Question

The area (in sq units) enclosed between the curves $y=x$ and $y=2x-x^2$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B

$\frac{1}{6}$

Explanation for correct answer option(s)

Step 1: Find the intersection points

Consider the given Equation of the curves as,

$y=x$

$y=2x-x^2$

Now, finding the point of intersection

$$\begin{gathered} \Rightarrow y=2x-x^2 \\ \Rightarrow x=2x-x^2\left[puty=x\right] \\ \Rightarrow x^2-x=0 \\ \Rightarrow x\left(x-1\right)=0 \\ \Rightarrow x=0,x=1 \end{gathered}$$

The graph is shown below for given curve

Step 2: Find the area (in sq units) enclosed between the curves.

Let,

$$\begin{gathered} y_1=x \\ {y}_2=2x-x^2 \end{gathered}$$

Now the required area is Equal to

$A={\int }_0^1\left(y_2-y_1\right)dx$

Then,

$$\begin{gathered} A={\int }_0^1\left(\left(2x-x^2\right)-x\right)dx \\ A={\int }_0^1\left(x-x^2\right)dx \\ A={\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1 \\ A=\left[\frac{1}{2}-\frac{1}{3}-\left(0-0\right)\right] \\ A=\left(\frac{1}{2}-\frac{1}{3}\right) \\ A=\frac{1}{6}\text{sq units} \end{gathered}$$

Hence, the area enclosed between the given curves is equal to $\frac{1}{6}\text{sq units}$

Therefore, the correct answer is Option B.