# Thread: Integral of a function

1. ## Integral of a function

I just want to see if this is right. I'm not sure I went down the right path.

Find function f(x) and constant a such that

I set x=a and found that a=3. Then I used the fundamental theorem of calculus to get rid of the integral. That left me with

So my final answer is f(x) = x^3 - 3, and a=3. Is this right?

2. Originally Posted by ban26ana
I just want to see if this is right. I'm not sure I went down the right path.

Find function f(x) and constant a such that

I set x=a and found that a=3. Then I used the fundamental theorem of calculus to get rid of the integral. That left me with $\leftarrow$ should be $\frac{f(x)}{x^2} = 2$

So my final answer is f(x) = x^3 - 3, and a=3. Is this right?
. . .

3. Thanks. Can I ask why that is? I'm a little more confused now.

4. to make it easy, let $g(t) = \frac{f(t)}{t^2}$ and we have $6 + \int_a^x g(t) ~dt = 2x$

differentiate both side and using FTC, we have

$\frac{d}{dx} \left[6 + \int_a^x g(t) ~dt \right]= \frac{d}{dx}\left[2x\right] \rightarrow g(x) = 2$