Integral function with a Gaussian and linear approximation

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Let $F(x)=\int_2^{x^2+x}e^{-t^2}\,dt,\qquad x\in\mathbb{R}.$ (a) Compute $F'(x)$. (b) Find the first-degree polynomial approximation of $F$ near $x=1$ and use it to compute $\lim_{x\to 1}\frac{F(x)}{x-1}.$

Solution

(a) Derivative. By the fundamental theorem of calculus with the chain rule, setting $u(x)=x^2+x$ ($u'(x)=2x+1$) and integrand $e^{-t^2}$, $F'(x)=e^{-(x^2+x)^2}\,(2x+1).$

(b) Linear approximation at $x=1$. At $x=1$ we have $x^2+x=2$, so the upper limit equals the lower one and $F(1)=\int_2^2 e^{-t^2}\,dt=0.$ Moreover $F'(1)=e^{-(2)^2}\,(2\cdot 1+1)=3e^{-4}.$ The first-degree (Taylor) approximation is $F(x)\approx F(1)+F'(1)(x-1)=3e^{-4}(x-1).$

Limit. Since $F(1)=0$, the ratio is the definition of the derivative: $\lim_{x\to 1}\frac{F(x)}{x-1}=\lim_{x\to 1}\frac{F(x)-F(1)}{x-1}=F'(1)=3e^{-4}\approx 0.0549.$

$\boxed{F'(x)=e^{-(x^2+x)^2}(2x+1),\qquad \lim_{x\to 1}\frac{F(x)}{x-1}=3e^{-4}\approx 0.0549}$