Taylor Series with integral notation

Question

I have a question regarding how to write this type of Taylor Series.

Let f be the function given by $$f(x) = \int_0^x cos\sqrt{t} \,dt$$ where $x\geq0$

Write the Taylor series of f about $x=0$

Answer 1

Hint: Evaluate $\cos(\sqrt{t})$ as a Taylor series and then integrate the polynomial (you can do that because the Taylor series is uniformly convergent for any $|t|<\infty$).

Answer 2

Well, you know that the "Taylor series" for function f(x), at x= 0, is given by $f(0)+ f'(0)x+ (f''(0)/2)x^2+ (f'''(0)/3!)x^3+ \cdot\cdot\cdot$, don't you?

Here, f(x) is defined as $f(x)= \int_0^x cos(\sqrt{t})dt$ so obviously f(0)= 0. Then, by the "fundamental theorem of Calculus", $f'(x)= cos(\sqrt{x})$ so that $f'(0)= cos(0)= 1$. The second derivative is the derivative of $cos(x^{1/2})$, $-\frac{1}{2}x^{-1/2}sin(x^{1/2})$ and $f''(0)= 0$, etc.