Mastering the Fundamental Theorem of Calculus for Competitive Exams

The Fundamental Theorem of Calculus (FTC) is a cornerstone of calculus, providing a powerful link between differentiation and integration. Understanding its two parts is crucial for solving a wide range of problems in competitive exams like JEE Mathematics.

Part 1: The Derivative of an Integral

The first part of the FTC states that if you define a function $F(x)$ as the integral of another function $f(t)$ from a constant $a$ to $x$ , then the derivative of $F(x)$ is simply $f(x)$ . Mathematically, if $F(x) = \int_{a}^{x} f(t) dt$ , then $F'(x) = f(x)$ .

The derivative of an area-accumulation function is the original function.

Imagine accumulating area under a curve $f(t)$ from a fixed point $a$ up to a variable point $x$ . The rate at which this accumulated area changes as $x$ moves is precisely the height of the curve at $x$ , which is $f(x)$ .

Let $F(x) = \int_{a}^{x} f(t) dt$ . This represents the signed area under the curve of $f(t)$ from $t=a$ to $t=x$ . To find the derivative $F'(x)$ , we can consider the change in $F(x)$ as $x$ increases by a small amount $\Delta x$ . The change in area, $\Delta F = F(x+\Delta x) - F(x)$ , is approximately the area of a thin rectangle with height $f(x)$ and width $\Delta x$ . Thus, $\Delta F \approx f(x) \Delta x$ . Dividing by $\Delta x$ , we get $\frac{\Delta F}{\Delta x} \approx f(x)$ . Taking the limit as $\Delta x \to 0$ , we arrive at $F'(x) = \lim_{\Delta x \to 0} \frac{F(x+\Delta x) - F(x)}{\Delta x} = f(x)$ .

G(x) = \int_{2}^{x} \sin(t^2) dt

, what is

G'(x)

$G'(x) = \sin(x^2)$

Part 2: Evaluating Definite Integrals

The second part of the FTC provides a method to calculate definite integrals. It states that if $F$ is an antiderivative of $f$ (meaning $F'(x) = f(x)$ ), then the definite integral of $f$ from $a$ to $b$ is the difference in the values of $F$ at the upper and lower limits. Mathematically, $\int_{a}^{b} f(x) dx = F(b) - F(a)$ .

The second part of the Fundamental Theorem of Calculus, $\int_{a}^{b} f(x) dx = F(b) - F(a)$ , is a powerful tool for evaluating definite integrals. It means that to find the exact area under a curve $f(x)$ between points $a$ and $b$ , you first find any function $F(x)$ whose derivative is $f(x)$ (an antiderivative). Then, you simply subtract the value of $F$ at the lower limit ( $a$ ) from its value at the upper limit ( $b$ ). This bypasses the need for Riemann sums, making calculations much more efficient. For example, to calculate $\int_{1}^{3} x^2 dx$ , we find an antiderivative of $x^2$ , which is $\frac{x^3}{3}$ . Then, we evaluate this at the limits: $\frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3}$ . This visual represents the area under the parabola $y=x^2$ from $x=1$ to $x=3$ .

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What is the relationship between a function

f(x)

and its antiderivative

F(x)

$F'(x) = f(x)$

This part of the theorem is incredibly useful for solving problems where you need to find the net change of a quantity given its rate of change. For instance, if you know the velocity of an object, you can find its displacement by integrating the velocity function.

Applications in Competitive Exams

In competitive exams, the FTC is applied in various contexts:

Evaluating Definite Integrals: Directly using $F(b) - F(a)$ for functions and their antiderivatives.
Problems involving Area and Volume: Calculating areas between curves or volumes of revolution.
Rate of Change Problems: Using the FTC to find total change from a rate of change (e.g., displacement from velocity, total production from production rate).
Properties of Integrals: Understanding how derivatives and integrals interact.

Remember that the antiderivative $F(x)$ is not unique; it can include an arbitrary constant $C$ . However, when evaluating definite integrals using $F(b) - F(a)$ , the constant $C$ cancels out: $(F(b) + C) - (F(a) + C) = F(b) - F(a)$ .

What is the definite integral of

f(x) = 2x

from

x=1

x=4

The antiderivative of $2x$ is $x^2$ . So, $\int_{1}^{4} 2x dx = [x^2]_{1}^{4} = 4^2 - 1^2 = 16 - 1 = 15$ .

Key Takeaways for JEE Mathematics

Mastering the Fundamental Theorem of Calculus involves understanding both its parts and practicing their application. Focus on identifying antiderivatives correctly and applying the evaluation formula. Be prepared for problems that combine differentiation and integration concepts.

Learning Resources

Khan Academy: The Fundamental Theorem of Calculus(video)

A clear video explanation of both parts of the Fundamental Theorem of Calculus with examples.

Brilliant.org: Fundamental Theorem of Calculus(blog)

An interactive explanation of the FTC, focusing on intuition and applications.

Paul's Online Math Notes: The Fundamental Theorem of Calculus(documentation)

Comprehensive notes covering both parts of the FTC, including proofs and examples.

MIT OpenCourseware: Calculus - Fundamental Theorem of Calculus(documentation)

Lecture notes and explanations from MIT's introductory calculus course.

YouTube: Fundamental Theorem of Calculus Part 1 - Proof and Examples(video)

A detailed explanation and proof of the first part of the FTC.

YouTube: Fundamental Theorem of Calculus Part 2 - Proof and Examples(video)

A detailed explanation and proof of the second part of the FTC.

Wikipedia: Fundamental Theorem of Calculus(wikipedia)

A detailed overview of the theorem, its history, and generalizations.

Art of Problem Solving: Fundamental Theorem of Calculus(documentation)

A resource focused on problem-solving techniques related to the FTC, often useful for competitive exams.

MathisFun: The Fundamental Theorem of Calculus(blog)

A simplified explanation of the FTC with clear examples and interactive elements.

StackExchange Mathematics: Applications of FTC(blog)

A collection of user-submitted questions and answers discussing various applications and nuances of the FTC.