Integral Calculus – JEE Mains Mathematics

1. Integral as an Anti-Derivative

Integration is the reverse operation of differentiation, i.e., the anti-derivative of a function.
If F'(x) = f(x), then ∫f(x) dx = F(x) + C, where C is the constant of integration.
Integration is used to find the area under curves, and solve problems involving rates of change.

2. Fundamental Integrals

Some basic integrals for algebraic, trigonometric, exponential, and logarithmic functions:

∫x^n dx = (x^(n+1))/(n+1) + C, for n ≠ -1
∫sin(x) dx = -cos(x) + C
∫cos(x) dx = sin(x) + C
∫e^x dx = e^x + C
∫1/x dx = ln|x| + C

3. Integration Techniques

3.1. Integration by Substitution

If a function is complicated, substitution is used to simplify the integral.
If u = g(x), then ∫f(g(x)) * g'(x) dx = ∫f(u) du
Example: ∫x * cos(x²) dx → let u = x², then du = 2x dx.

3.2. Integration by Parts

Integration by parts is based on the product rule of differentiation:
∫u dv = uv - ∫v du
It is useful when the integral involves a product of two functions.
Example: ∫x * e^x dx → u = x, dv = e^x dx, then du = dx, v = e^x.

3.3. Integration by Partial Fractions

Used to integrate rational functions by expressing them as a sum of simpler fractions.
Example: ∫(1/(x² - 1)) dx can be decomposed into partial fractions.
The technique involves expressing the denominator as a product of linear factors and solving for constants.

3.4. Integration Using Trigonometric Identities

Some integrals can be simplified using trigonometric identities, such as:

sin²(x) = (1 - cos(2x))/2
cos²(x) = (1 + cos(2x))/2
Using these, integrals like ∫sin²(x) dx or ∫cos²(x) dx can be simplified and solved.

4. The Fundamental Theorem of Calculus

It establishes the connection between differentiation and integration.
The first part states that if f(x) is continuous on [a, b], then the function F(x) = ∫[a, x] f(t) dt is differentiable, and its derivative is f(x).
The second part states that if F(x) is the antiderivative of f(x), then ∫[a, b] f(x) dx = F(b) - F(a).

5. Properties of Definite Integrals

If f(x) is continuous on [a, b], then:

∫[a, b] f(x) dx = F(b) - F(a), where F(x) is the antiderivative of f(x).
∫[a, b] f(x) dx = -∫[b, a] f(x) dx (reverse limits).
∫[a, b] [f(x) ± g(x)] dx = ∫[a, b] f(x) dx ± ∫[a, b] g(x) dx.
∫[a, b] c f(x) dx = c ∫[a, b] f(x) dx (constant factor rule).

6. Evaluation of Definite Integrals

To evaluate a definite integral, find the antiderivative of the function and then substitute the limits of integration.
Example: ∫[0, 1] (2x) dx = [x²] from 0 to 1 = 1² - 0² = 1.

7. Determining Areas of Regions Bounded by Curves

Definite integrals can be used to find areas between curves.
If a curve is defined by y = f(x) and bounded by x = a and x = b, the area between the curve and the x-axis is given by:

Area = ∫[a, b] f(x) dx

For regions bounded by two curves y = f(x) and y = g(x), the area between the curves is:

Area = ∫[a, b] [f(x) - g(x)] dx