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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!

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  • $\begingroup$ you are on the right track. $\endgroup$ – Anurag A Dec 9 '14 at 4:04
  • $\begingroup$ Hint: To find $a$, what happens at $x=4$? $\endgroup$ – Daryl Dec 9 '14 at 4:16
  • $\begingroup$ @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? $\endgroup$ – Devin Crossman Dec 9 '14 at 4:19
  • $\begingroup$ @Devin Since $a$ is constant, $a=4$ for all $x$ as described in the answer below. $\endgroup$ – Daryl Dec 9 '14 at 5:25
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Your reasoning is correct. Differentiate both sides of the given equation (using the Fundamental Theorem of Calculus to differentiate the term with the integral:

\begin{align} a+\int_{4}^{x}\frac{f(t)}{t^2}dt&=2\sqrt{x}\tag{1}\\ {f(x)\over x^2}&=x^{-1/2}\\ f(x)&=x^{3/2}. \end{align}

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: $$ a+\int_4^4 {f(t)\over t^2}\,dt=2\sqrt{4}\implies a+0=2\cdot 2\implies a=4. $$

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