Introduction

Integration is a fundamental concept in mathematics that involves finding the antiderivative of a function. It is the reverse process of differentiation. In simpler terms, integration is used to find the area under a curve. In this topic, we will explore various methods and techniques used in integration.

Fundamental Theorem of Calculus

The fundamental theorem of calculus states that if a function $f(x)$ is continuous on the interval $[a, b]$, and $F(x)$ is an antiderivative of $f(x)$, then the definite integral of $f(x)$ from $a$ to $b$ is given by: $$\int_{a}^{b} f(x) dx = F(b) - F(a)$$

Example

Find $\int_{1}^{3} (2x + 1) dx$. Given $f(x) = 2x + 1$, the antiderivative of $f(x)$ is $F(x) = x^2 + x + C$. Using the fundamental theorem of calculus: $$\int_{1}^{3} (2x + 1) dx = [x^2 + x]_{1}^{3} = (3^2 + 3) - (1^2 + 1) = 11$$

Methods of Integration

1. Substitution

Substitution is a method used to simplify integrals by substituting a variable in place of a more complex expression.

Example

Evaluate $\int 2x(x^2 + 1)^3 dx$ using substitution. Let $u = x^2 + 1$, then $du = 2x dx$. Substituting the values: $$\int 2x(x^2 + 1)^3 dx = \int u^3 du = \frac{u^4}{4} + C = \frac{(x^2 + 1)^4}{4} + C$$

2. Integration by Parts

Integration by parts is a method used to integrate the product of two functions by applying the formula: $$\int u dv = uv - \int v du$$

Example

Find $\int x \cos x dx$ using integration by parts. Let $u = x$ and $dv = \cos x dx$. Then, $du = dx$ and $v = \sin x$. Substituting into the formula: $$\int x \cos x dx = x \sin x - \int \sin x dx = x \sin x + \cos x + C$$

Common Mistakes

• Forgetting to add the constant of integration (C) when evaluating indefinite integrals.
• Incorrect application of integration rules and formulas.
• Confusing between differentiation and integration processes.

Key Points

• Integration is the reverse process of differentiation.
• The fundamental theorem of calculus relates integration to differentiation.
• Methods such as substitution and integration by parts are used to simplify integrals.
• Always remember to include the constant of integration when evaluating indefinite integrals.

Practice Questions

1. Calculate $\int_{0}^{\pi} x \sin x dx$.

Solution: Using integration by parts with $u = x$ and $dv = \sin x dx$, we get: $$\int x \sin x dx = -x \cos x + \int \cos x dx = -x \cos x + \sin x + C$$ Evaluating from 0 to $\pi$: $$[-\pi \cos \pi + \sin \pi] - [0 - 0] = \pi$$

2. Evaluate $\int \frac{1}{x^2} dx$.

Solution: The integral of $\frac{1}{x^2}$ is $- \frac{1}{x} + C$.

3. Find $\int e^x \sin x dx$.

Solution: Applying integration by parts with $u = e^x$ and $dv = \sin x dx$: $$\int e^x \sin x dx = e^x (-\cos x) - \int -\cos x e^x dx = -e^x \cos x - e^x \sin x + C$$

4. Determine $\int \frac{2x}{x^2 + 1} dx$.

Solution: Using substitution with $u = x^2 + 1$, we have $du = 2x dx$: $$\int \frac{2x}{x^2 + 1} dx = \int \frac{1}{u} du = \ln|u| + C = \ln|x^2 + 1| + C$$

5. Solve $\int \frac{x^2 + 3x + 2}{x + 2} dx$.

Solution: Divide the polynomial by $x + 2$ to get $x + 1$: $$\int \frac{x^2 + 3x + 2}{x + 2} dx = \int (x + 1) dx = \frac{x^2}{2} + x + C$$