Integration and areas

Maths: Applications and Interpretation · Calculus. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Integration is the reverse of differentiation. If you know the derivative, integrating takes you back to the function. For a power of x, raise the power by one and divide by the new power. Because every constant differentiates to zero, you must add a constant of integration, C. A known point on the curve, called a boundary condition, lets you find C.

A definite integral has a lower and an upper limit and gives a number, not a function. When the curve is above the x-axis between the limits, this number is the area between the curve and the x-axis. You may evaluate definite integrals on your calculator.

Rules to remember

∫ axⁿ dx = ax^{n + 1} ÷ (n + 1) + C, for n ≠ −1
∫ a dx = ax + C (a constant integrates to ax)
Integrate a sum one term at a time
To find C: substitute the given point (x, y) into the integrated function
Definite integral from a to b: [F(x)] = F(b) − F(a), with no + C
Area under y = f(x) from x = a to x = b, where f(x) > 0, is the definite integral of f(x) from a to b
Integrating a rate gives the total change, e.g. ∫ dV/dt dt gives volume

Common mistake

Students divide by the old power instead of the new one, or forget + C. Check by differentiating your answer: you must get back the function you started with.

Worked example

Find the area between the curve y = 3x² + 2, the x-axis and the lines x = 1 and x = 3.

The curve is above the x-axis, so area = ∫ (3x² + 2) dx from 1 to 3
Integrate: x³ + 2x
At x = 3: 27 + 6 = 33; at x = 1: 1 + 2 = 3
Subtract: 33 − 3 = 30

Answer: Area = 30 square units.

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Topic 5: Calculus

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