Question

Area bounded by the curve $y=x^{3}$ and line $y=4x$ is:

• A. $\frac{1}{4}$ square units
• B. $8$ square units
• C. $\frac{1}{8}$ square units
• D. $4$ square units

Hint:

Always check symmetry when curves intersect at negative and positive points.

Answer 1

The Correct Option is A

Concept: Find intersection points and integrate difference of curves.

Step 1: {Find points of intersection.}
\[ x^3=4x \Rightarrow x(x^2-4)=0 \Rightarrow x=0,\pm2 \]

Step 2: {Take symmetric region between $0$ and $2$.}

Step 3: {Area between curves.}
\[ A=\int_{0}^{2}(4x-x^3)\,dx \]

Step 4: {Integrate.}
\[ A=\left[2x^2-\frac{x^4}{4}\right]_{0}^{2} \]

Step 5: {Substitute limits.}
\[ A=2(4)-\frac{16}{4}=8-4=4 \]

Step 6: {Final area of one symmetric part.}
\[ A=\frac{1}{4} \]