Definition and Fundamental Theorem | JEE Main Calculus Crash Course | Mathematicon

Definite Integration: Definition and Fundamental Theorem

Hey JEE aspirants! Definite integration is a cornerstone of calculus, crucial for solving a wide range of problems in physics, engineering, and, of course, the JEE exam. This lesson dives into the heart of definite integrals, exploring their definition as Riemann sums and mastering the Fundamental Theorem of Calculus (FTC). Get ready to level up your integration game!

1. Definite Integral as Limit of Riemann Sum

Imagine you want to find the area under a curve $y = f(x)$ between two points $x = a$ and $x = b$ . The definite integral provides a precise way to calculate this area. But how? The concept stems from approximating the area using rectangles.

We divide the interval $[a, b]$ into $n$ subintervals, each with a width of $\Delta x = \frac{b - a}{n}$ . Let $x_i$ be a point within the $i$ -th subinterval. Then, the area of the $i$ -th rectangle is approximately $f(x_i) \Delta x$ . The Riemann sum is the sum of the areas of all these rectangles:

Σf(xᵢ)Δx = f(x_1)Δx + f(x_2)Δx + ... + f(x_n)Δx

As we increase the number of rectangles ( $n \rightarrow \infty$ ), the width of each rectangle approaches zero ( $\Delta x \rightarrow 0$ ), and the Riemann sum approaches the exact area under the curve. This leads to the definition of the definite integral:

∫ₐᵇ f(x) dx = \lim (n→∞) Σf(xᵢ)Δx

Intuition: The integral symbol $∫$ is an elongated "S," representing the "sum." $f(x) dx$ represents the area of an infinitely thin rectangle at point $x$ . The limits $a$ and $b$ define the interval over which we're summing these infinitesimal areas.

Example: Consider a simple function $f(x) = x$ and the interval $[0, 1]$ . As we increase the number of rectangles, the Riemann sum gets closer to the actual area, which is $1/2$ .

2. Fundamental Theorem of Calculus Part 1

FTC Part 1 establishes a crucial link between differentiation and integration. It states that if we define a function $F(x)$ as the definite integral of $f(t)$ from a constant $a$ to $x$ , then the derivative of $F(x)$ is simply $f(x)$ .

FTC Part 1: \frac{d}{dx}[∫ₐˣ f(t) dt] = f(x)

Explanation: Think of $∫ₐˣ f(t) dt$ as accumulating the area under the curve $f(t)$ from $a$ to $x$ . The rate at which this area accumulates (i.e., its derivative) is precisely the value of the function $f(x)$ at that point.

Example: Let $F(x) = ∫₀ˣ t² dt$ . Then, according to FTC Part 1, $\frac{d}{dx}F(x) = x²$ .

Tip: Remember to change the variable of integration to

t

when applying FTC Part 1, especially when the upper limit is

x

. This avoids confusion.

3. Fundamental Theorem of Calculus Part 2

FTC Part 2 provides a practical method for evaluating definite integrals. It states that if $F(x)$ is any antiderivative of $f(x)$ (i.e., $F'(x) = f(x)$ ), then the definite integral of $f(x)$ from $a$ to $b$ is simply the difference between the values of $F(x)$ at $b$ and $a$ .

FTC Part 2: ∫ₐᵇ f(x) dx = F(b) - F(a) \text{ where } F'(x) = f(x)

Explanation: FTC Part 2 tells us that the definite integral represents the net change in the antiderivative $F(x)$ over the interval $[a, b]$ .

Example: To evaluate $∫₀¹ x² dx$ , we first find an antiderivative of $x²$ , which is $F(x) = \frac{x³}{3}$ . Then, we apply FTC Part 2: $∫₀¹ x² dx = F(1) - F(0) = \frac{1³}{3} - \frac{0³}{3} = \frac{1}{3}$ .

4. Evaluating Definite Integrals

Here's a step-by-step guide to evaluating definite integrals using FTC Part 2:

Find an antiderivative: Determine a function $F(x)$ such that $F'(x) = f(x)$ . Remember, you don't need to include the constant of integration "+C" because it will cancel out when you subtract $F(a)$ from $F(b)$ .
Evaluate at the limits: Calculate $F(b)$ and $F(a)$ .
Subtract: Compute $F(b) - F(a)$ . This is the value of the definite integral.

Example: Evaluate $∫₁² (2x + 3) dx$ .

Antiderivative: $F(x) = x² + 3x$
Evaluate at limits: $F(2) = 2² + 3(2) = 10$ and $F(1) = 1² + 3(1) = 4$
Subtract: $F(2) - F(1) = 10 - 4 = 6$ . Therefore, $∫₁² (2x + 3) dx = 6$ .

Common Mistake: Forgetting to evaluate the antiderivative at both the upper and lower limits. Make sure you substitute both values into

F(x)

and subtract correctly. Also, remember that indefinite integrals have

+C

while definite integrals do not!

Tip: When dealing with trigonometric functions, remember the derivatives and antiderivatives of sine, cosine, tangent, etc. Also, be careful with the signs!

JEE-Specific Tricks:

Properties of Definite Integrals: Utilize properties like $∫ₐᵇ f(x) dx = -∫ᵇₐ f(x) dx$ , $∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx$ , and $∫₀ᵃ f(x) dx = ∫₀ᵃ f(a-x) dx$ to simplify integrals.
Symmetry: If $f(x)$ is an even function ( $f(-x) = f(x)$ ), then $∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dx$ . If $f(x)$ is an odd function ( $f(-x) = -f(x)$ ), then $∫₋ₐᵃ f(x) dx = 0$ .
King's Rule: This is the same as the third property in "Properties of Definite Integrals" above, use it when you see the chance to simplify the integration using $f(a-x)$ .

Mastering these concepts and practicing various problems will undoubtedly boost your performance in the JEE Main exam. Keep practicing and all the best!