Question

Fundamental theorem of calculus

`g(s)=\int\limits^s_6 {(t-t^4)^6} \, dt`

Answer 1

`\displaystyle g'(s) = (s-s^4)^6`

Explanation:

The Fundamental Theorem of Calculus states that:

`\displaystyle (d)/(dx)\left[ \int_a^x f(t)\, dt \right] = f(x)`

Where a is some constant.

We can let:

`\displaystyle g(t) = (t-t^4)^6`

By substitution:

`\displaystyle g(s) = \int_6^s g(t)\, dt`

Taking the derivative of both sides results in:

`\displaystyle g'(s) = (d)/(ds)\left[ \int_6^s g(t)\, dt\right]`

Hence, by the Fundamental Theorem:

`\displaystyle \begin{aligned} g'(s) & = g(s) \\ \\ & = (s-s^4)^6\end{aligned}`