First Fundamental Theorem of Calculus Domain Suppose $F:[-1,4]\to\Bbb R$, $F(x):=\int_{-3}^{2x^4+x^2+1}{e^{-t}}$, $x \in [-1,4]$. I want to find a formula for the derivative without the integral symbol.
So I know how to do this by splitting the integral up to give: $F(x):=\int_{0}^{2x^4+x^2+1}{e^{-t}} =\int_{0}^{2x^4+x^2+1}{e^{-t}} +  \int_{-3}^{0}{e^{-t}}$. Then the second integral has derivative $0$ and then you define a composite function for the first integral with upper limit $x$ and a function mapping $x$ to $2x^4+x^2+1$. My question is that I know I have a problem because I need to prove the functions are continuous on $[-1,4]$ but the lower limit $-3$ is outside the domain.
So what do I do ? Do I say the integral can't be evaluated since $-3$ is outside the domain, or do I split the $\int_{-3}^{0}{e^{-t}} = \int_{-1}^{0}{e^{-t}}+\int_{-3}^{-1}{e^{-t}}$ and ignore the $\int_{-3}^{-1}{e^{-t}}$ ?
Cheers.
 A: Let $G:[-1,1]\to\Bbb R$, $G(x):=$ $\int_{-5}^{2x^4+x^2+1}{e^{-t^4}} \;dt$, for $x \in [-1,1].$
Solution: Let $C:[-1,1]\to\Bbb R$ be given by $C(x):=\int_{-5}^{x}{e^{-t^4}}$ for all $x \in [-1,1]$. The function $c:[-1,1]\to\Bbb R$ given by $c(t)=e^{-t^4}$ for all $t \in [-1,1]$. The function $c$ is continuous as a composition of $t\mapsto e^t$ and $t\mapsto -t^4$ where first is an elementary function and second is a polynomial.
Hence, apply Fundamental Theorem of Calculus for continuous functions, which implies that $C$ is differentiable with $C'(x) = c(x) = e^{-x^4}$ for all $x\in[-1,1]$.
Now, observe that $G(x) = C(2x^4+x^2+1)$, so $G=C\circ g$, where $g(x):=2x^4+x^2+1$ is differentiable as a polynomial. Also, $C$ is differentiable by previous discussion using Fundamental Theorem of Calculus, so the chain rule implies that $G$ is differentiable with derivative:
$G'(x) = C'(g(x))g'(x)$ = $({e^{-(2x^4+x^2+1)^4}})(8x^3+2x)$ for all $x$ $\in [-1,1]$ as required. $\square$
Question: $-5$ is not in the domain $[-1,1]$, so what do I do ?
A: Find a formula for the derivative function of $G:[-1,1]\to\Bbb R$, $G(x):=$ $\int_{-5}^{2x^4+x^2+1}{e^{-t^4}} \;dt$, for $x \in [-1,1].$
Solution: Let $C:\Bbb R\to\Bbb R$ be given by $C(x):=\int_{-5}^{x}{e^{-t^4}}$ for all $x \in \Bbb R$. The function $c:\Bbb R\to\Bbb R$ given by $c(t)=e^{-t^4}$ for all $t \in \Bbb R$. The function $c$ is continuous as a composition of $t\mapsto e^t$ and $t\mapsto -t^4$ where first is an elementary function and second is a polynomial.
Hence, apply Fundamental Theorem of Calculus for continuous functions, which implies that $C$ is differentiable with $C'(x) = c(x) = e^{-x^4}$ for all $x\in \Bbb R$.
Now, observe that $G(x) = C(2x^4+x^2+1)$, so $G=C\circ g$, where $g(x):=2x^4+x^2+1$ is differentiable as a polynomial. Also, $C$ is differentiable by previous discussion using Fundamental Theorem of Calculus, so the chain rule implies that $G$ is differentiable with derivative:
$G'(x) = C'(g(x))g'(x)$ = $({e^{-(2x^4+x^2+1)^4}})(8x^3+2x)$ for all $x$ $\in [-1,1]$ as required. $\square$
**@AnneBauval you misunderstood the original question I think before because the domain of $G$ is already defined in the question to be [-1,1]. Is it okay to define $C$ and $c$ to have domains of $\Bbb R$ since I have defined them myself? *
