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Integration techniques

lesson · about 3 minutes

Substitution, integration by parts, and improper integrals.

Pen and paper is fine · no calculator needed why?

Opens at level 44. You're level 1. You can read and practice here now.

the idea

Some integrals need rearranging before the power rule can finish them.

Substitution undoes the chain rule: when the integrand holds an inside function and that inside's derivative, rename the inside u. Integration by parts undoes the product rule: ∫ u dv = uv − ∫ v du trades a product for something simpler.

An improper integral runs to infinity; the antiderivative's value as x grows stands in for the top limit.

techniques

Substitute, and change the limits

An inside function times its derivative, or a multiple of it, like 2x(x² + 1)³ or x(x² + 3)².

Let u be the inside, and find du: for u = x² + 1, du = 2x dx.
If the front factor is short, divide by the missing number: x dx = du/2.
Change the limits: put each x-limit into u.
Integrate in u between the new limits: uⁿ becomes uⁿ⁺¹/(n + 1).

worked example

Example: What is ∫₀¹ 3x²(x³ + 1)² dx? Enter a fraction or a whole number.

Let u = x³ + 1, so du = 3x² dx.
x = 0 gives u = 1. x = 1 gives u = 2.
Integrate u² from 1 to 2: (8 − 1)/3 = 7/3.

Answer: 7/3

Integration by parts

x·eˣ, x·sin(x) or x·cos(x), or ln(x) on its own.

Let u be the part that gets simpler when differentiated, x or ln(x). Let dv be the rest.
Find du and v, and use ∫ u dv = uv − ∫ v du. Then take top limit minus bottom.
Useful values: sin π = 0, cos π = −1, sin(π/2) = 1, ln(e) = 1, ln(1) = 0.

worked example

Example: What is ∫ x·cos(x) dx from x = 0 to x = π?

u = x and dv = cos(x) dx, so du = dx and v = sin(x).
Integral: x·sin(x) − ∫ sin(x) dx = x·sin(x) + cos(x).
At π: 0 + (−1) = −1. At 0: 0 + 1 = 1. So −1 − 1 = −2.

Answer: −2

Improper integrals to infinity

A top limit of ∞, with A/xᵖ (p above 1) or e^(−kx) (k above 0).

Find the antiderivative: A/xᵖ becomes −A/((p − 1)xᵖ⁻¹), and e^(−kx) becomes −e^(−kx)/k.
As x → ∞, both of those shrink to 0.
Answer: 0 minus the antiderivative at the lower limit.

worked example

Example: What is ∫₁^∞ 6/x⁴ dx? Enter a fraction or a whole number.

6/x⁴ is 6x⁻⁴, so an antiderivative is −2/x³.
As x → ∞ that goes to 0. At x = 1 it is −2.
0 − (−2) = 2.

Answer: 2

watch out for

Switching to u but keeping the old x-limits. Put each limit into u first.
Stopping after the uv part. The − ∫ v du part still has to be done.
Sign slips: the antiderivative of sin(x) is −cos(x), and it is always top minus bottom.

practice

Substitution

worked example

What is ∫₀¹ 2x(x² + 2)³ dx? Enter a fraction or a whole number.

Answer: 65/4

Let u = x² + 2; then du = 2x dx. x = 0 gives u = 2 and x = 1 gives u = 3.
The integral becomes ∫ u³ du from u = 2 to u = 3.
That is (3⁴ − 2⁴)/4 = (81 − 16)/4 = 65/4.

Integration by parts

worked example

What is ∫ 6x·cos(x) dx from x = 0 to x = π?

Answer: -12

By parts with u = x and dv = cos(x) dx: ∫ x·cos(x) dx = x·sin(x) + cos(x).
At π: 0 + (−1) = −1. At 0: 0 + 1 = 1.
So the integral is 6 × (−1 − 1) = −12.

Improper integrals

worked example

What is ∫₁^∞ 9/x³ dx? Enter a fraction or a whole number.

Answer: 9/2

9/x³ = 9x⁻³, so an antiderivative is 9x⁻²/(−2) = −9/(2x²).
As x → ∞ it goes to 0; at x = 1 it is −9/2.
So the integral is 0 − (−9/2) = 9/2.