int_{-1}^2 x f(1-x) d x$

$=\int_{-1}^2(1-y) f(y) d y$

$=\int_{-1}^2 f(y) d y-\int_{-1}^2 y f(y) d y$

$\therefore \quad R_1=R_2-R_1 \Rightarrow 2 R_1=R_2$

Find the area of the region

$(x, y)\in R^2 :$

$y \geqslant \sqrt{|x+3|},$

$5 y \leq x+9 \leq 15$.

Solution: $\quad 5 y \leq x+9 \leq 15$

(1) $5 y \leq x+9 and x \leq 6$

(2) $y \leq \frac{x+9}{5}$ and $x \leq 6$ :

Let, $f(x)=\sqrt{|x+3|}$

=$\sqrt{x+3}\quad \text {if} x \geqslant-3$

$\sqrt{-(x+3)} \quad \text{ if } x<-3$

$y=\sqrt{|x+3|}$ intersects the line

$x=6$ at $(6,3)$

$y=\sqrt{(x+1)}$ intersects the line

$y=\frac{x+9}{5}$ at

the points $(-4,1)$ and $(1,2)$.

So, the region is as below

Area of the region

= Area of trapezium $A B C D-\int_{-4}^{-3} \sqrt{-(x+3)} d x$

Area of the region

$\rightarrow$ Area of trapezium ABCD

$= \int_{-4}^{-3} \sqrt{-(x+3)} d x$ $-\int_{-3}^1 \sqrt{x+3} d x$

$\rightarrow\int_{-3}^1 \sqrt{x+3} d x$

$=\frac{2}{3}(x+3)^{3 / 2}\mid_{-3} ^1$

$=\frac{2}{3}(8-0)=\frac{16}{3}$

$\int_{-4}^{-3} \sqrt{-x-3} d x$

$=\frac{-2}{3}(-x-3)^{3 / 2}\mid_{-4} ^{-3}$ $=\frac{-2}{3}(0-1)=\frac{2}{3}$

Area of trapezium ABCD

$=\frac{1}{2} \times(A D+B C) \times C D$ $=\frac{1}{2}(1+2) \times 5=\frac{15}{2}$

$\therefore$ Area of the region $=\frac{15}{2}-(\frac{16}{3}+\frac{2}{3})$ $=\frac{15}{2}-6=\frac{3}{2}$

Let, $F(x)=\int_x^{x^2+ \frac{\pi}{6}} 2 \cos ^2 t d t,$ $x \in R$ and

$f:[0, \frac{1}{2}] \rightarrow[0, \infty)$ be a continuous function.

For each $a \in[0, \frac{1}{2}]$, if $F^{\prime}(a)+2$ is the area of the region bounded by $x=0, y=0,$

$y=f(x)$ and $x=a$, then find the value of $f(0).$

Solution: Given,

$F^{\prime}(a)+2=\int_0^a f(x) d x$ $F(x) =\int_x^{x^2+\frac{\pi}{6}} 2 \cos ^2 t d t$

$\Rightarrow F^{\prime}(x) =2 \cos ^2(x^2+\frac{\pi}{6}) \cdot 2 x-2 \cos ^2 x \cdot 1$

$=4 x \cos ^2(x^2+\frac{\pi}{6})-2 \cos ^2 x$

$\therefore \quad F^{\prime}(a)+2$ $=4 a \cos ^2(a^2+\frac{\pi}{6})-2 \cos ^2 a+2$

$=4 a \cos ^2\left(a^2+\frac{\pi}{6}\right)+2 \sin ^2 a$

$\therefore \quad \int_0^a f(x) d x$

$=4 a \cos ^2(a^2+\frac{\pi}{6})+2 \sin ^2 a$

differentiating w.r.t. a,

$f(a)=4 \cos ^2(a^2+\frac{\pi}{6})$

$+4 a \cdot 2 \cos (a^2+\frac{\pi}{6}) \cdot \sin (a^2+\frac{\pi}{6}) \cdot 2 a$ $+4 \sin a \cos a$

Put $a=0$ :

$f(0) =4 \cos ^2 \frac{\pi}{6}+0+0$ $=4(\frac{\sqrt{3}}{2})^2=3$

$\therefore \quad f(0)=3$