TheIBAxolotl
Undoing the derivative
If differentiation answers "what is the gradient?", integration answers the reverse: "which function has this gradient?". That function is the antiderivative. Since \(\frac{d}{dx}(x^3) = 3x^2\), an antiderivative of \(3x^2\) is \(x^3\).
But \(x^3 + 5\) and \(x^3 - 2\) also have derivative \(3x^2\): constants disappear when differentiating. So the general answer carries a constant of integration \(C\):
\[\int 3x^2 \, dx = x^3 + C.\]
The reverse power rule
To integrate \(ax^n\), do the opposite steps of the power rule in the opposite order: add one to the power, then divide by the new power.
\[\int ax^n \, dx = \frac{a x^{n+1}}{n + 1} + C, \qquad n \ne -1.\]
(The case \(n = -1\) needs \(\ln x\), covered later.) Integrate sums term by term, and rewrite fractions as powers first, just like differentiation: \(\int \frac{4}{x^3} dx = \int 4x^{-3} dx = -2x^{-2} + C\).
Worked example: finding C
The gradient of a curve is \(f'(x) = 6x^2 - 4\), and the curve passes through \((1, 5)\). Find \(f(x)\).
- Integrate: \(f(x) = 2x^3 - 4x + C\).
- Use the point: \(f(1) = 2 - 4 + C = 5\), so \(C = 7\).
- \(f(x) = 2x^3 - 4x + 7\).
The given point is called a boundary condition; without it, \(C\) cannot be found.
Definite integrals and area
A definite integral has limits:
\[\int_a^b f(x)\,dx.\]
When \(f(x) \ge 0\) on \([a, b]\), it equals the area between the curve, the \(x\)-axis and the lines \(x = a\) and \(x = b\). Picture slicing the region into thin rectangles of width \(dx\) and height \(f(x)\) and adding them up: the integral sign is a stretched S for "sum".
Example. The area under \(y = 4 - x^2\) from \(x = 0\) to \(x = 2\) (where the curve is above the axis) is
\[\int_0^2 (4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}.\]
Using technology
On Paper 2, the GDC evaluates definite integrals numerically: enter the function and the limits. This is the expected method when the function has no easy antiderivative, and a good check otherwise.
Common mistakes
- Forgetting \(+C\) on indefinite integrals.
- Dividing by the old power instead of the new one.
- Integrating a product by integrating each factor separately; expand first.
- Using \(\int_a^b\) for area when part of the curve is below the axis (see SL 5.11).
Exam tips
- Differentiate your answer mentally to check it gives back the integrand.
- Show the boundary-condition substitution as a separate line.
- On calculator papers, write the integral you entered, with limits, before the value.