Question

The area bounded by the curves $y=x$ and $y=x^3$ is equal to

• A. $0$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is C

$\frac{1}{2}$

Explanation for the correct option:

Step 1: Intersection points and diagram:

Let us first draw the diagram showing both curves and line with their intersection points found as

put $y=x$ in $y=x^3$we get

$$\begin{gathered} x=x^3 \\ \Rightarrow x(1-x^2)=0 \end{gathered}$$

so, $x=0,+1,-1$

$\Rightarrow y=0,+1,-1$ as shown in graph

Step 2: Bounded area:

From the symmetrical nature of graph, we will calculate are in first quadrant and then double the amount of this area will be actual area of shaded region so, required area A will be calculated as

$$\begin{gathered} A={\int }_0^1(x-x^3)dx \\ A={{\left[\left(\frac{x^2}{2}\right)-\left(\frac{x^4}{4}\right)\right]}^1}_0\left\{{\mathbf{\int }}_{}^{}{\mathit{x}}^{\mathbf{n}}\mathit{d}\mathit{x}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{n}\mathbf{+}\mathbf{1}}}{\mathbf{n}\mathbf{+}\mathbf{1}}\right\} \\ A=\frac{1}{2}-\frac{1}{4} \\ A=\frac{1}{4} \end{gathered}$$

and we see that, actual area will be twice of this area so,

Required area $$\begin{gathered} A\text{'}=2A \\ A\text{'}=\frac{1}{2}uni{t}^2 \end{gathered}$$

Hence, the correct answer is option (C)