UK K12 GCSE A-Level Year 13 A-Level Upper Sixth A-Level Mathematics Pure Integration By Parts Substitution

Learning Objectives

By the end of this topic, students will be able to:

• Apply the method of integration by parts to find definite integrals
• Use the substitution method to solve a wide range of integrals
• Recognize and apply the appropriate integration technique for a given integral
• Evaluate definite integrals using the substitution method
• Apply the chain rule and the product rule to integration by parts
• Solve integrals involving trigonometric functions, exponential functions, and rational functions using substitution and integration by parts

Core Concepts

Integration by Parts

Integration by parts is a method used to find the definite integral of a product of two functions. It is based on the product rule for differentiation, which states that if we have two functions u(x) and v(x), then the derivative of their product is given by:

d(uv)/dx = u(dv/dx) + v(du/dx)

To apply integration by parts, we need to choose two functions u(x) and v(x) such that the product of their derivatives is easy to integrate. The formula for integration by parts is:

∫u(x)v'(x)dx = u(x)v(x) - ∫v(x)(du/dx)dx

We can choose u(x) and v(x) in a variety of ways, but a common choice is to let u(x) be the function that is more complicated to integrate, and v(x) be the function that is easier to integrate.

Substitution Method

The substitution method is another technique used to find the definite integral of a function. It involves substituting a new variable into the function, and then using the chain rule to find the derivative of the new variable. The formula for the substitution method is:

∫f(g(x))g'(x)dx = ∫f(u)du

where u = g(x)

The substitution method can be used to solve a wide range of integrals, including integrals involving trigonometric functions, exponential functions, and rational functions.

Chain Rule and Product Rule

The chain rule and the product rule are two important rules in calculus that are used to find the derivative of a function. The chain rule states that if we have a composite function f(g(x)), then the derivative of the function is given by:

d(f(g(x))/dx = f'(g(x))g'(x)

The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by:

d(uv)/dx = u(dv/dx) + v(du/dx)

These rules are used extensively in integration by parts and the substitution method.

Worked Examples

Example 1: Integration by Parts

Find the definite integral of x^2 sin(x) from 0 to π.

Let u(x) = x^2 and v'(x) = sin(x). Then du/dx = 2x and v(x) = -cos(x).

Using the formula for integration by parts, we get:

∫x^2 sin(x)dx = -x^2 cos(x) + 2∫x cos(x)dx

Using integration by parts again, we get:

∫x cos(x)dx = x sin(x) + ∫sin(x)dx

Evaluating the integral, we get:

∫x^2 sin(x)dx = -x^2 cos(x) + 2x sin(x) - 2 cos(x)

Evaluating the integral from 0 to π, we get:

∫x^2 sin(x)dx from 0 to π = -π^2 cos(π) + 2π sin(π) - 2 cos(π) - (-2 cos(0) + 2 sin(0))

Simplifying, we get:

∫x^2 sin(x)dx from 0 to π = 2π

Example 2: Substitution Method

Find the definite integral of (x^2 + 1)/(x^2 - 1) from 1 to 2.

Let u(x) = x^2 - 1. Then du/dx = 2x and u(1) = 0, u(2) = 3.

Using the substitution method, we get:

∫(x^2 + 1)/(x^2 - 1)dx = ∫(1 + 2x)/(2x)dx

Let v(x) = 1 + 2x. Then dv/dx = 2.

Using the formula for the substitution method, we get:

∫(x^2 + 1)/(x^2 - 1)dx = ∫v(x)dx/2

Evaluating the integral, we get:

∫(x^2 + 1)/(x^2 - 1)dx = (1/2)∫(1 + 2x)dx

Evaluating the integral, we get:

∫(x^2 + 1)/(x^2 - 1)dx = (1/2)(x + x^2) from 1 to 2

Simplifying, we get:

∫(x^2 + 1)/(x^2 - 1)dx = (1/2)(2 + 4) - (1/2)(1 + 1)

Simplifying, we get:

∫(x^2 + 1)/(x^2 - 1)dx = 3

Common Misconceptions

• Students may confuse the product rule and the chain rule, and use the wrong rule to find the derivative of a composite function.
• Students may forget to use the chain rule when finding the derivative of a composite function.
• Students may confuse the substitution method with the integration by parts method, and use the wrong method to find the definite integral of a function.
• Students may forget to evaluate the definite integral of a function from the lower limit to the upper limit.

