Question

The area (in sq. units) of the region $\left\{(x,y):0\le y\le x^2+1,0\le y\le x+1,\frac{1}{2}\le x\le 2\right\}$ is

• A. $\frac{79}{16}$
• B. $\frac{23}{6}$
• C. $\frac{79}{24}$
• D. $\frac{23}{16}$

Answer 1

The correct option is C

$\frac{79}{24}$

Explanation for the correct option:

Finding area within the curves :

Given,

$$\begin{gathered} y=x^2+1 \\ y=x+1 \end{gathered}$$

Plotting the curves,

Solving both the above equations,

$$\begin{gathered} \Rightarrow x+1=x^2+1 \\ \Rightarrow x^2-x=0 \\ \Rightarrow x(x-1)=0 \\ \Rightarrow x=0,1 \end{gathered}$$

Substituting the value of $x$ in $y=x+1$ to get point of intersection.

Now the points of intersection are $(0,1),(1,2)$.

Hence, the required area is,

$\Rightarrow A={\int }_{\frac{1}{2}}^2{x}^2+1dx+{\int }_1^2(x+1)dx$

$$\begin{gathered} ={[\frac{x^3}{3}+x]}_{\frac{1}{2}}^1+{[\frac{x^2}{2}+x]}_1^2 \\ =\frac{1}{3}+1-\frac{1}{24}-\frac{1}{2}+2+2-\frac{1}{2}-1 \\ =\frac{4}{3}-\frac{1}{24}-\frac{1}{2}+3-\frac{1}{2} \\ =\frac{13}{3}-\frac{1}{24}-1 \\ =\frac{79}{24}\text{sq.units} \end{gathered}$$

Hence, option (C) is the correct answer.