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How to Calculate Integrals
Integration undoes differentiation. Here are the rules and techniques that solve most textbook problems.
An integral answers the question: which function has this derivative? That reverse function is called the antiderivative. Integrals also measure area: the definite integral of f(x) from a to b is the area between the graph and the x-axis over that interval, with area below the axis counted as negative.
Indefinite integrals and the + C
Because the derivative of any constant is zero, x², x² + 5 and x² − 100 all have the derivative 2x. So every indefinite integral ends with + C, meaning “plus any constant.” Forgetting the C is one of the most common ways to lose marks.
The power rule for integrals
Add one to the exponent, then divide by the new exponent:
Example: Integrate 3x² + 2x.
3·x³/3 + 2·x²/2 = x³ + x² + C. Check by differentiating: you get 3x² + 2x back.
Try it in CasCalc →The one exception is n = −1: the integral of 1/x is ln|x| + C.
Common integrals
| Function | Integral |
|---|---|
| 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 |
u-substitution
Substitution is the chain rule in reverse. Look for a function and its derivative appearing together. Call the inside function u, replace du, and integrate the simpler expression.
Example: Integrate 2x·cos(x²).
Let u = x². Then du = 2x dx, which is exactly the rest of the expression. The integral becomes ∫ cos(u) du = sin(u) + C.
Put x back in: sin(x²) + C.
Try it in CasCalc →Integration by parts
For a product of two different kinds of function, such as x times ex, use the product rule in reverse:
Example: Integrate x·ex.
Choose u = x (it gets simpler when differentiated) and dv = ex dx. Then du = dx and v = ex.
∫ x ex dx = x ex − ∫ ex dx = x ex − ex + C = (x − 1)ex + C.
Try it in CasCalc →Definite integrals
To evaluate a definite integral, find the antiderivative F(x), then subtract its values at the two limits:
Example: Evaluate the integral of x² from 0 to 1.
The antiderivative is x³/3, so the answer is 1³/3 − 0³/3 = 1/3. That is the area under the parabola between x = 0 and x = 1.
Try it in CasCalc →Example: Evaluate the integral of sin(x) from 0 to π.
The antiderivative is −cos(x), so the answer is −cos(π) − (−cos(0)) = 1 + 1 = 2. One arch of the sine wave has an area of exactly 2.
Try it in CasCalc →There is no + C in a definite integral, because the constant cancels when you subtract.
Always check by differentiating
Integration is harder than differentiation, but checking is easy: differentiate your answer and make sure you get the original function back.
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