Let f(x) = ∫ x³√(3-x²) dx. If 5f(√2) = -4, then f(1) is equal to

Q: Why did we choose $3-x^2 = t^2$ instead of just $3-x^2 = t$?

Excellent question! If you set $3-x^2 = t$, then $dt = -2x dx$, which leaves you with a $\sqrt{t}$ term that is harder to manage. By setting it to $t^2$, the square root $\sqrt{t^2}$ simplifies perfectly to $t$, eliminating the radical entirely. It is about choosing the path of least resistance to make the algebra work for you, not against you.

Q: Why can't we just use definite integration limits?

You absolutely could! However, the problem gives us an indefinite integral structure $f(x)$ and a specific condition $5f(\sqrt{2}) = -4$. This implies we need the general form of the function, including the constant of integration $C$. Finding $C$ is a vital part of the problem, so treat it as an indefinite integral first to capture the full family of curves.

Q: Is there a trigonometric substitution method for this?

Yes, you could use $x = \sqrt{3} \sin \theta$. It is a valid path, but it often leads to complex trigonometric identities like $\int \sin^3 \theta \cos^2 \theta d\theta$. While elegant, the algebraic substitution $3-x^2 = t^2$ is generally faster and less prone to calculation errors in a high-pressure exam environment. Always choose the method that minimizes the number of steps.

Question

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

Options:

(A) $-\frac{2\sqrt{2}}{5}$

(B) $-\frac{8\sqrt{2}}{5}$

(C) $-\frac{4\sqrt{2}}{5}$

(D) $-\frac{6\sqrt{2}}{5}$

Sigma Percentile · Accepted Answer

### The Art of the Substitution: Unlocking the Integral

Welcome, future engineer. Today, we are going to dissect a problem that, at first glance, might seem like a tangled mess of radicals and powers. We are looking at the function $f(x) = \int x^3 \sqrt{3-x^2} dx$.

It is easy to feel intimidated by the $x^3$ sitting outside the square root, but in the world of JEE Advanced, intimidation is just a sign that you are about to learn a beautiful technique. Let us peel back the layers of this problem together.

### Phase 1: The Strategic Vision

When you see an integral involving $\sqrt{a^2 - x^2}$, your brain should immediately flag it for substitution. The radical is the 'bottleneck' of the expression; if we can simplify that, the rest of the integral often falls into place.

We choose the substitution $3-x^2 = t^2$. We use $t^2$ instead of $t$ because we want to eliminate the square root entirely, resulting in $\sqrt{3-x^2} = t$. It is a clean, surgical removal of the radical.

### Phase 2: The Algebraic Transformation

Now, we must account for the $x^3$ and the $dx$. We split the $x^3$ into $x^2 \cdot x$.

Because our substitution $3-x^2 = t^2$ implies $x^2 = 3-t^2$, and differentiating $3-x^2 = t^2$ gives $-2x dx = 2t dt$, we find that $x dx = -t dt$.

Look at the elegance of the transformation:
1. The radical $\sqrt{3-x^2}$ becomes $t$.
2. The $x^2$ becomes $(3-t^2)$.
3. The $x dx$ becomes $-t dt$.

Our integral, which looked like a nightmare, now becomes:
$$f(x) = \int (3-t^2) \cdot t \cdot (-t dt)$$

### Phase 3: The Simplification

Let us distribute that $-t$ into the parenthesis. We get $\int (3-t^2)(-t^2) dt$, which simplifies to $\int (-3t^2 + t^4) dt$.

This is the moment of truth. We have moved from a complex radical integral to a simple polynomial integral. Applying the power rule, $\int t^n dt = \frac{t^{n+1}}{n+1}$, we get:

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

Don't forget the constant $C$! In indefinite integration, $C$ is the signature of the family of functions we are working with.

### Phase 4: Returning to Reality

We cannot leave our answer in terms of $t$. We must return to the world of $x$. Since $t = \sqrt{3-x^2}$, we substitute this back:

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

Now, we use the boundary condition provided: $5f(\sqrt{2}) = -4$. When $x = \sqrt{2}$, $x^2 = 2$, so $3-x^2 = 1$.

Substituting $x = \sqrt{2}$ into our expression, we get:

$$5 \left[ \frac{1^{5/2}}{5} - 1^{3/2} + C \right] = -4$$

$$5 \left[ \frac{1}{5} - 1 + C \right] = -4 \Rightarrow 1 - 5 + 5C = -4$$

$$-4 + 5C = -4 \Rightarrow 5C = 0 \Rightarrow C = 0$$

### The Final Victory

With $C=0$, our function is fully defined: $f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2}$. To find $f(1)$, we simply plug in $x=1$:

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

Since $2^{5/2} = 4\sqrt{2}$ and $2^{3/2} = 2\sqrt{2}$, we have:

$$f(1) = \frac{4\sqrt{2}}{5} - 2\sqrt{2} = \frac{4\sqrt{2} - 10\sqrt{2}}{5} = \mathbf{-\frac{6\sqrt{2}}{5}}$$

And there you have it. You have navigated the substitution, handled the algebra, solved for the constant, and arrived at the final answer. This is not just math; this is the art of problem-solving.

Animated Solution for Mathematics - Indefinite Integration: Let $f (x) = \int x^{3} 3 - x^{2} d x$ . If $5 f (2) = - 4$ , then $f (1)$ is equal to

Select Answer:

Visualized Solution

Analyze the Integral Structure

Choosing the Substitution

Differentiating the Substitution

Rewriting the Integral

Simplifying the Integrand

Executing the Integration

Back-substitution to x

Applying the Initial Condition

Solving for Constant C

Calculating f(1)

Final Conclusion

The Sigma Insight: Integration by Substitution

The Art of the Substitution

Phase 1

Phase 2

Phase 3

Phase 4

The Final Victory

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Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∫x33−x2​dx. If 5f(2​)=−4, then f(1) is equal to