Calculus II: Use a power series to approximate.

[BlueKai]

I've been stuck on this problem for the longest. Can anybody help me solve this:

Problem: Use a power series to approximate the value of the integral with an error of less than 0.0001. Find the number of terms needed to produce a value less than 0.0001

. . . . .\(\displaystyle \displaystyle{ \int_0^1 \, e^{-x^3}\, dx }\)

Referring to the Power Series for Elemental Functions Table I used e^x = x^n/n!

I substituted -x^3 to get: -x^3n/n! then integrated that to get

. . . . .\(\displaystyle \displaystyle{ \frac{-(1)^{3n+1}}{(3n\, +\, 1)n!} }\)

But that summation is wrong according to this Pearson Lab

How do I approximate this correctly.

 

[Ishuda]

It looks like you were headed in the correct direction
\(\displaystyle \int_0^1 e^{-x^3} dx = \int_0^1 \Sigma (-1)^n \frac{x^{3n}}{n!} dx \)
\(\displaystyle = \Sigma \frac{(-1)^{3n}}{n!} \int_0^1 x^{3n} dx \)
\(\displaystyle = \Sigma \frac{(-1)^n}{n!} \frac{x^{3n+1}}{3n+1}|_0^1 \)
\(\displaystyle = \Sigma \frac{(-1)^n}{n!} \frac{1}{3n+1} \)
Since this is an alternating series with decreasing terms, the error at stopping at the nth term is less than or equal to the magnitude of the first dropped term so we want
\(\displaystyle \frac{1}{n!} \frac{1}{3n+1} < 0.0001 \)

 

[Ishuda]

I've been stuck on this problem for the longest. Can anybody help me solve this:

Problem: Use a power series to approximate the value of the integral with an error of less than 0.0001. Find the number of terms needed to produce a value less than 0.0001

. . . . .\(\displaystyle \displaystyle{ \int_0^1 \, e^{-x^3}\, dx }\)

Referring to the Power Series for Elemental Functions Table I used e^x = x^n/n!

I substituted -x^3 to get: -x^3n/n! then integrated that to get

. . . . .\(\displaystyle \displaystyle{ \frac{-(1)^{3n+1}}{(3n\, +\, 1)n!} }\)

But that summation is wrong according to this Pearson Lab

How do I approximate this correctly.

Your expression looks correct to me although I would simplify it as I did above

 

[BlueKai]

Your expression looks correct to me although I would simplify it as I did above

Thank you so much Ishuda !
I was stuck on that one problem for hours