Question

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

Answer 1

Given; $\left\{\right(x,y):0\le y\le x^2+1,0\le y\le x+1,0\le x\le 2\}$

$A:0\le y\le x^2+1$

$B:0\le y\le x+1$

$C:0\le x\le 2$

$0\le y\le x^2+1_{y=x^2+1i.e.x^2=y-1\text{So, it is a parabola}}^{y\ge 0\text{So it is above}x-axis}$

$0\le y\le x+1_{y=x+1\text{It is a straight line}}^{y\ge 0\text{So it is above}x-axis}$

Here, P and Q are intersection of parabola and line

Solving $y=x^2+1\&y=x+1$

$\Rightarrow x^2+1+x+1$

$\Rightarrow x^2-x=0$

$\Rightarrow x(x-1)=0$

$\to x=0,1$

$$\begin{array}{cc}\text{For}x=0& \text{For}x=1\\ y=x+1& y=x+1\\ y=0+1=1& y=1+1=2\\ So,P(0,1)1& So,Q(1,2)\end{array}$$


$=\frac{4}{3}=[2+2-\frac{3}{2}]$

$=4=\frac{4}{3}-\frac{3}{2}$

$=4+\frac{8-9}{6}$

$=44-\frac{1}{6}=\frac{23}{6}$

Final answer:
Therefore, required area $\frac{23}{6}$ square units.