Express the definite integral as an infinite series and find its value to within an error of at most 10^-4

the integral of e^(-x^3) dx from 0 to 1. It has something to do with Taylor Series. I don't know how to bond the error.

Start with e^t = Σ(n = 0 to ∞) t^n/n!.

Let t = -x^3:
e^(-x^3) = Σ(n = 0 to ∞) (-1)^n x^(3n)/n!.

Integrate both sides from 0 to 1:
∫(x = 0 to 1) e^(-x^3) dx
= Σ(n = 0 to ∞) (-1)^n x^(3n+1)/[(3n+1) n!] {for x = 0 to 1}
= Σ(n = 0 to ∞) (-1)^n/[(3n+1) n!].
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Since this is an alternating series, the error after n terms is bounded above by the absolute value of the following (n+1)th term.

So, we need 1/[(3n+4) (n+1)!] < 10^(-4)
==> (3n+4) (n+1)! > 10^4
==> n = 6 (or higher).

So, ∫(x = 0 to 1) e^(-x^3) dx
≈ Σ(n = 0 to 6) (-1)^n/[(3n+1) n!]
≈ 0.8075.

I hope this helps!

Compute the taylor series for the function. To *bound the error, use enough terms in the taylor series to give you an error less than 0.0001, compare the taylor series expansion with the computed integral.

Seek a tutor, practice easier problems and work your way up.