# Thread: urgent hw help on calculus

1. ## urgent hw help on calculus

find a function f and a number a such that 6+ integral of t=x to t=a f(t)/t^2dt = 2square root x for all x>0.

2. Originally Posted by twilightstr
find a function f and a number a such that 6+ integral of t=x to t=a f(t)/t^2dt = 2square root x for all x>0.
$\displaystyle 6 + \int_x^a \frac{f(t)}{t^2} \, dt = 2\sqrt{x}$

hint ...

$\displaystyle \frac{d}{dx}\left[\int_a^x f(z) \, dz\right] = f(x)$

3. ## still need help

i still dont know what to do from there. srry

4. Originally Posted by twilightstr
i still dont know what to do from there. srry
Re-arrange: $\displaystyle \int_a^x \frac{f(t)}{t^2} \, dt = 6 - 2 \sqrt{x}$.

Differentiate both sides wrt x: $\displaystyle \frac{f(x)}{x^2} = -\frac{1}{\sqrt{x}} \Rightarrow f(x) = -x^{3/2}$.

Substitute this expression into $\displaystyle \int_a^x \frac{f(t)}{t^2} \, dt = 6 - 2 \sqrt{x}$ and solve for a.

5. its actually -6+ 2rootx but no worries i still understand it

thanks

6. Originally Posted by twilightstr
its actually -6+ 2rootx but no worries i still understand it

thanks
No it's not. Study my reply carefully - I also reversed the integral terminals (so that the Fundamental Theorem of Calculus could be used to differentiate the integral).

7. in the given problem, the upper bound is x and the lower bound is a.

8. Originally Posted by twilightstr
in the given problem, the upper bound is x and the lower bound is a.
I do realise that.

Are you familiar with the following property:

$\displaystyle \int_{\alpha}^{\beta} f(x) \, dx = - \int^{\alpha}_{\beta} f(x) \, dx$.

That's the property I used.

If you're familiar with the Fundamental Theorem of Calculus (see post #2) then you will understand why the integral terminals in your question need to be reversed before the integral can be differentiated.

9. the answer in the book is f(x)= x^3/2 and a=9