Properties of Definite Integrals

Properties of Definite Integrals: Your JEE Main Advantage

Hello students! Welcome to a crucial lesson in definite integration. Mastering the properties of definite integrals is not just about memorizing formulas; it's about unlocking powerful techniques that can drastically simplify complex JEE Main problems. These properties act as shortcuts, allowing you to solve seemingly impossible integrals with ease. So, buckle up, and let's dive in!

Conceptual Understanding and Intuition

Before we jump into the formulas, let's build a solid conceptual foundation. Remember, a definite integral represents the signed area under a curve between two limits. Thinking in terms of areas will give you an intuitive grasp of these properties.

1. ∫ₐᵃ f(x) dx = 0

This one's simple! If the upper and lower limits of integration are the same, you're essentially calculating the area of a line, which is zero. Imagine trying to find the area under a curve from point $a$ to the same point $a$. There's no width, hence no area!

2. ∫ₐᵇ f(x) dx = -∫ᵇₐ f(x) dx

Here, we're reversing the limits of integration. Geometrically, this means we're considering the area from right to left instead of left to right. This flips the sign of the area because we're changing the direction in which we're accumulating it. Think of it as walking the same path forward and backward – the distance is the same, but the direction is reversed. Consequently, the sign changes.

3. ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx

This property allows you to split an integral into multiple parts. If $c$ lies between $a$ and $b$, you can calculate the area from $a$ to $c$ and then from $c$ to $b$, and add them up to get the total area from $a$ to $b$. Imagine you want to calculate the area under a curve from $a = 0$ to $b = 5$, and you choose $c = 2$. You could integrate from 0 to 2, and then from 2 to 5, adding those two results together, to obtain the definite integral from 0 to 5.

4. King's Property: ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a+b-x) dx

This is a REALLY important one for JEE! It's often called the "King's Property" (or sometimes Queen's!). It states that you can replace $x$ with $a + b - x$ without changing the value of the integral. The substitution $x = a + b - t$ (and $dx = -dt$) essentially reflects the function around the midpoint of the interval $[a, b]$. The area remains the same, even though the function is transformed.

5. Even/Odd Function Properties

These properties are incredibly useful when dealing with integrals over symmetric intervals (from $-a$ to $a$). Remember:

• An even function satisfies $f(-x) = f(x)$ (symmetric about the y-axis, like $x^2$ or $\cos(x)$).
• An odd function satisfies $f(-x) = -f(x)$ (symmetric about the origin, like $x^3$ or $\sin(x)$).

Important Formulas and Their Derivations

Explanation: Since even functions are symmetric about the y-axis, the area from $-a$ to $0$ is identical to the area from $0$ to $a$. Therefore, we can just double the area from $0$ to $a$ to get the total area.

Derivation: $$∫₋ₐᵃ f(x) dx = ∫₋ₐ⁰ f(x) dx + ∫₀ᵃ f(x) dx$$ Let $x = -t$ in the first integral, so $dx = -dt$. When $x = -a$, $t = a$, and when $x = 0$, $t = 0$. So, $$∫₋ₐ⁰ f(x) dx = ∫ₐ⁰ f(-t) (-dt) = ∫₀ᵃ f(-t) dt$$ Since $f$ is even, $f(-t) = f(t)$. Therefore, $$∫₋ₐ⁰ f(x) dx = ∫₀ᵃ f(t) dt = ∫₀ᵃ f(x) dx$$ $$∫₋ₐᵃ f(x) dx = ∫₀ᵃ f(x) dx + ∫₀ᵃ f(x) dx = 2∫₀ᵃ f(x) dx$$

Explanation: For odd functions, the area from $-a$ to $0$ is the negative of the area from $0$ to $a$. The positive and negative areas cancel each other out, resulting in a total area of zero.

Derivation: $$∫₋ₐᵃ f(x) dx = ∫₋ₐ⁰ f(x) dx + ∫₀ᵃ f(x) dx$$ Let $x = -t$ in the first integral, so $dx = -dt$. When $x = -a$, $t = a$, and when $x = 0$, $t = 0$. So, $$∫₋ₐ⁰ f(x) dx = ∫ₐ⁰ f(-t) (-dt) = ∫₀ᵃ f(-t) dt$$ Since $f$ is odd, $f(-t) = -f(t)$. Therefore, $$∫₋ₐ⁰ f(x) dx = ∫₀ᵃ -f(t) dt = -∫₀ᵃ f(x) dx$$ $$∫₋ₐᵃ f(x) dx = -∫₀ᵃ f(x) dx + ∫₀ᵃ f(x) dx = 0$$

Explanation: This is a special case of King's property, where $b = 0$. Geometrically, this means the area remains unchanged when we replace $x$ with $a-x$. This substitution effectively reflects the function around the vertical line $x = a/2$.

Derivation: Let $x = a - t$, so $dx = -dt$. When $x = 0$, $t = a$, and when $x = a$, $t = 0$. So, $$∫₀ᵃ f(x) dx = ∫ₐ⁰ f(a-t) (-dt) = ∫₀ᵃ f(a-t) dt = ∫₀ᵃ f(a-x) dx$$

Explanation: This property states that if $f(2a - x) = f(x)$, the integral from $0$ to $2a$ is twice the integral from $0$ to $a$. The condition $f(2a - x) = f(x)$ implies that the function has a kind of symmetry about the line $x = a$.

Derivation: $$∫₀²ᵃ f(x) dx = ∫₀ᵃ f(x) dx + ∫ₐ²ᵃ f(x) dx$$ Let $x = 2a - t$ in the second integral, so $dx = -dt$. When $x = a$, $t = a$, and when $x = 2a$, $t = 0$. So, $$∫ₐ²ᵃ f(x) dx = ∫ₐ⁰ f(2a-t) (-dt) = ∫₀ᵃ f(2a-t) dt$$ Since $f(2a-t) = f(t)$, $$∫ₐ²ᵃ f(x) dx = ∫₀ᵃ f(t) dt = ∫₀ᵃ f(x) dx$$ $$∫₀²ᵃ f(x) dx = ∫₀ᵃ f(x) dx + ∫₀ᵃ f(x) dx = 2∫₀ᵃ f(x) dx$$

Let's consider a quick example. Evaluate: $$I = ∫₋π/₂^{π/₂} \sin⁵(x) dx$$ Since $\sin⁵(x)$ is an odd function, the integral evaluates to zero directly! No integration needed. Boom!

Another illustrative example utilizing King's Property:

$$I = ∫₀^{π/₂} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$

Applying King's Property, substitute $x$ with $\frac{π}{2} - x$:

$$I = ∫₀^{π/₂} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} dx = ∫₀^{π/₂} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$$

Adding both equations:

$$2I = ∫₀^{π/₂} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = ∫₀^{π/₂} 1 dx = \frac{π}{2}$$

Thus, $$I = \frac{π}{4}$$. Notice how the integrand simplified beautifully after applying King's Property!

Mastering these properties requires practice, so keep solving problems! Good luck, and keep learning!