Derivative of a definite integral: $F(t) = \int_0^t \sqrt{1-x^8} dx$ I'm preparing for my Calculus 1 exam and I've stumbled across the following exercise, which I am not able to solve. Any help will be appreciated.
a) Find the domain and the derivative of the following function: $\displaystyle  F(t) = \int \limits_0^t \sqrt{1-x^8} dx $ .
b) Find first three nonzero terms of the Maclaurin sequence that is equal to F(t). 
 A: The fundamental theorem of calculus states that if $f$ is a continuous, real-valued funtion defined on $[a,b]$, and $F(x)=\int_{a}^{x} f(t)dt$, then $F'(x)=f(x)$. Then, the derivative of your function should be $\sqrt{1-t^8}$. Your function is only defined on the closed interval $[-1,1] \in \mathbb R$, since any other values render the integrand indeterminate.
The Maclaurin series is a Taylor series centered at 0. We have $$\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$ So, find the first, second, and third derivatives at $a=0$, and plug them into this equation. You already have the first derivative, so set $t=0$. Then, $f^{(1)}(0)=1$, and the first term of the series is $\frac{1}{1!}(t-0)^1=t$. This is the term for $n=1$. The term for $n=0$ is simply zero, since $\int_{0}^{0}\sqrt{1-x^8}dx=0$. I will let you find the terms for $n=2$ and $n=3$.
A: the domain of $\int_0^t \sqrt{1-x^8} \, dx$ is the domain of $\sqrt{1-x^8}$ which is the closed interval $[-1,1].$ by the fundamental theorem of calculus, the derivative $F'(t) = \sqrt{1-t^8}.$
$$F(t) = \int_0^x (1-x^8)^{1/2}\, dx = \int_0^t \left(1 + \frac12 (-x^8) + \dfrac{\frac12 \frac{-1}2}{2}(-x^8)^2 + \dots\right)\, dx\\
=t-\frac1{18}t^9+\frac1{136}t^{17}+\dots $$
A: Try:
\begin{align}
F\left(t\right)&=\int_0^t f\left(x\right)\:dx\tag{1}\\
&= F\left(t\right)-F\left(0\right)\tag{2}\\
&=F\left(t\right)-\int_0^0f\left(x\right)\:dx\tag{3}\\
&=F\left(t\right).\tag{4}
\end{align}
Therefore,
\begin{align}
F'\left(t\right)&=\frac{d}{dt}\int_0^t f\left(x\right)\:dx\tag{5}\\
&=\frac{d}{dt}\left[F\left(t\right)-F\left(0\right)\right]\tag{6}\\
&=\frac{d}{dt}\left[F\left(t\right)-\int_0^0f\left(x\right)\:dx\right]\tag{7}\\
&=\frac{d}{dt}F\left(t\right)\tag{8}\\
&=F'\left(t\right).\tag{9}
\end{align}
In your case, let $f\left(x\right)=\sqrt{1-x^8}$, which gives us
\begin{align}
F\left(t\right)&=\int_0^t \sqrt{1-x^8}\:dx\tag{10}\\
&=\int_0^t f\left(x\right)\:dx,\tag{11}
\end{align}
and therefore
\begin{align}
F'\left(t\right)&=\frac{d}{dt}\int_0^t\sqrt{1-x^8}\:dx\tag{12}\\
&=\frac{d}{dt}\left[F\left(t\right)-\int_0^0\sqrt{1-x^8}\:dx\right]\tag{13}\\
&=f\left(t\right)\tag{14}\\
&=\sqrt{1-t^8}.\tag{15}
\end{align}
