Sigma Percentile
JEE Advanced 2001
LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let and . If , then equals

Select Answer:

Visualized Solution

Visualizing the Integral

  • We are given a function , which represents the area under the curve from to .
  • Here, we are interested in the area up to a variable limit , represented as .

Fundamental Theorem of Calculus (FTC)

  • Recall the Fundamental Theorem of Calculus (Part 1):
  • If , then its derivative with respect to is simply the integrand:
  • This establishes a direct link between the rate of change of the area function and the height of the curve.

Expanding the Given Equation

  • We are given the relation:
  • Let's expand the right-hand side to make differentiation easier:
  • This equation relates the accumulated area up to to a simple polynomial in .

Differentiating with Respect to

  • To extract from , we must differentiate both sides of the equation with respect to :
  • The right-hand side is straightforward to differentiate using the power rule.

Applying the Chain Rule on LHS

  • On the left-hand side, we have a composite function . We must apply the Chain Rule:
  • Since , we get:

Substituting the Integrand Function

  • Now, we use our FTC relation: .
  • Replacing with gives:
  • This brings us directly to the function that we need to evaluate!

Isolating the Function

  • To solve for , we divide both sides by (since , ):
  • Simplifying the fraction:

Substituting for

  • We want to find the value of .
  • To get from , we set the argument .
  • Since the domain is , we take the positive root: .
  • Substitute into our simplified equation:

Final Calculation of

  • Evaluating the expression:
  • Thus, the value of is .
  • This matches Option 3 (or the third option, which is ).

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

Imagine you are standing on the edge of a vast, unknown landscape. You are given a function , which represents the area under the curve from to .
In our problem, we are looking at . This is the area under the curve up to a variable limit . It is a dynamic, shifting boundary, and our goal is to uncover the hidden function that defines this area.

The Fundamental Theorem of Calculus

Our Logic Bridge
How do we connect the area function to the function ? This is where the Fundamental Theorem of Calculus (FTC) becomes our most powerful tool.
It tells us that the derivative of an area function with respect to its upper limit is simply the value of the function at that limit. Mathematically, this is expressed as:
This is the bridge we need. It allows us to transform an integral equation into a differential one, which is much easier to handle.

Expanding the Given Equation

We are given the relation . Before we dive into the calculus, let us simplify the right-hand side.
By distributing , we get:
This simple algebraic step is crucial. It transforms a product into a sum, making the subsequent differentiation much cleaner and less prone to errors.

The Chain Rule

The Core Difficulty
Now, we must differentiate both sides with respect to . We are differentiating with respect to .
Since the inner function is , we must use the Chain Rule:
The derivative of is . On the right side, the derivative of is .
Putting it all together, we get:

Substituting and Solving

Now, we use our FTC relation: . Substituting this into our equation, we get:
To isolate , we divide both sides by . Since the domain is , is never zero, so this division is perfectly valid.
We obtain:

The Final Calculation

We are asked to find . To get this, we set the argument . Since must be positive, we take .
Substituting into our expression, we get:
The final answer is 4. This problem is a perfect example of how the Fundamental Theorem of Calculus and the Chain Rule work in harmony to solve complex problems.

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