## Definite Integrals

The integrals that we have calculated up until this point have all been indefinite integrals as they contain an arbitrary constant and are therefore not fully defined.

The definite integral is used when estimating the area under a curve. Since this process involves the subtraction of two points, the arbitrary constant is cancelled out and is unimportant.

The formula for the definite integral is: $\int\limits_{a}^{b}{f(x)}dx=\lim_{\delta x\to\infty}\sum\limits_{i=1}^{n}\times\delta x_{i}$

The first part of this equation is the definite integral. The second part shows what the definite integral does which is essentially partition the area under the curve f(x) into n equal sub intervals with the length δx. Each of these sub intervals are then added up as thin rectangles to calculate the area under the curve.  The smaller the sub intervals get the greater number of rectangles and the more accurate the estimate of area under the curve

### Definite Integral Example

Question 2 b from exam 1 in 2010.

Find p given that $\int\limits_{2}^{3}{\frac{1}{1-x}}=log_{e}(p)$

Step 1 we integrate the equation using our power integration, making the denominator the same as what is in the log brackets and divide by the derivative of the denominator which in this case is negative one. $\int\limits_{2}^{3}{\frac{1}{1-x}}\newline = [-log_{e}(|1-x|)]^{3}_{2}$

Make sure to leave the absolute signs within the log like the formula says to; otherwise we will get logs of negatives and no result. Also don’t forget to keep the 3 & 2 at the end of the brackets as this reminds us we need to sub in these values and minus them from each other, making it a definite integral.

For this same reason we can also leave out any arbitrary constants as they would just cancel out when we subtract the two values from one another. Doing this substitution and subtraction phase gives us: $(-log_{e}(|1-3|)-(-log_{e}(|1-2|)\newline=-log_{e}(|-2|)-0 \newline=log_{e}(2^{-1})=log_{e}(\frac{1}{2})$

Then our answer is what is inside the brackets: $p=\frac{1}{2}$

You can see that to complete a question like this we need to have a very good handle on a lot of different basics even to just complete one question. In this question we needed to know logarithmic integration, definite integrals and log laws to get the correct answer. It is important to know all the topics not just a few!