 # Chris Fox's Engineering Section

## Calculus: Integration

The integral of a function f(x) is denoted in the form ∫ f(x) dx. f(x) dx is the area of an infinitessimally thin rectangle under the graph at a point x, and the symbol ∫ (archaic S) denotes that the areas of a continuous series of these rectangles are to be summed.

Imagine that the function to be integrated is the derivative of another function y(x); in other words, dy/dx. The area of each rectangle would therefore be (dy/dx)dx, or dy. Summing them gives y. Differentiation and integration are therefore mutually inverse processes.

Because parallel lines have the same gradient, an integral will have an infinite number of results. A constant, denoted by c, is therefore added to the result. To find the value of the constant, insert the values of the variables for a known point.

The constant is not needed in the case of a definite integral, which shows the values of x between which the function is to be integrated (An integral that omits those values is an indefinite integral). A definite integral is written in the form ∫ab f(x) dx, where a is the lower value (a and b should be in line). On an x-y graph where y is the integral, the difference between the values of y corresponding to two values of x is fixed, regardless of the position of the line along the y-axis.

The First Principles method can also be used for integration. Consider a rectangle under the graph between x and x+h. If the areas under the graph to these points, from x=0, are given by A(x) and A(x+h) respectively, the area of the rectangle can be expressed as either A(x+h)-A(x) or hy(x), where y(x) is the function to be integrated. Therefore, {A(x+h)-A(x)}/h=y(x).

The product of two functions can be integrated using the Integration By Parts method, the inverse of the Product Rule. If the functions are u and dv/dx, then ∫u(dv/dx)dx=uv-∫v(du/dx)dx. Note that dv/dx needs to be integrated to obtain v.

The function u should be one where x disappears on differentiation, for example a power of x. v(du/dx) should then be a single function of x. If it is a compound, repeat the process on this part of the formula until only a single function remains to be integrated.

Integration By Parts can be used to integrate inverse functions. Simply introduce 1 as a new factor; that then becomes the factor to be integrated.

If the function to be integrated is complex, it can be simplified by introducing a new variable t, where t=f(x). dt is inserted in place of (dt/dx)dx, and the function integrated. Remember to re-introduce x.