Answer
\begin{aligned}
& \text{Question: Let } F(x) = \int_{2}^{x} e^{t^2} dt \text{, find the equation of the tangent line to the curve } y = F(x) \text{ at the point } x = 0. \\
& \text{Answer: } y =e^4(x-2)
\end{aligned}
Work Step by Step
\begin{aligned}
& \text{The given question is like that:} \\
& \text{Let } F(x) = \int_{2}^{x} e^{t^2} dt \text{, find the equation of the tangent line to the curve } y = F(x) \text{ at the point } x = 0. \\
& \text{So let's start by noticing that when } x \text{ is } 2 \text{, } F(2) = 0 \text{ because integral from } 2 \text{ to } 2 \text{ is } 0. \\
& \text{So the tangent line equation is like that:} \\
& (y-y_0)=F'(x_0)(x-x_0) \\
& \text{So here } y_0 = F(2) = 0 \text{ and } x_0 = 2, \\
& \text{and by fundamental theorem of calculus, the derivative of that integral is } e^{x^2} \\
& \text{so derivative at the point } x = 2 \text{ is equal to } e^4. \\
& \text{And when we plug all this in tangent line equation we get:} \\
y &=e^4(x-2)
\end{aligned}
