urgent hw help on calculus

**twilightstr** · November 29th 2008, 03:08 PM

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.

**skeeter** · November 29th 2008, 03:19 PM

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.

$6 + \int_x^a \frac{f(t)}{t^2} \, dt = 2\sqrt{x}$

hint ...

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

**twilightstr** · November 29th 2008, 06:41 PM

i still dont know what to do from there. srry

**mr fantastic** · November 29th 2008, 10:32 PM

Originally Posted by twilightstr

i still dont know what to do from there. srry

Re-arrange: $\int_a^x \frac{f(t)}{t^2} \, dt = 6 - 2 \sqrt{x}$ .

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

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

**twilightstr** · November 29th 2008, 11:48 PM

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

thanks

**mr fantastic** · November 29th 2008, 11:55 PM

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).

**twilightstr** · November 30th 2008, 12:47 AM

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

**mr fantastic** · November 30th 2008, 02:22 AM

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:

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

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.

**twilightstr** · November 30th 2008, 11:39 AM

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

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