Integration is the reverse of differentiation. A derivative takes a function and gives its rate of change. An integral takes a rate of change and gives back the total. If you know a car’s speed at every moment, the integral tells you how far it went. This guide assumes you know the basic derivative rules.
Antiderivatives
An antiderivative of f(x) is a function whose derivative is f(x). Since the derivative of x³ is 3x², an antiderivative of 3x² is x³.
But x³ + 5 and x³ − 12 also have derivative 3x², because the derivative of a constant is 0. So every antiderivative comes with a + C, a constant that could be anything:
∫ 3x² dx = x³ + C
The ∫ sign means “integrate”, and dx says x is the variable. This is called an indefinite integral.
The basic rules
| Function | Integral |
|---|---|
| k (a constant) | kx + C |
| xⁿ (n ≠ −1) | xn+1/(n + 1) + C |
| 1/x | ln∣x∣ + C |
| eˣ | eˣ + C |
| sin x | −cos x + C |
| cos x | sin x + C |
| sec² x | tan x + C |
Two more rules let you handle sums and constants:
- ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx
- ∫ k·f(x) dx = k ∫ f(x) dx
The power rule is the one you will use most: add 1 to the exponent, then divide by the new exponent.
Example Find ∫ (4x³ − 6x + 5) dx
- 4x³ becomes 4 × x⁴/4 = x⁴.
- −6x becomes −6 × x²/2 = −3x².
- 5 becomes 5x.
- Answer: x⁴ − 3x² + 5x + C.
- Check by differentiating: 4x³ − 6x + 5. ✓
Example Find ∫ (√x + 2/x²) dx
- Rewrite with exponents: x1/2 + 2x−2. The exponent rules guide explains why this works.
- x1/2 becomes x3/2/(3/2) = (2/3)x3/2.
- 2x−2 becomes 2 × x−1/(−1) = −2/x.
- Answer: (2/3)x3/2 − 2/x + C.
There is no product rule for integrals
∫ x(x + 1) dx is not (x²/2)(x²/2 + x). Multiply out first: ∫ (x² + x) dx = x³/3 + x²/2 + C.
Definite integrals and area
A definite integral has limits, a lower one and an upper one:
∫ab f(x) dx
When f(x) is positive, this equals the area between the graph and the x-axis from x = a to x = b. Where the graph is below the axis, that area counts as negative.
The Fundamental Theorem of Calculus
The Fundamental Theorem connects the two ideas. To evaluate a definite integral, find any antiderivative F(x), then subtract:
∫ab f(x) dx = F(b) − F(a)
The + C cancels in the subtraction, so you can leave it out.
Example Evaluate ∫03 x² dx
- Antiderivative: x³/3.
- F(3) − F(0) = 27/3 − 0 = 9.
- The area under y = x² from 0 to 3 is 9 square units.
Example Evaluate ∫14 (2x + 1) dx
- Antiderivative: x² + x.
- F(4) = 16 + 4 = 20 and F(1) = 1 + 1 = 2.
- 20 − 2 = 18.
Example A car’s velocity is v(t) = 3t² + 2 meters per second. How far does it travel from t = 0 to t = 2?
- Distance is the integral of velocity: ∫02 (3t² + 2) dt.
- Antiderivative: t³ + 2t.
- (8 + 4) − 0 = 12 meters.
Area between two curves
To find the area between two curves, integrate the top one minus the bottom one between where they cross.
Example Find the area between y = x and y = x² from x = 0 to x = 1
- On this interval, x is above x² (at x = 0.5, x = 0.5 and x² = 0.25).
- ∫01 (x − x²) dx = [x²/2 − x³/3] from 0 to 1 = 1/2 − 1/3 = 1/6.
U-substitution
U-substitution undoes the chain rule. Use it when the integral contains a function and also (a multiple of) that function’s derivative.
Example Find ∫ 2x(x² + 1)⁵ dx
- Let u = x² + 1. Then du = 2x dx.
- The integral becomes ∫ u⁵ du = u⁶/6 + C.
- Substitute back: (x² + 1)⁶/6 + C.
Example Find ∫ x·cos(x²) dx
- Let u = x². Then du = 2x dx, so x dx = du/2.
- ∫ cos u × du/2 = (1/2) sin u + C.
- Answer: (1/2) sin(x²) + C.
The skill is picking u. Look for the “inside” piece of a composite function whose derivative also appears outside.
Practice problems
Find ∫ (6x² + 4x − 7) dx.
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2x³ + 2x² − 7x + C.
Find ∫ (3/x + eˣ) dx.
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3 ln|x| + eˣ + C.
Evaluate ∫02 (x³ + 1) dx.
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6. The antiderivative is x⁴/4 + x, and (4 + 2) − 0 = 6.
Evaluate ∫0π sin x dx.
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2. [−cos x] from 0 to π is −cos π − (−cos 0) = 1 + 1.
Find ∫ 3x²(x³ − 4)⁴ dx.
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(x³ − 4)⁵/5 + C. Let u = x³ − 4, so du = 3x² dx.
Find the area between y = 4 − x² and the x-axis.
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32/3. The curve crosses the axis at x = ±2, and ∫−22 (4 − x²) dx = (8 − 8/3) − (−8 + 8/3) = 32/3.