NIMCET 2007 — Mathematics PYQ

To find the area bounded between the two curves, we need to determine their points of intersection and integrate the difference between the upper and lower functions.

Step 1: Find the points of intersection

Equate the two equations to find the values of $x$:

$$x^3=4x^2$$

$$x^3-4x^2=0$$

$$x^2(x-4)=0$$

This gives the limits of integration:

$$x=0\text{and}x=4$$

Step 2: Determine the upper curve

In the interval $[0,4]$, let us pick a value, say $x=1$:

• For $y=4x^2\Rightarrow y=4(1{)}^2=4$

• For $y=x^3\Rightarrow y=(1{)}^3=1$

Since 4 > 1, the curve $y=4x^2$ lies above $y=x^3$ in the region from $x=0$ to $x=4$.

Step 3: Set up and evaluate the definite integral

The bounded area $A$ is given by the formula:

$$A={\int }_a^b(y_{\text{upper}}-y_{\text{lower}})dx$$

Substituting our functions and limits:

$$A={\int }_0^4(4x^2-x^3)dx$$

Now, integrate each term individually:

$$A={\left[\frac{4x^3}{3}-\frac{x^4}{4}\right]}_0^4$$

Substitute the upper limit $x=4$ and lower limit $x=0$:

$$A=\left(\frac{4(4{)}^3}{3}-\frac{(4{)}^4}{4}\right)-(0)$$

$$A=\frac{256}{3}-\frac{256}{4}$$

$$A=\frac{256}{3}-64$$

Take a common denominator to subtract:

$$A=\frac{256-192}{3}$$

$$A=\frac{64}{3}$$

Conclusion

The total area enclosed between the given curves is $\frac{64}{3}$ square units.

Correct Option: (B) $\frac{64}{3}$ sq. units