NIMCET 2020 — Mathematics PYQ

We need to evaluate the integral:

$$I=\int x^{5}f(x^3)dx$$

Given that:

$$\int f(x)dx=g(x)⟹\frac{d}{dx}[g(x)]=f(x)$$

Step 1: Integration by Substitution

First, let's simplify the inner function argument $x^3$ by substituting it with a new variable $t$.

Let:

$$t=x^3$$

Differentiating both sides with respect to $x$:

$$dt=3x^{2}dx⟹x^{2}dx=\frac{1}{3}dt$$

Now, rewrite the original integral $I$ by splitting $x^5$ into $x^3\cdot x^2$:

$$I=\int x^3\cdot f(x^3)\cdot (x^{2}dx)$$

Substitute $x^3=t$ and $x^{2}dx=\frac{1}{3}dt$:

$$I=\int t\cdot f(t)\cdot \left(\frac{1}{3}dt\right)$$

$$I=\frac{1}{3}\int t\cdot f(t)dt$$

Step 2: Integration by Parts

To integrate $\int t\cdot f(t)dt$, we will use the Integration by Parts formula:

$$\int udv=uv-\int vdu$$

Let's select the parts using the ILATE rule:

• First function, $u=t⟹du=dt$

• Second function, $dv=f(t)dt⟹v=\int f(t)dt=g(t)$

Applying the formula:

$$\int t\cdot f(t)dt=t\cdot g(t)-\int g(t)dt$$

Substitute this back into our expression for $I$:

$$I=\frac{1}{3}\left[t\cdot g(t)-\int g(t)dt\right]$$

$$I=\frac{1}{3}tg(t)-\frac{1}{3}\int g(t)dt+c$$

Step 3: Substitute Back $t=x^3$

Now replace the temporary variable $t$ back with $x^3$:

$$I=\frac{1}{3}{x}^{3}g(x^3)-\frac{1}{3}\int g(x^3)\cdot (3x^{2}dx)+c$$

Simplify the second term by canceling out the constants ($\frac{1}{3}\times 3=1$):

$$I=\frac{1}{3}{x}^{3}g(x^3)-\int x^{2}g(x^3)dx+c$$

Conclusion

The evaluated expression matches option B.

Therefore, the correct option is B) $\frac{1}{3}{x}^{3}g(x^3)-\int x^{2}g(x^3)dx+c$.