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Integration and areas
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Integration reverses differentiation. A definite integral gives the area under a curve.
Integrating powers
∫xn dx = xn + 1(n + 1) + c (n ≠ −1)
Add 1 to the power, divide by the new power, add c.
∫(6x2 + 4x) dx = 2x3 + 2x2 + c
Add 1 to the power, divide by the new power, add c.
∫(6x2 + 4x) dx = 2x3 + 2x2 + c
sin, cos and e
∫cos ax dx = 1a sin ax + c
∫sin ax dx = −1a cos ax + c
∫eax dx = 1aeax + c
∫sin ax dx = −1a cos ax + c
∫eax dx = 1aeax + c
Definite integrals and area
∫ab f(x) dx = [F(x)]ab = F(b) − F(a)
The area between a curve, the x-axis and x = a, x = b is ∫ab y dx.
Area below the x-axis comes out negative, so find those parts separately.
The area between a curve, the x-axis and x = a, x = b is ∫ab y dx.
Area below the x-axis comes out negative, so find those parts separately.
Find the area enclosed by y = 4x − x2 and the x-axis.
- Roots: x = 0 and x = 4
- ∫04 (4x − x2) dx = 2x2 − x3<span>304
- 32 − 643 = 323
Answer: 323 square units
Add 1 to the power and divide by it. cos ax → 1a sin ax, sin ax → −1a cos ax, eax → 1aeax. Area = ∫ab y dx.
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