Let f be a continuous function satisfying integral from 0 to t^2 of (f(x) + x^2) dx = (4/3)t^3, ∀t

Q: Wait, why do we differentiate the integral? Doesn't that make it harder?

That is a great question! Think of the integral as a 'container' holding the function $f(x)$. As long as $f(x)$ is inside the integral, we cannot access its value at a specific point. By differentiating, we use the Fundamental Theorem of Calculus to 'open' the container, allowing us to isolate $f(t^2)$ and solve for it algebraically. It is the most direct path to the solution!

Q: Why is the derivative of the upper limit $2t$?

This comes from the Chain Rule. The Newton-Leibniz rule states that $\frac{d}{dt} \int_0^{h(t)} F(x) \, dx = F(h(t)) \cdot h'(t)$. Here, our upper limit is $h(t) = t^2$. The derivative of $t^2$ with respect to $t$ is $2t$. If you forget this $2t$, the entire equation will be off by a factor, so always be careful with the upper limit!

Q: Why do we choose $t = \frac{\pi}{2}$ and not $-\frac{\pi}{2}$?

Excellent catch! When we solve $t^2 = \frac{\pi^2}{4}$, we get $t = \pm \frac{\pi}{2}$. However, the problem explicitly states that $t > 0$. In JEE problems, domain constraints are often the 'trap' designed to test your attention to detail. Always check the initial conditions before proceeding with your final substitution.

Question

Let $f$ be a continuous function satisfying $\int_0^{t^2} (f(x) + x^2) \, dx = \frac{4}{3} t^3, \forall t > 0$. Then $f\left(\frac{\pi^2}{4}\right)$ is equal to

Options:

(A) $\pi^2(1 - \frac{\pi^2}{16})$

(B) $-\pi(1 + \frac{\pi^2}{16})$

(C) $\pi(1 - \frac{\pi^2}{16})$

(D) $-\pi^2(1 + \frac{\pi^2}{16})$

Sigma Percentile · Accepted Answer

### The Mystery of the Trapped Function

Have you ever felt like a function was playing hide-and-seek with you? In this problem, we are given an integral equation:

$$\int_0^{t^2} (f(x) + x^2) \, dx = \frac{4}{3} t^3$$

Our target is to find $f\left(\frac{\pi^2}{4}\right)$. The function $f(x)$ is buried deep inside an integral, and our job is to extract it. This is not just a calculation; it is a detective story where we use the tools of calculus to uncover the truth.

### The Newton-Leibniz Key

When you see a variable in the limit of an integral, your first instinct should be the **Newton-Leibniz rule**. This theorem is our master key.

It states that if we have an integral $\int_0^{h(t)} F(x) \, dx$, its derivative with respect to $t$ is the integrand evaluated at the upper limit, multiplied by the derivative of that limit. Mathematically, we write this as:

$$\frac{d}{dt} \int_{a}^{h(t)} F(x) \, dx = F(h(t)) \cdot h'(t)$$

Imagine you are standing on a wedge, looking at the area under the curve $y = f(x) + x^2$. As $t$ changes, the area changes. By differentiating, we are essentially looking at the "rate of change" of this area, which brings us directly to the function values at the boundary.

### The Differentiation Dance

Let us apply this to our equation. We differentiate both sides with respect to $t$:

$$\frac{d}{dt} \int_0^{t^2} (f(x) + x^2) \, dx = \frac{d}{dt} \left( \frac{4}{3} t^3 \right)$$

On the left side, our inner function is $F(x) = f(x) + x^2$, and our upper limit is $h(t) = t^2$. Substituting $x = t^2$ into the inner function gives us $f(t^2) + (t^2)^2 = f(t^2) + t^4$.

Applying the chain rule, we multiply by the derivative of the upper limit, $\frac{d}{dt}(t^2) = 2t$. Thus, the left side becomes:

$$(f(t^2) + t^4) \cdot 2t$$

On the right side, the derivative is a straightforward power rule application:

$$\frac{d}{dt} \left( \frac{4}{3} t^3 \right) = \frac{4}{3} \cdot 3t^2 = 4t^2$$

Equating the two sides, we obtain:

$$(f(t^2) + t^4) \cdot 2t = 4t^2$$

### The Algebraic Resolution

Since the problem guarantees $t > 0$, we can divide both sides by $2t$ without fear of division by zero. This simplifies our equation beautifully:

$$f(t^2) + t^4 = 2t$$

Now, the function $f$ is finally free. We isolate it to get our master equation:

$$f(t^2) = 2t - t^4$$

### The Final Act

We need to find $f\left(\frac{\pi^2}{4}\right)$. By comparing this to our expression $f(t^2)$, it is obvious that we must set $t^2 = \frac{\pi^2}{4}$.

Taking the square root, we get $t = \pm \frac{\pi}{2}$. Recalling the condition $t > 0$, we confidently select $t = \frac{\pi}{2}$.

Now, we substitute this value back into our isolated function equation:

$$f\left(\frac{\pi^2}{4}\right) = 2\left(\frac{\pi}{2}\right) - \left(\frac{\pi}{2}\right)^4$$

Simplifying this, we get $\pi - \frac{\pi^4}{16}$. Expressing this in a factored form, we arrive at the **final answer**:

$$f\left(\frac{\pi^2}{4}\right) = \pi \left( 1 - \frac{\pi^3}{16} \right)$$

Animated Solution for Mathematics - Definite Integration: Let $f$ be a continuous function satisfying $\int_{0}^{t^{2}} (f (x) + x^{2}) d x = \frac{4}{3} t^{3}, \forall t > 0$ . Then $f (\frac{π ^{2}}{4})$ is equal to

Select Answer:

Visualized Solution

The Integral Equation

The Area Function

Newton-Leibniz Theorem

Differentiating the LHS

LHS Derivative

Differentiating the RHS

Equating and Simplifying

Isolating $f (t^{2})$

Setting the Target Input

Solving for $t$

Substituting $t$

Final Calculation

The Sigma Insight: Newton-Leibniz & Reduction Formulas

The Mystery of the Trapped Function

The Newton-Leibniz Key

The Differentiation Dance

The Algebraic Resolution

The Final Act

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