JEE Advanced 2004

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: If $f (x)$ is differentiable and $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}$ , then $f (4/25)$ equals

Q: Wait, why do we differentiate with respect to $t$ instead of $x$?

Great question! The integral is defined with respect to $x$, but the limits and the result are functions of $t$. Since $t$ is the independent variable that controls the boundary of our area, differentiating with respect to $t$ is the only way to 'unlock' the function $f(x)$ trapped inside the integral.

Q: Why does the lower limit $0$ just disappear when we differentiate?

The Newton-Leibniz rule states that $\frac{d}{dt} \int_{a(t)}^{b(t)} g(x) dx = g(b(t)) \cdot b'(t) - g(a(t)) \cdot a'(t)$. Since our lower limit $a(t) = 0$ is a constant, its derivative $a'(t)$ is $0$. Therefore, the entire term $g(a(t)) \cdot a'(t)$ becomes zero and vanishes.

Q: Why do we take the positive root when solving $t^2 = 4/25$?

In the context of this problem, $t^2$ represents the upper limit of an integral. While $t$ could mathematically be negative, the problem implies a standard setup where the interval $[0, t^2]$ is well-defined. Taking the positive root $t = 2/5$ is the standard approach for these types of problems unless specified otherwise.

Select Answer:

Visualized Solution

The Integral as an Area

The integral $\int_{0}^{t^{2}} x f (x) d x$ represents the area under the curve $y = x f (x)$ .
The limits of integration are from $x = 0$ to $x = t^{2}$ .
This area is given as $\frac{2}{5} t^{5}$ .

Differentiating the Area

To find $f (x)$ , we need to remove the integral sign.
We apply differentiation on both sides with respect to $t$ .
This requires the Newton-Leibniz Rule for differentiating under the integral sign.

The Newton-Leibniz Rule

$\frac{d}{d t} \int_{a (t)}^{b (t)} g (x) d x = g (b (t)) \cdot b^{'} (t) - g (a (t)) \cdot a^{'} (t)$
Here, $g (x) = x f (x)$ .
Upper limit $b (t) = t^{2}$ , Lower limit $a (t) = 0$ .

Differentiating the Left Hand Side

Substitute $x = t^{2}$ into $x f (x)$ : $(t^{2}) f (t^{2})$ .
Multiply by the derivative of $t^{2}$ : $\frac{d}{d t} (t^{2}) = 2 t$ .
Lower limit is $0$ , so its derivative is $0$ .
LHS becomes: $t^{2} f (t^{2}) \cdot 2 t$ .

Differentiating the Right Hand Side

The RHS is $\frac{2}{5} t^{5}$ .
Differentiating with respect to $t$ : $\frac{d}{d t} (\frac{2}{5} t^{5})$ .
Apply power rule: $\frac{2}{5} \cdot 5 t^{4} = 2 t^{4}$ .

Equating the Derivatives

Equating the differentiated LHS and RHS:
$t^{2} f (t^{2}) \cdot 2 t = 2 t^{4}$
Simplify the LHS: $2 t^{3} f (t^{2}) = 2 t^{4}$

Solving for $f (t^{2})$

We have $2 t^{3} f (t^{2}) = 2 t^{4}$ .
Divide both sides by $2 t^{3}$ (assuming $t \neq = 0$ ):
$f (t^{2}) = \frac{2 t ^{4}}{2 t ^{3}}$
$f (t^{2}) = t$

Finding the Target Value

We have the function rule: $f (t^{2}) = t$ .
The question asks for the value of $f (\frac{4}{25})$ .
We need to match the input $t^{2}$ with $\frac{4}{25}$ .

Calculating $t$

Set the inputs equal: $t^{2} = \frac{4}{25}$ .
Take the square root of both sides.
$t = \frac{4}{25} = \frac{2}{5}$ .
(We take the positive root as $t^{2}$ represents the upper limit of an area in the positive quadrant).

Final Conclusion

Substitute $t = \frac{2}{5}$ back into our function rule $f (t^{2}) = t$ .
$f (\frac{4}{25}) = \frac{2}{5}$ .
The correct option is 2/5.

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The Sigma Insight: Newton-Leibniz & Reduction Formulas

Solution Diagram

Analyzing the Setup

You are presented with the integral equation:

\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}

This equation defines the relationship between an unknown function

f (x)

and the variable

t

. Our objective is to isolate

f (x)

and evaluate it at a specific point.

The Newton-Leibniz Key

To extract

f (x)

from the integral, we employ the Newton-Leibniz rule (Leibniz Integral Rule). This theorem allows us to differentiate an integral with respect to its variable limit:

\frac{d}{d t} \int_{a (t)}^{b (t)} g (x) d x = g (b (t)) \cdot b^{'} (t) - g (a (t)) \cdot a^{'} (t)

In this problem, we identify

g (x) = x f (x)

, the upper limit

b (t) = t^{2}

, and the lower limit

a (t) = 0

The Differentiation Dance

We differentiate both sides of the original equation with respect to

t

. Applying the rule to the left-hand side:

\frac{d}{d t} \int_{0}^{t^{2}} x f (x) d x = (t^{2}) f (t^{2}) \cdot \frac{d}{d t} (t^{2}) - (0) f (0) \cdot 0

This simplifies to:

2 t^{3} f (t^{2})

Now, we differentiate the right-hand side,

\frac{2}{5} t^{5}

, with respect to

t

using the power rule:

\frac{d}{d t} (\frac{2}{5} t^{5}) = 2 t^{4}

The Elegant Cancellation

Equating the derivatives from both sides, we obtain:

2 t^{3} f (t^{2}) = 2 t^{4}

Assuming $t eq 0$, we divide both sides by

2 t^{3}

to isolate the function:

f (t^{2}) = t

This result reveals that for any input

u = t^{2}

, the function behaves as

f (u) = u

The Final Calculation

We are tasked with finding the value of

f (4/25)

. Given our derived rule

f (t^{2}) = t

, we set

t^{2} = 4/25

Solving for

t

, we find

t = 2/5

. Substituting this into our expression for

f (t^{2})

f (\frac{4}{25}) = \frac{2}{5}

The final result is $2/5$ .

