Integration

1.8 Integration

Reverse differentiation

Integration undoes differentiation. Add 1 to the power and divide by the new power.

The constant $c$ appears because any constant differentiates to zero. Rewrite roots and fractions as powers first, exactly as for differentiation.

$\int \left(6x^2 - 4x^{-3} + x^{\frac{1}{2}}\right)\mathrm{d}x = 2x^3 + 2x^{-2} + \frac{2}{3}x^{\frac{3}{2}} + c$

Integrating $(ax + b)^n$

The extra division by $a$ undoes the factor the chain rule would bring out.

$\frac{(2x + 1)^{\frac{3}{2}}}{2 \times \frac{3}{2}} + c = \frac{1}{3}(2x + 1)^{\frac{3}{2}} + c$

Finding the constant

If you know a point on the curve, substitute it to find $c$.

$y = 2x^3 - 2x^2 + c$. At $(1, 5)$: $5 = 2 - 2 + c$, so $c = 5$ and $y = 2x^3 - 2x^2 + 5$.

Definite integrals

$\left[\frac{1}{3}(2x + 1)^{\frac{3}{2}}\right]_0^4 = \frac{1}{3}\left(9^{\frac{3}{2}} - 1^{\frac{3}{2}}\right) = \frac{1}{3}(27 - 1) = \frac{26}{3}$

Area under a curve

The area between a curve, the $x$-axis and the lines $x = a$ and $x = b$ is $\displaystyle\int_a^b y\,\mathrm{d}x$. Using $\displaystyle\int_c^d x\,\mathrm{d}y$ gives the area between a curve and the $y$-axis.

$\int_1^3 (x^2 + 1)\,\mathrm{d}x = \left[\frac{x^3}{3} + x\right]_1^3 = (9 + 3) - \left(\frac{1}{3} + 1\right) = \frac{32}{3}$

Area between two curves

$\text{Area} = \int_a^b (\text{upper} - \text{lower})\,\mathrm{d}x$

This works wherever the region lies relative to the axis.

They meet where $4x - x^2 = x$, so $x = 0$ or $x = 3$. Between these, the curve is above the line:

$\int_0^3 (3x - x^2)\,\mathrm{d}x = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3 = \frac{27}{2} - 9 = \frac{9}{2}$

Volumes of revolution

Rotating a region through 360° about an axis gives a solid. Each thin slice is a disc of radius $y$ (or $x$).

$V = \pi\int_0^4 x\,\mathrm{d}x = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi$

$x^2 = y$, so

$V = \pi\int_1^4 y\,\mathrm{d}y = \pi\left[\frac{y^2}{2}\right]_1^4 = \pi\left(8 - \frac{1}{2}\right) = \frac{15\pi}{2}$