# Domain Integration Calculus

Suppose $$G:[-1,1]\to\Bbb R$$, $$G(x):=\int_{-5}^{2x^4+x^2+1}{e^{-t^4}}$$, $$x \in [-1,1]$$. I want to find a formula for the derivative without the integral symbol.

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 $$C:[-5,4]\to\Bbb R$$ be given by $$C(x):=\int_{-5}^{x}{e^{-t^4}}$$ for all $$x \in [1,4]$$. 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 $$x\in \Bbb R$$.

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. $$\square$$

I am really confused about the domain. I know the upper endpoint of the integral is for $$[1,4]$$. Can someone please help me and edit my solution where the domain / $$x$$-values are wrong. I think I am getting confused between the domain of the function and values of the integral. It's really annoying me because I can't seem to understand for two weeks. Can I set all $$x\in\cdots$$ and $$[]$$ to $$\Bbb R$$??? I don't think I can because in the question $$G$$ is defined for $$[-1,1]$$ where $$x \in [-1,1]$$.

My definition for the Fundamental Theorem of Calculus. If $$f\colon[a,b] \to \Bbb R$$ is continuous and $$F\colon [a,b] \to \Bbb R$$ is given by $$F(x):=\int_{a}^{x}f$$ for all $$x\in[a,b]$$, then $$F$$ is differentiable with derivative $$F'(x)=f(x)$$ for all $$x \in [a,b]$$.

• @JonathanZsupportsMonicaC Edited. It should be $G$ the whole way through with the lower limit being $-5$.
– user1071088
Commented Nov 27, 2022 at 20:17
• Why do you care about the domain ? You can extend the function $G$ to the function $\tilde{G}(x)$ defined on the whole real line by the same formula. Then $(\tilde{G})'(x)=e^{-x^4}(8x^3+2x)$ for any $x.$ Hence $G'(x)=(\tilde{G})'(x)$ for $-1<x<1$ and $G'_-(1)=(\tilde{G})'(1),$ $G'_+(-1)=(\tilde{G})'(-1).$ Commented Nov 27, 2022 at 21:36
• @RyszardSzwarc Because look at my definition for the fundamental theorem of calculus which I must use in the question. The question already defines the domain of $G$ to be $[-1,1]$. I must use this
– user1071088
Commented Nov 27, 2022 at 21:50
• The function $C$ in your answer can be defined for every $x.$ Then $G(x)=C(2x^4+x^2+1)$ for $-1\le x\le 1.$ Let $H(x)=C(2x^4+x^2+1)$ for $x\in\mathbb{R}.$ Then $G(x)=H(x)$ for $-1\le x\le 1.$ Hence $G'(x)=H'(x)$ for $-1<x<1.$ By the fundamental theorem of calculus $H'(x)=e^{-x^2}(8x^3+2x).$ Hence $G'(x)=e^{-x^2}(8x^3+2x)$ for $-1<x<1.$ Commented Nov 27, 2022 at 22:02
• @RyszardSzwarc okay can please copy my post with your edits for a complete answer so I can accept it. It would also make more sense to me (also you used strict inequalities sometimes)
– user1071088
Commented Nov 27, 2022 at 23:30

The function $$G:[-1,1]\to \mathbb{R}$$ is defined by the formula $$G(x)=\int\limits_{-5}^{2x^4+x^2+1}e^{-t^4}\,dt$$ The function $$G$$ is not defined off the interval $$[-1,1],$$ or it may defined there by the same or by another formula. It will be irrelevant for the solution.
As we are dealing with the closed interval, we may calculate $$G'(x)$$ for $$-1 and $$G'_+(-1)$$ and $$G'_-(1).$$ We have to use one sided derivatives at the endpoints of the interval, as we have no information on the behaviour of $$G$$ for $$x<-1$$ and for $$x>1.$$
Consider the function $$H:\mathbb{R}\to \mathbb{R}$$ defined by the same formula as $$G$$ $$H(x)=\int\limits_{-5}^{2x^4+x^2+1}e^{-t^4}\,dt$$ but with no restriction on $$x.$$ By the fundamental theorem of calculus and the chain rule, we get $$H'(x)=e^{-(2x^4+x^2+1)^4}(8x^3+2x)$$ Clearly $$G(x)=H(x),$$ for $$|x|\le 1.$$ Therefore $$G'(x)=H'(x),\ |x|<1,\ G'_+(-1)=H'(-1),\quad G'_-(1)=H'(1)$$ Thus $$G'(x)=e^{-(2x^4+x^2+1)^4}(8x^3+2x),\ -1 and $$G'_+(-1)=-10e^{-64}$$,$$\$$ $$G'_-(1)=10e^{-64}$$