Question

Area enclosed by$0\leq y\leq x^{2} +1 , 0\leq y \leq x+1,\frac{1}{2} \leq x\leq 2 \: is$

Answer 1

We're asked to find area enclosed by,

• $y\geq0$

• $y\leq x^2+1$

• $y\leq x+1$

• $x\geq\dfrac{1}{2}$

• $x\leq2$

So the area is to be found in the interval $\left[\dfrac{1}{2},\ 2\right]$ since it's enclosed by $x\geq\dfrac{1}{2}$ and $x\leq2.$

Let us evaluate this inequality.

$\longrightarrow x^2+1\leq x+1$

$\longrightarrow x^2\leq x$

$\longrightarrow x^2-x\leq 0$

$\longrightarrow x(x-1)\leq 0$

$\Longrightarrow x\in[0,\ 1]$

This implies, the region whose area is to be found, consists of region enclosed by $0\leq y\leq x^2+1$ in the interval $\left[\dfrac{1}{2},\ 1\right]$ and that enclosed by $0\leq y\leq x+1$ in the interval $[1,\ 2].$

Therefore, area of the total region is,

$\displaystyle\longrightarrow A=\int\limits_{\frac{1}{2}}^1(x^2+1)\ dx+\int\limits_1^2(x+1)\ dx$

$\displaystyle\longrightarrow A=\left[\dfrac{x^3}{3}+x\right]_{\frac{1}{2}}^1+\left[\dfrac{x^2}{2}+x\right]_1^2$

$\displaystyle\longrightarrow A=\left(\dfrac{4}{3}-\dfrac{13}{24}\right)+\left(4-\dfrac{3}{2}\right)$

$\displaystyle\longrightarrow\underline{\underline{A=\dfrac{79}{24}}}$