Common Traps & FAQs

Wait, why do we differentiate with respect to

t

instead of

x

Great question! The integral is defined with respect to

x

, but the limits and the result are functions of

t

. Since

t

is the independent variable that controls the boundary of our area, differentiating with respect to

t

is the only way to 'unlock' the function

f (x)

trapped inside the integral.

Why does the lower limit

0

just disappear when we differentiate?

The Newton-Leibniz rule states that

\frac{d}{d t} \int_{a (t)}^{b (t)} g (x) d x = g (b (t)) \cdot b^{'} (t) - g (a (t)) \cdot a^{'} (t)

. Since our lower limit

a (t) = 0

is a constant, its derivative

a^{'} (t)

0

. Therefore, the entire term

g (a (t)) \cdot a^{'} (t)

becomes zero and vanishes.

Why do we take the positive root when solving

t^{2} = 4/25

In the context of this problem,

t^{2}

represents the upper limit of an integral. While

t

could mathematically be negative, the problem implies a standard setup where the interval

[0, t^{2}]

is well-defined. Taking the positive root

t = 2/5

is the standard approach for these types of problems unless specified otherwise.

Animated Solution for Mathematics - Definite Integration: If $f (x)$ is differentiable and $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}$ , then $f (4/25)$ equals

Select Answer:

Visualized Solution

The Integral as an Area

Differentiating the Area

The Newton-Leibniz Rule

Differentiating the Left Hand Side

Differentiating the Right Hand Side

Equating the Derivatives

Solving for $f (t^{2})$

Finding the Target Value

Calculating $t$

Final Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Newton-Leibniz Key

The Differentiation Dance

The Elegant Cancellation

The Final Calculation

Similar Questions

If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

Let $f : R \to R$ be a differentiable function and $f (1) = 4$ . Then the value of $lim_{x \to 1} \int_{4}^{f (x)} \frac{2 t}{x - 1} d t$ is

Let $f : (0, \infty) \to R$ and $F (x) = \int_{0}^{x} f (t) d t$ . If $F (x^{2}) = x^{2} (1 + x)$ , then $f (4)$ equals

If $f : R \to R$ is a differentiable function and $f (2) = 6$ , then $lim_{x \to 2} \frac{\int _{6}^{f (x)} 2 t d t}{x - 2}$ is :-

Let $f : R \to R$ be a differentiable function having $f (2) = 6, f^{'} (2) = \frac{1}{48}$ . Then $lim_{x \to 2} \int_{6}^{f (x)} \frac{4 t ^{3}}{x - 2} d t$ equals

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

$lim_{x \to π /4} \frac{\int _{2}^{s e c^{2} x} f ( t ) d t}{x ^{2} - \frac{π ^{2}}{16}}$ equals

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

Let $f$ be a twice differentiable function on $R$ . If $f^{'} (0) = 4$ and $f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x$ , then $(2 a + 1)^{5} a^{2}$ is equal to

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) is differentiable and ∫0t2​xf(x)dx=52​t5, then f(4/25) equals

Select Answer:

Visualized Solution

The Integral as an Area

Differentiating the Area

The Newton-Leibniz Rule

Differentiating the Left Hand Side

Differentiating the Right Hand Side

Equating the Derivatives

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t2)

Finding the Target Value

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Final Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Newton-Leibniz Key

The Differentiation Dance

The Elegant Cancellation

The Final Calculation

Similar Questions

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Animated Solution for Mathematics - Definite Integration: If $f (x)$ is differentiable and $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}$ , then $f (4/25)$ equals

Solving for $f (t^{2})$

Calculating $t$

If $\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t$ , then $f^{'} (1/2)$ is :

Let $f : R \to R$ be a differentiable function and $f (1) = 4$ . Then the value of $lim_{x \to 1} \int_{4}^{f (x)} \frac{2 t}{x - 1} d t$ is

Let $f : (0, \infty) \to R$ and $F (x) = \int_{0}^{x} f (t) d t$ . If $F (x^{2}) = x^{2} (1 + x)$ , then $f (4)$ equals

If $f : R \to R$ is a differentiable function and $f (2) = 6$ , then $lim_{x \to 2} \frac{\int _{6}^{f (x)} 2 t d t}{x - 2}$ is :-

Let $f : R \to R$ be a differentiable function having $f (2) = 6, f^{'} (2) = \frac{1}{48}$ . Then $lim_{x \to 2} \int_{6}^{f (x)} \frac{4 t ^{3}}{x - 2} d t$ equals

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is

$lim_{x \to π /4} \frac{\int _{2}^{s e c^{2} x} f ( t ) d t}{x ^{2} - \frac{π ^{2}}{16}}$ equals

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

Let $f$ be a twice differentiable function on $R$ . If $f^{'} (0) = 4$ and $f (x) + \int_{0}^{x} (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x$ , then $(2 a + 1)^{5} a^{2}$ is equal to

For $x \in (0, 5 π /2)$ , define $f (x) = \int_{0}^{x} t sin t d t$ . Then $f$ has