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This unit looks at how to calculate the area bounded by a curve using integration.

The second major component of the Calculus is called integration. This may be introduced as a means of finding areas using summation and limits. We shall adopt this approach in the present Unit. In later units, we shall also see how integration may be related to differentiation.

This unit explain integration as the reverse of differentiation.

A special rule, integration by parts, is available for integrating products of two functions. This unit derives and illustrates this rule with a number of examples.

This unit explain integration by substitution.

The derivative of ln x is 1/x. As a consequence, if we reverse the process, the integral of 1/x is ln x+c. In this unit we generalise this result and see how a wide variety of integrals result in logarithm functions.

We may regard integration as the reverse of differentiation. So if we have a table of derivatives, we can read it backwards as a table of anti-derivatives. When we do this, we often need to deal with constants which arise in the process of differentiation.

Sometimes the integral of an algebraic fraction can be found by first expressing the algebraic fraction as the sum of its partial fractions. In this unit we will illustrate this idea. We will see that it is also necessary to draw upon a wide variety of other techniques such as completing the square, integration by substitution, using standard forms, and so on.

In this unit we are going to look at how we can integrate some more algebraic fractions. We shall concentrate on the case where the denominator of the fraction involves an irreducible quadratic factor. The case where all the factors of the denominator are linear has been covered in the first unit on integration using partial fractions.

This unit explains how trig identities and trig substitutions can help when finding integrals.