Calculus 1 Integrals FTC

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Nov 23rd 2008, 08:57 AM
littlejodo

Calculus 1 Integrals FTC

Find a function f and a number "a" such that

2 + int_(x at top, a at bottom) f(t)/t^3 dt = 6x^-3

?? I've been learning the fundamental theorem of calculus but I've just been staring at this question. I really don't know where to begin. I didn't think the lower limit, a, should come into play with the FTC... I don't know.

Please help. Thanks!!
Nov 23rd 2008, 02:13 PM
littlejodo

Even if someone could show me an example of another problem like this and how to solve it that would be helpful! I just don't know the process and I can't find a similar problem in my textbook or online.
Nov 23rd 2008, 02:43 PM
skeeter

Quote:

Originally Posted by littlejodo

Find a function f and a number "a" such that

2 + int_(x at top, a at bottom) f(t)/t^3 dt = 6x^-3

?? I've been learning the fundamental theorem of calculus but I've just been staring at this question. I really don't know where to begin. I didn't think the lower limit, a, should come into play with the FTC... I don't know.

Please help. Thanks!!

$2 + \int_x^a \frac{f(t)}{t^3} \, dt = 6x^{-3}$

$2 - \int_a^x \frac{f(t)}{t^3} \, dt = 6x^{-3}$

$\frac{d}{dx} \left[2 - \int_a^x \frac{f(t)}{t^3} \, dt = 6x^{-3}\right]$

$-\frac{f(x)}{x^3} = -18x^{-4}$

$f(x) = \frac{18}{x}$
Nov 23rd 2008, 03:15 PM
littlejodo

So you never have to take the constant 2 into consideration?

And what does it mean to find an "a" value in this problem? It doesn't seem like the value of a mattered at all.
Nov 23rd 2008, 03:32 PM
skeeter

Quote:

Originally Posted by littlejodo

So you never have to take the constant 2 into consideration?

the derivative of a constant = 0

Quote:

And what does it mean to find an "a" value in this problem? It doesn't seem like the value of a mattered at all.

it didn't when taking the derivative, but you can find the value of a now that you have f(x) ...

$2 - \int_a^{x} \frac{18}{t^4} \, dt = 6x^{-3}$

$2 - \left[-\frac{6}{t^3}\right]_a^x = 6x^{-3}$

$2 - \left[-\frac{6}{x^3} + \frac{6}{a^3}\right] = 6x^{-3}$

$2 + \frac{6}{x^3} - \frac{6}{a^3} = 6x^{-3}$

$a^3 = 3$

$a = \sqrt[3]{3}$