Home / Notes / Class 12 / Mathematics / Integration
Integration as the reverse of differentiation, standard integrals, substitution, integration by parts, partial fractions, definite integrals with properties, and the area under a curve.
Integration is the reverse of differentiation. If d/dx F(x) = f(x), then F(x) is an anti-derivative of f(x), and we write ∫ f(x) dx = F(x) + C. This is the indefinite integral. C is the constant of integration, because the derivative of any constant is zero.
Basic rules:
Standard integrals (x in radians):
Integration of a linear function inside: if ∫ f(x) dx = F(x) + C, then ∫ f(ax + b) dx = F(ax + b)/a + C. For example, ∫ cos(2x + 1) dx = sin(2x + 1)/2 + C.
Integration by substitution: choose a part of the integrand as u so that du appears too. Replace everything in terms of u, integrate, and then return to x. A very common form is ∫ f′(x)/f(x) dx = ln |f(x)| + C. For example, ∫ 2x/(x² + 1) dx = ln(x² + 1) + C, since the top is the derivative of the bottom.
Useful trigonometric identities for integration are sin² x = (1 − cos 2x)/2 and cos² x = (1 + cos 2x)/2. So ∫ sin² x dx = x/2 − (sin 2x)/4 + C.
Integration by parts is used for a product of two different kinds of function. The formula is ∫ u dv = uv − ∫ v du. Choose u in the order of ILATE: inverse trigonometric, logarithmic, algebraic, trigonometric, exponential. For ∫ x eˣ dx take u = x and dv = eˣ dx, so the answer is x eˣ − eˣ + C.
Partial fractions are used to integrate a rational function whose denominator can be factorised. For example, 1/[(x − 1)(x + 2)] = (1/3)[1/(x − 1) − 1/(x + 2)], so its integral is (1/3) ln |(x − 1)/(x + 2)| + C.
The definite integral from a to b is ∫ₐᵇ f(x) dx = F(b) − F(a), where F is any anti-derivative. There is no constant C because it cancels. This is the fundamental theorem of calculus.
Properties of definite integrals:
Area under a curve: the area between y = f(x), the x-axis, and the lines x = a and x = b is ∫ₐᵇ f(x) dx when f(x) ≥ 0 on [a, b]. If the curve lies below the x-axis, the integral is negative, so take its absolute value for the area. The area between two curves is ∫ₐᵇ [upper − lower] dx. When substituting in a definite integral, change the limits to match the new variable.
Study tip: Learn the standard integrals as a table, and always differentiate your answer to check it, because this takes only a few seconds. For substitution, write the choice of u and du clearly, since marks are given for the method. In area questions, sketch the curve first, mark the limits, and give the answer in square units.
Say the meaning first, then open the card to check.
A function whose derivative is the given function.
The general anti-derivative ∫ f(x) dx = F(x) + C.
The arbitrary constant C added to an indefinite integral.
The function that is being integrated.
The number ∫ₐᵇ f(x) dx = F(b) − F(a), found between two limits.
The numbers a and b at the bottom and top of the integral sign.
A method that replaces part of the integrand with a new variable u.
The method based on ∫ u dv = uv − ∫ v du.
| Name | Formula | What it means |
|---|---|---|
| Power rule | ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C | Valid for n ≠ −1. |
| Reciprocal | ∫ 1/x dx = ln |x| + C | The case n = −1. |
| Log form | ∫ f′(x)/f(x) dx = ln |f(x)| + C | Top is the derivative of the bottom. |
| By parts | ∫ u dv = uv − ∫ v du | Choose u by ILATE. |
| Definite integral | ∫ₐᵇ f(x) dx = F(b) − F(a) | F is an anti-derivative of f. |
| Area | A = ∫ₐᵇ f(x) dx | For f(x) ≥ 0 on [a, b]; result in square units. |
These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.
The derivative of a constant is zero, so many functions that differ only by a constant have the same derivative. The constant C represents this unknown constant.
∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C. It fails when n = −1, because the denominator would be zero. In that case ∫ 1/x dx = ln |x| + C.
Since ∫ cos u du = sin u + C and the inside is 3x, divide by 3. So ∫ cos(3x) dx = (sin 3x)/3 + C.
The formula is ∫ u dv = uv − ∫ v du. Write ln x as ln x × 1. Take u = ln x and dv = dx, so du = dx/x and v = x. Then ∫ ln x dx = x ln x − ∫ x (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C.
To find the area between y = f(x), the x-axis and the lines x = a and x = b, first make sure f(x) is not negative on the interval. Then the area is the definite integral ∫ₐᵇ f(x) dx = F(b) − F(a), where F is an anti-derivative of f. The answer is in square units. If part of the curve is below the x-axis, find that part separately and take its absolute value.
The top is the derivative of the bottom, so the integral is ln(x² + 1) + C. No modulus is needed because x² + 1 is always positive.
Answer: x³ + x² − 5x + C
Answer: 14/3
Answer: (x − 1) eˣ + C
Answer: (1/3)(x² + 4)^(3/2) + C
Answer: 9 square units