Approximating Integrals with Taylor Series

Evaluate the integral .

solution
This integral cannot be evaluated exactly. We have to approximate the value with a Taylor polynomial, and we will use the substitution method.

Let and .

which gives:

So,

Compare this to the actual value (to 6 d.p) of .

Evaluate the integral

solution
There is no antiderivative for . We have to approximate with a Taylor polynomial, and we will use the substitution method

then let

So,

What's our error?

where

Since we have , the largest value we can get is , so that's what we will use.

however, this is the error of the polynomial. We want the error of the integral.

Going back to our integral:

For example, evaluating at :

compare this to the actual value of .

In fact, we overestimated the error (true error is ).

Suppose we have a circle whose are is defined by the equation .

The area of the shaded part is:

Using integrals:

putting it all together: