Question

Fundamental theorem of calculus %20%5C%2C%20dt" id="TexFormula1" title="g(s)=\int\limits^s_6 {(t-t^4)^6} \, dt" alt="g(s)=\int\limits^s_6 {(t-t^4)^6} \, dt" align="absmiddle" class="latex-formula">

Answer 1

Answer:

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

Step-by-step explanation:

The Fundamental Theorem of Calculus states that:
$\displaystyle \frac{d}{dx}\left[ \int_a^x f(t)\, dt \right] = f(x)$

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:
$\displaystyle g'(s) = \frac{d}{ds}\left[ \int_6^s g(t)\, dt\right]$

Hence, by the Fundamental Theorem:

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