Let f(x) = x^3 3-x^2 dx. If 5f( 2) = -4, then f(1) is equal to — Solution | JEE Mathematics

MCQMediumJEE 2025Integration Techniques (Substitution, Parts, Partial Fractions)

Let $f(x) = \int x^3 \sqrt{3-x^2} \, dx$ . If $5f(\sqrt{2}) = -4$ , then $f(1)$ is equal to

Answer

Correct answer:D

Standard Method

Given: $f(x) = \int x^3 \sqrt{3-x^2} \, dx$ and $5f(\sqrt{2}) = -4$ .

Find: $f(1)$ .

Use substitution. Let

u = 3 - x^2

Then

du = -2x \, dx

x \, dx = -\frac{1}{2} \, du

and since $x^2 = 3-u$ , we get

f(x) = \int x^3 \sqrt{3-x^2} \, dx = \int x(3-u)\sqrt{u}\left(-\frac{1}{2}\right) \, du

Therefore,

f(x) = -\frac{1}{2}\int (3-u)\sqrt{u} \, du = -\frac{1}{2}\left(\int 3u^{1/2} \, du - \int u^{3/2} \, du\right)

Now,

\int 3u^{1/2} \, du = 2u^{3/2}

and

\int u^{3/2} \, du = \frac{2}{5}u^{5/2}

So,

f(x) = -\frac{1}{2}\left(2u^{3/2} - \frac{2}{5}u^{5/2}\right) + C

that is,

f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2} + C

Using $5f(\sqrt{2}) = -4$ ,

5\left(\frac{(3-2)^{5/2}}{5} - (3-2)^{3/2} + C\right) = -4

5\left(\frac{1}{5} - 1 + C\right) = -4

1 - 5 + 5C = -4

5C = 0

C = 0

Hence,

f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2}

Now evaluate at $x=1$ :

f(1) = \frac{(3-1)^{5/2}}{5} - (3-1)^{3/2}

f(1) = \frac{2^{5/2}}{5} - 2^{3/2}

f(1) = 2^{3/2}\left(\frac{2}{5} - 1\right)

f(1) = 2^{3/2}\left(-\frac{3}{5}\right)

f(1) = -\frac{6\sqrt{2}}{5}

Therefore, the correct option is D.

Alternative Substitution

Given: $f(x) = \int x^3 \sqrt{3-x^2} \, dx$ and $5f(\sqrt{2}) = -4$ .

Find: $f(1)$ .

Let

3-x^2 = t^2

Then

-2x \, dx = 2t \, dt

x \, dx = -t \, dt

Also, $x^2 = 3-t^2$ . Hence,

f(x) = \int x^2\sqrt{3-x^2} \, x \, dx

f(x) = \int (3-t^2) \cdot t \cdot (-t \, dt) + C

f(x) = \int (t^4 - 3t^2) \, dt + C

f(x) = \frac{t^5}{5} - t^3 + C

f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2} + C

Now use the condition:

5\left(\frac{(3-2)^{5/2}}{5} - (3-2)^{3/2} + C\right) = -4

5\left(\frac{1}{5} - 1 + C\right) = -4

-4 + 5C = -4

C = 0

Thus,

f(1) = \frac{(3-1)^{5/2}}{5} - (3-1)^{3/2} = \frac{2^{5/2}}{5} - 2^{3/2} = -\frac{6\sqrt{2}}{5}

Therefore, the correct option is D.

Forgetting the constant of integration. Since $f(x)$ is an indefinite integral, a constant $C$ must be included first. Use the condition $5f(\sqrt{2})=-4$ to determine $C$ before evaluating $f(1)$ .
Using the substitution incorrectly. If $u=3-x^2$ , then $du=-2x\,dx$ , not $-2\,dx$ . One factor of $x\,dx$ must be retained while rewriting $x^3$ as $x\cdot x^2$ .
Making an exponent error while simplifying powers of $2$ . Note that $2^{5/2}=4\sqrt{2}$ and $2^{3/2}=2\sqrt{2}$ . Factor $2^{3/2}$ carefully to avoid choosing the wrong option.

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