Exam Tips

• Make sure to choose the correct function u(x) and v(x) when applying integration by parts.
• Make sure to use the chain rule and the product rule correctly when finding the derivative of a composite function.
• Make sure to evaluate the definite integral of a function from the lower limit to the upper limit.
• Make sure to check your work and simplify your answer.

MCQs with Explanations

MCQ 1: [F] Integration by Parts

What is the formula for integration by parts?

A) ∫u(x)v'(x)dx = u(x)v(x) - ∫v(x)(du/dx)dx B) ∫u(x)v'(x)dx = u(x)v(x) + ∫v(x)(du/dx)dx C) ∫u(x)v'(x)dx = u(x)v(x) - ∫u(x)(dv/dx)dx D) ∫u(x)v'(x)dx = u(x)v(x) + ∫u(x)(dv/dx)dx

Correct answer: A) ∫u(x)v'(x)dx = u(x)v(x) - ∫v(x)(du/dx)dx

Why the distractors fail:

• Option B is incorrect because it uses a plus sign instead of a minus sign.
• Option C is incorrect because it uses u(x) and dv/dx in the wrong places.
• Option D is incorrect because it uses a plus sign instead of a minus sign.

MCQ 2: [H] Substitution Method

What is the formula for the substitution method?

A) ∫f(g(x))g'(x)dx = ∫f(u)du B) ∫f(g(x))g'(x)dx = ∫f(u)du + ∫f'(u)du C) ∫f(g(x))g'(x)dx = ∫f(u)du - ∫f'(u)du D) ∫f(g(x))g'(x)dx = ∫f(u)du - ∫f(u)du

Correct answer: A) ∫f(g(x))g'(x)dx = ∫f(u)du

Why the distractors fail:

• Option B is incorrect because it adds an extra term to the formula.
• Option C is incorrect because it subtracts an extra term from the formula.
• Option D is incorrect because it subtracts the same term from itself.

MCQ 3: [F] Integration by Parts

What is the product rule?

A) d(uv)/dx = u(dv/dx) - v(du/dx) B) d(uv)/dx = u(dv/dx) + v(du/dx) C) d(uv)/dx = u(du/dx) - v(dv/dx) D) d(uv)/dx = u(du/dx) + v(dv/dx)

Correct answer: B) d(uv)/dx = u(dv/dx) + v(du/dx)

Why the distractors fail:

• Option A is incorrect because it uses a minus sign instead of a plus sign.
• Option C is incorrect because it uses u(du/dx) and v(dv/dx) in the wrong places.
• Option D is incorrect because it uses a plus sign instead of a minus sign.

MCQ 4: [H] Substitution Method

What is the chain rule?

A) d(f(g(x))/dx = f'(g(x))g'(x) B) d(f(g(x))/dx = f'(g(x)) - g'(x) C) d(f(g(x))/dx = f'(g(x)) + g'(x) D) d(f(g(x))/dx = f'(g(x)) - f'(x)

Correct answer: A) d(f(g(x))/dx = f'(g(x))g'(x)

Why the distractors fail:

• Option B is incorrect because it subtracts g'(x) from f'(g(x)).
• Option C is incorrect because it adds g'(x) to f'(g(x)).
• Option D is incorrect because it subtracts f'(x) from f'(g(x)).

MCQ 5: [F] Integration by Parts

What is the formula for integration by parts?

A) ∫u(x)v'(x)dx = u(x)v(x) + ∫v(x)(du/dx)dx B) ∫u(x)v'(x)dx = u(x)v(x) - ∫v(x)(du/dx)dx C) ∫u(x)v'(x)dx = u(x)v(x) + ∫u(x)(dv/dx)dx D) ∫u(x)v'(x)dx = u(x)v(x) - ∫u(x)(dv/dx)dx

Correct answer: B) ∫u(x)v'(x)dx = u(x)v(x) - ∫v(x)(du/dx)dx

Why the distractors fail:

• Option A is incorrect because it uses a plus sign instead of a minus sign.
• Option C is incorrect because it uses u(x) and dv/dx in the wrong places.
• Option D is incorrect because it uses a minus sign instead of a plus sign.

Short-answer Questions

Question 1

Find the definite integral of x^2 sin(x) from 0 to π using integration by parts.

Question 2

Find the definite integral of (x^2 + 1)/(x^2 - 1) from 1 to 2 using the substitution method.

Question 3

Find the derivative of the composite function f(g(x)) using the chain rule.

Question 4

Find the derivative of the product of two functions u(x) and v(x) using the product rule.

Question 5

Find the definite integral of the function f(x) = x^3 sin(x) from 0 to π using the substitution method.