Question

Estimate `int_0^1 e^(-2x) dx` within an error of 0.01?

So I know how to set up a Taylor Polynomial w/a given number of derivatives and find the error...but not sure how to go the other way around. How do you figure out which derivative to go till?

Answer 1

See below.

Explanation:

`e^(-2x) = sum_(0)^oo (-1)^k((2x)^k)/(k!)` then

and `int_0^x e^(-2x) dx = -1/2 (e^(-2x)-1) = -1/2(sum_(0)^oo (-1)^k((2x)^k)/(k!)-1) = -1/2sum_1^oo (-1)^k((2x)^k)/(k!)`

This series is an alternate convergent series so the truncation error is less or equal to the last term considered.

Now making

`1/2(2xx1)^k/(k!) le 0.01`

we have

`k = 8` because `2^7/(7!) = 0.02539` and `2^8/(8!) = 0.00635`

Then the approximate integral can be computed as

`-1/2sum_(k=1)^8(-1)^k2^k/(k!) = 136/315` and

`int_0^1 e^(-2x) dx -136/315 = 0.000586327`