# use fundamental theorem of calculus to find a function $f(x)$ and a number $a$

I thought I understood the fundamental theorem of calculus but I'm confused on the following problem..

Use the Fundamental Theorem of Calculus to find a function $f(x)$ and a number $a$ so that $a+\int_{4}^{x}\frac{f(t)}{t^2}dt=2\sqrt{x}$ for all $x>0$.

I don't have the answer to check but what I did was take the derivative of both sides of the equation which means $a$ could be any number because it's derivative will be zero and the derivative of the integral is $\frac{f(x)}{x^2}$ and the derivative of the right hand side is $\frac{1}{\sqrt{x}}$ so I determined $f(x)$ would have to equal $x^{3/2}$

Am I correct? or could someone please explain the method to solve this problem. Thanks!

• you are on the right track. Dec 9, 2014 at 4:04
• Hint: To find $a$, what happens at $x=4$? Dec 9, 2014 at 4:16
• @Daryl when x is 4 the R.H.S. is 4 and the integral is 0 so a would have to be 4? What if x is not 4 then wouldn't a be something else? Dec 9, 2014 at 4:19
• @Devin Since $a$ is constant, $a=4$ for all $x$ as described in the answer below. Dec 9, 2014 at 5:25

To determine the value of the constant $a$, revisit the original equation $(1)$, now knowing that $f(t)=t^{3/2}$: \begin{align} a+\int_{4}^{x}\frac{t^{3/2}}{t^2}\,dt&=2\sqrt{x}\\ a+\int_{4}^{x}t^{-1/2}\,dt&=2\sqrt{x}\\ a+2t^{1/2}\Big|_{4}^{x}&=2\sqrt{x}\\ a+2\left(x^{1/2}-4^{1/2}\right)&=2\sqrt{x}\\ a&=4 \end{align}
An alternative to this last method is simply to evaluate $(1)$ at $x=4$ since this makes the integral term vanish: $$a+\int_4^4 {f(t)\over t^2}\,dt=2\sqrt{4}\implies a+0=2\cdot 2\implies a=4.$$