Question

The area (in sq units) enclosed between the parabola $y=x^2-x+2$ and the line $y=x+2$ equals

• A. $\frac{8}{3}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{4}{3}$

Answer 1

The correct option is D

$\frac{4}{3}$

Explanation for correct option:

Finding the area enclosed between the parabola and the line:

Given,

Equation of parabola $y=x^2-x+2......\left(1\right)$

Equation of line $y=x+2......\left(2\right)$

Substitute the value of Equation $\left(2\right)$ in the Equation $\left(1\right)$ we get,

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

Points of intersection are at $x=0$ and $x=2$

The graph is shown below

Now required area is equal to

$\begin{array}{rcl}A& =& {\int }_0^2\left(\left(x+2\right)-\left(x^2-x+2\right)\right)dx\\ & =& {\int }_0^2\left(x+2-x^2+x-2\right)dx\\ & =& {\int }_0^2\left(2x-x^2\right)dx\\ & =& {\left[2\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^2\\ & =& {\left[x^2-\frac{x^3}{3}\right]}_0^2\\ & =& \left[\left(4-\frac{8}{3}\right)-\left(0-0\right)\right]\\ & =& 4-\frac{8}{3}\\ A& =& \frac{4}{3}\text{sq units}\end{array}$

Hence, the value of required area is equal to $\frac{4}{3}\text{sq units}$

Therefore, option (D) is the correct answer.