Definite integrals by substitution
When you change money into another currency, you change the amounts at both ends. In a definite integral, when x is replaced by t, the two limits must also be replaced.
Put t = g(x) and change the limits using the same relation: the new limits are g(a) and g(b). Then there is no need to return to x at the end.
Formula: ∫ from a to b of f(g(x)) g′(x) dx = ∫ from g(a) to g(b) of f(t) dt
Rule 1: Change the limits with the variable — Compute the new limits first.
- Easy: ∫ from 0 to 1 of 2x (x² + 1)³ dx: t = x² + 1 gives limits 1 and 2, so [t⁴/4] from 1 to 2 = 15/4.
- Medium: ∫ from 0 to π/2 of sin x cos x dx: t = sin x gives limits 0 and 1, so the value is ½.
- Exam-type: ∫ from 0 to 1 of x e^(x²) dx = [e^t/2] from 0 to 1 = (e − 1)/2.
Rule 2: Logarithms and roots — Put the inner part equal to t.
- Easy: ∫ from 1 to e of (ln x)/x dx = [t²/2] from 0 to 1 = ½.
- Medium: ∫ from 0 to π/4 of tan x dx = [−ln|cos x|] from 0 to π/4 = ½ ln 2.
- Exam-type: ∫ from 0 to 2 of x √(x² + 1) dx = [(x² + 1)^(3/2)/3] from 0 to 2 = (5√5 − 1)/3.
Write the new limits next to the new variable and never mix the old and new limits.
Using the old limits with the new variable t.
Change the variable, change the limits.
Practice questions
- 1Evaluate ∫ from 0 to 1 of 2x (x² + 1)³ dx.
- 2Evaluate ∫ from 0 to π/2 of sin x cos x dx.
- 3Evaluate ∫ from 0 to 1 of x e^(x²) dx.
- 4Evaluate ∫ from 1 to e of (ln x)/x dx.
- 5Evaluate ∫ from 0 to π/4 of tan x dx.
- 6Evaluate ∫ from 0 to 2 of x √(x² + 1) dx.
- 7Evaluate ∫ from 0 to 1 of dx/(1 + x²).
- 8Evaluate ∫ from 0 to 1 of x/(1 + x²) dx.
Show answers (try them yourself first)
- 115/4
- 21/2
- 3(e − 1)/2
- 41/2
- 5½ ln 2
- 6(5√5 − 1)/3
- 7tan⁻¹1 − tan⁻¹0 = π/4
- 8[½ ln(1 + x²)] from 0 to 1 = ½ ln 2
Properties of definite integrals
A walk of 3 km east and then 3 km west ends where it started. Even and odd symmetry, reversal and splitting make many definite integrals nearly free.
A definite integral can be reversed, split at an inner point, and simplified by symmetry. The "king" property replaces x by a + b − x. For even functions the two halves of the graph about the y-axis are equal, and for odd functions they cancel.
Formula: ∫ from −a to a of f(x) dx = 2 ∫ from 0 to a of f(x) dx if f is even, and 0 if f is odd
Rule 1: Reverse and split — Reversing the limits changes the sign.
- Easy: ∫ from 2 to 0 of x dx = −∫ from 0 to 2 of x dx = −2.
- Medium: If ∫ from 0 to 2 of f = 5 and ∫ from 2 to 5 of f = 3, then ∫ from 0 to 5 of f = 8.
- Exam-type: ∫ from 0 to 4 of |x − 2| dx = ∫ from 0 to 2 of (2 − x) dx + ∫ from 2 to 4 of (x − 2) dx = 2 + 2 = 4.
Rule 2: Even, odd and the king property — Test f(−x) first, and f(a + b − x) next.
- Easy: ∫ from −2 to 2 of x³ dx = 0, because x³ is odd.
- Medium: ∫ from −1 to 1 of x² dx = 2 ∫ from 0 to 1 of x² dx = 2/3.
- Exam-type: ∫ from 0 to π/2 of sin x/(sin x + cos x) dx = π/4. The king property gives the same integral with cos in the numerator, and the sum of the two is π/2.
An odd function over symmetric limits gives zero. An even function gives twice the half-range integral.
Calling x² + x an even function, or writing 0 for an even function over symmetric limits.
Even doubles, odd cancels.
Practice questions
- 1Evaluate ∫ from −2 to 2 of x³ dx.
- 2Evaluate ∫ from −1 to 1 of x² dx.
- 3If ∫ from 0 to 2 of f = 5 and ∫ from 2 to 5 of f = 3, find ∫ from 0 to 5 of f.
- 4Evaluate ∫ from −1 to 1 of |x| dx.
- 5Evaluate ∫ from −π/2 to π/2 of x⁵ cos x dx.
- 6Evaluate ∫ from 0 to π/2 of sin x/(sin x + cos x) dx.
- 7Evaluate ∫ from 2 to 5 of √x/(√x + √(7 − x)) dx.
- 8Evaluate ∫ from 0 to 4 of |x − 2| dx.
Show answers (try them yourself first)
- 10 (odd function)
- 22 ∫ from 0 to 1 of x² dx = 2/3
- 35 + 3 = 8
- 42 ∫ from 0 to 1 of x dx = 1
- 50, because x⁵ cos x is odd
- 6π/4
- 7(5 − 2)/2 = 3/2 (a + b = 7)
- 84