Substituition of the integral limit in the first fundemantal theorem of calculus

Question

As I was reading G. Strang's Calculus book, I came across an example problem I couldn't understand. The topic is the First Fundamental Theorem of Calculus. The form proposed for the theorem is (Picture): $$F( x) \ =\ \int ^{x}_{a} f( t) dt\\$$ After proving this, there is an example problem. $$Let\ F( x) \ =\ \int ^{\sqrt{x}}_{1} \sin(t)dt. Find\ F'( t)$$Book solves the question by (Picture) $$u(x) = \sqrt{x}\ then\ F(x) = \int ^{u(x)}_{1}\sin(t)dt\\F'( x) \ = \sin( u( x))\frac{du}{dx}\\ =\sin( u( x)) \cdot \left(\frac{1}{2} x^{-\frac{1}{2}}\right)\\ =\frac{\sin\sqrt{x}}{2\sqrt{x}} \cdot \\$$My question is that, in it's original form, there is an $x$ in the upper limit of the integral. When we substitute it with $u(x)$, shouldn't we change $F(x)$ with $F(u(x))$ also? How does it stay unchanged?

Answer 1

Well, it’s exactly what you said, just write it in the following way: $F(x) = \int_1^{\sqrt{x}} \sin t \, dt =: G(\sqrt{x})$. Then by the Fundamental Theorem of Calculus, you get that $G’(\sqrt{x}) = \sin(\sqrt{x})$. At the same time, by the chain rule we get that $F’(x) = G’(\sqrt{x}) \frac{d \, \sqrt{x}}{d \, x} = \frac{\sin \sqrt{x}}{2 \sqrt{x}}.$ I hope this sheds some light on the issue. :)

Answer 2

Since $F(x)=\int_1^{\sqrt x}\sin t\,\mathrm dt$, $F\bigl(u(x)\bigr)=\int_1^{\sqrt{\sqrt x}}\sin t\,\mathrm dt$. But that's not the function that you want to differentiate. Instead, it's $F(x)$.