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.

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- Nov 29th 2008, 03:08 PMtwilightstrurgent 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.

- Nov 29th 2008, 03:19 PMskeeter
- Nov 29th 2008, 06:41 PMtwilightstrstill need help
i still dont know what to do from there. srry

- Nov 29th 2008, 10:32 PMmr fantastic
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. - Nov 29th 2008, 11:48 PMtwilightstr
its actually -6+ 2rootx but no worries i still understand it

thanks - Nov 29th 2008, 11:55 PMmr fantastic
- Nov 30th 2008, 12:47 AMtwilightstr
in the given problem, the upper bound is x and the lower bound is a.

- Nov 30th 2008, 02:22 AMmr fantastic
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. - Nov 30th 2008, 11:39 AMtwilightstr
the answer in the book is f(x)= x^3/2 and a=9