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?

1 Answer
Dec 24, 2017

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#