Antiderivatives

Section 1: Introduction to Antiderivatives

1.1 Concept of Antiderivative

In differentiation, we find the derivative of a given function. The inverse process of finding a function whose derivative is given is called anti-differentiation or integration.

If $\frac{d}{dx} F(x) = f(x)$, then $F(x)$ is called the antiderivative (or primitive) of $f(x)$. Mathematically, we write: $\int f(x) dx = F(x) + C$ Where $C$ is the Constant of Integration. We add $C$ because the derivative of a constant is zero; hence, multiple functions can have the same derivative.

Example: Since $\frac{d}{dx} (x^2) = 2x$ and $\frac{d}{dx} (x^2 + 5) = 2x$, we write: $\int 2x , dx = x^2 + C$

1.2 Basic Integration Formulae (Power Rule & Standard Integrals)

Here is a list of fundamental formulas derived directly from differentiation rules. (Memorize these carefully).

1. Power Rule (General Form)

$\int x^n , dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$

2. Special Case (Reciprocal)

$\int \frac{1}{x} , dx = \ln|x| + C$

3. Exponential and Logarithmic

$\int e^x , dx = e^x + C$ $\int a^x , dx = \frac{a^x}{\ln a} + C$

4. Trigonometric Formulae

$\int \sin x , dx = -\cos x + C$ $\int \cos x , dx = \sin x + C$ $\int \sec^2 x , dx = \tan x + C$ $\int \csc^2 x , dx = -\cot x + C$ $\int \sec x \tan x , dx = \sec x + C$ $\int \csc x \cot x , dx = -\csc x + C$

Section 2: Integration by Substitution Method

2.1 Substitution Method (Change of Variable)

Sometimes, integrals are not in a standard form. We can simplify them by substituting a part of the integrand with a new variable.

Rule: If $\int f(g(x)) \cdot g'(x) , dx$, we put $g(x) = t$. Then $g'(x) dx = dt$, and the integral reduces to $\int f(t) , dt$.

Example 1 (Linear Substitution): Find $\int (2x + 3)^5 , dx$. Solution: Let $t = 2x + 3 \implies dt = 2 , dx \implies dx = \frac{dt}{2}$. The integral becomes: $\int t^5 \cdot \frac{dt}{2} = \frac{1}{2} \cdot \frac{t^6}{6} + C = \frac{(2x+3)^6}{12} + C$

2.2 Trigonometric Substitution

This is a specialized technique used when the integrand involves expressions like $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$.

Key Substitutions:

1. For $\sqrt{a^2 - x^2}$: Put $x = a \sin \theta$ or $x = a \cos \theta$.
2. For $\sqrt{a^2 + x^2}$: Put $x = a \tan \theta$.
3. For $\sqrt{x^2 - a^2}$: Put $x = a \sec \theta$.

Example 2: Evaluate $\int \frac{dx}{\sqrt{9 - x^2}}$. Solution: Let $x = 3 \sin \theta \implies dx = 3 \cos \theta , d\theta$. $\sqrt{9 - x^2} = \sqrt{9 - 9\sin^2 \theta} = \sqrt{9\cos^2 \theta} = 3 \cos \theta$ The integral becomes: $\int \frac{3 \cos \theta , d\theta}{3 \cos \theta} = \int 1 , d\theta = \theta + C = \sin^{-1}\left(\frac{x}{3}\right) + C$

Section 3: Integration by Parts

3.1 The Formula

This method is used to integrate the product of two functions. It is derived from the product rule of differentiation.

If $u$ and $v$ are functions of $x$, then: $\int u \cdot v , dx = u \int v , dx - \int \left( \frac{du}{dx} \int v , dx \right) dx$

3.2 Choosing $u$ (The ILATE Rule)

To decide which function should be $u$ (the first function), we use the priority order I-L-A-T-E:

1. I - Inverse Trigonometric (e.g., $\sin^{-1}x$)
2. L - Logarithmic (e.g., $\ln x$)
3. A - Algebraic (e.g., $x^2$)
4. T - Trigonometric (e.g., $\sin x$)
5. E - Exponential (e.g., $e^x$)

The function with higher priority becomes $u$.

Example: Evaluate $\int x \cdot e^x , dx$. Solution: Here, $x$ (Algebraic) has higher priority than $e^x$ (Exponential). Let $u = x$ and $dv = e^x dx$. Then $du = dx$ and $v = e^x$. Applying the formula: $\int x e^x , dx = x \cdot e^x - \int e^x \cdot dx = x e^x - e^x + C = e^x (x - 1) + C$

Section 4: Definite Integral & Area Under the Curve

4.1 Definite Integral

A definite integral has upper and lower limits. It represents a number (not a family of functions). The Constant of Integration ($C$) is omitted in definite integrals.

If $\int f(x) dx = F(x)$, then: $\int_a^b f(x) , dx = \left[ F(x) \right]_a^b = F(b) - F(a)$

Example: Evaluate $\int_1^2 x^2 , dx$. $\int_1^2 x^2 , dx = \left[ \frac{x^3}{3} \right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$

4.2 Definite Integral as an Area Under the Curve

Geometrically, the definite integral $\int_a^b f(x) , dx$ gives the net area bounded by the curve $y = f(x)$, the x-axis, and the vertical lines $x=a$ and $x=b$.

Area Calculation:

• If $f(x) \ge 0$ on $[a, b]$, the area is simply $\int_a^b f(x) , dx$.
• If the curve crosses the x-axis, you must split the integral at the zeros to find the actual (total) area.
• Area between two curves: If $y = f(x)$ (upper curve) and $y = g(x)$ (lower curve), then: $\text{Area} = \int_a^b \left[ f(x) - g(x) \right] , dx$

Example: Find the area under the curve $y = x^2$ from $x=0$ to $x=2$. $\text{Area} = \int_0^2 x^2 , dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3} \text{ square units}$

Practice Problems for Students

1. Evaluate: $\int (3x^2 + 2x - 1) , dx$.
2. Evaluate: $\int \frac{\sin x}{1 + \cos x} , dx$. (Hint: Put $t = 1 + \cos x$).
3. Evaluate: $\int x \sin x , dx$.
4. Evaluate: $\int_0^{\pi/2} \sin 2x , dx$.
5. Find the area bounded by $y = x + 1$, the x-axis, and the lines $x=0$ and $x=2$.