Question

integral e^(-3t^(4))

a=0 and b=x. find the maclaurin polynomial of degree 5.

Answer 1

First, formulate a series expansion.
\[e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \Rightarrow e^{-3t^4}=\sum_{n=0}^\infty \frac{(-3t^4)^n}{n!}=\sum_{n=0}^\infty \frac{(-3)^nt^{4n}}{n!}\]Next, integrate term-by-term.\[\int e^{-3t^4}dt = \int \left( \sum_{n=0}^\infty \frac{(-3)^nt^{4n}}{n!} \right) dt = \sum_{n=0}^\infty \int \frac{(-3)^nt^{4n}}{n!} dt = \boxed{\displaystyle C+\sum_{n=0}^\infty \frac{(-3)^nt^{4n+1}}{n!(4n+1)}}\]I'm not sure what you mean by \(a=0\) and \(b=x\), but this is a start, I hope.

Answer 2

thank you so much. I meant this is definite integral with a=0 and b=x