Integral calculus

Integral calculus is the study of areas under curves. If the area under a curve is split into strips, each of length , so the area of one of the strips is . Then the total area under the curve from to is

$$\lim_{\delta x \to 0} \sum_{a}^{b} y_i \delta x$$

In calculus notation, this is written

$$\int_{a}^{b} y \mathrm{d}x$$

The above is a definite integral which gives the area between the curve and the -axis. Integration is, in fact, the opposite of differentiation, and an indefinite integral gives the antiderivative of the gradient function, i.e. , where is a constant. The constant of integration is necessary since a constant always differentiates to . Also . As differentiation, integration is linear, i.e.

$$\int \biggl( a f(x) + b g(x) \biggr) \mathrm{d}x = a \int f(x) \mathrm{d}x + b \int g(x) \mathrm{d}x$$

To evaluate a definite integral, we use the fundamental theorem of calculus which states that if ,

$$\int_{a}^{b} f(x) \mathrm{d}x = \Biggl [ \int f(x) \mathrm{d}x \Biggr ]_{a}^{b} = F(b) - F(a)$$

The indefinite integral evaluated at gives the area between and , where $p$ is a point depending on the integrand. Hence,

$$\int_{a}^{b} f(x) \mathrm{d}x = \int_{p}^{b} f(x) \mathrm{d}x - \int_{p}^{a} f(x) \mathrm{d}x = F(b) - F(a)$$

This justifies the previous statement. (This is obvious geometrically.)

To derive standard integrals, we simply reverse the differentiation process. For example

$$\frac{\mathrm{d}}{\mathrm{d}x} x^n = nx^{n-1}$$

Integrating gives

$$x^n = n \int x^{n-1} \mathrm{d}x + c$$

(since integration is also linear) so that

$$\int x^n \mathrm{d}x = \frac{x^{n+1}}{n+1} + c$$

Other results can be derived in a similar way. To find the area between the curve , the -axis, and the lines and , we need to evaluate

$$\int_{0}^{\pi/2} \sin{x} \mathrm{d}x$$

This is done as follows

$$\begin{eqnarray*} \int_{0}^{\pi/2} \sin{x} \mathrm{d}x & = & \Biggl [ -\cos{x} \Biggr ]_{0}^{\pi/2} & = & -\cos{\frac{\pi}{2}} + \cos{0} = 1 \end{eqnarray*}$$