Question

Find the area of the region $\{(x,y):0\le y\le x^2+1,0\le y\le x+1,0\le x\le 2\}$.

Answer 1

Consider the following expression.

$\left(i\right)0\le y\le x^2+1$

$\left(ii\right)0\le y\le x+1$

$\left(iii\right)0\le x\le 2$

Therefore, required region will be as shown in the figure.

We can see that, we have

$y=x^2+1$ and $y=x+1$

Therefore,

$x^2+1=x+1$

$x^2-x=0$

$x(x-1)=0$

$x=0,x=1$

Therefore, points of intersection will be,

$P(0,1)$ and $Q(1,2)$

Thus,

Area required $=$ Area $OPQRST$ $=$ Area $OPQT$ $+$ Area $QRST$

So, area $OPQT$ will be,

$\Rightarrow \underset{0}{\overset{1}{\int }}(x^2+1)dx$

$\Rightarrow {[\frac{x^3}{3}+x]}_0^1$

$\Rightarrow \frac{4}{3}$

Also, area $QRST$ will be,

$\Rightarrow \underset{1}{\overset{2}{\int }}(x+1)dx$

$\Rightarrow {[\frac{x^2}{2}+x]}_1^2$

$\Rightarrow \frac{5}{2}$

Therefore,

Area required $=\frac{4}{3}+\frac{5}{2}=\frac{23}{6}sq.units$

Hence, this is the required result.