# Function integrate

• March 1st 2008, 05:24 AM
taurus
Function integrate
I have to find the value of:

INTEGRATION[ f'(x) ((f(x))^3 -3f(x)-5) dx ]

with upper boud 6 and lower bound 4

given that
f(4) = 1, f '(4) = -7, f(6) = 2 and f '(6) = 2

So i tried the following which i dont think is right?
2((2)^3 - 3 * 2 - 5) - (-7) - 3 * 1 - 5)
• March 1st 2008, 06:36 AM
CaptainBlack
Quote:

Originally Posted by taurus
I have to find the value of:

INTEGRATION[ f'(x) ((f(x))^3 -3f(x)-5) dx ]

with upper boud 6 and lower bound 4

given that
f(4) = 1, f '(4) = -7, f(6) = 2 and f '(6) = 2

So i tried the following which i dont think is right?
2((2)^3 - 3 * 2 - 5) - (-7) - 3 * 1 - 5)

Observe:

$\frac{d}{dx}\left[\frac{(f(x))^4}{4} -\frac{3f(x)^2}{2}-5f(x)+C\right]=f'(x) [(f(x))^3 -3f(x)-5]$

RonL
• March 1st 2008, 11:05 PM
taurus
i dont get it?
• March 1st 2008, 11:23 PM
CaptainBlack
Quote:

Originally Posted by taurus
i dont get it?

Your integrand is easily seen to be the derivative of the
given polynomial in $f(x)$. Then the Fundamental
Theorem of Calculus tells us that the integral is that
polynomial in $f(x)$.

That is you use the result that:

$\int \frac{d}{dx}g(x) ~dx=g(x)+C$

or for a definite integral:

$\int_a^b \frac{d}{dx}g(x) ~dx=g(b)-g(a)$

RonL
• March 2nd 2008, 12:14 AM
taurus
• March 2nd 2008, 03:53 AM
wingless
$\int^{6}_{4}f'(x)[f^3(x)-3f(x)-5]~dx$

$\int^{6}_{4}f^3(x)f'(x)-3f(x)f'(x)-5f'(x)~dx$

$\int^{6}_{4}f^3(x)f'(x)~dx-3\int^{6}_{4}f(x)f'(x)~dx-5\int^{6}_{4}f'(x)~dx$

Use $f'(x)~dx = d(f(x))$

$\int^{6}_{4}f^3(x)~d(f(x))-3\int^{6}_{4}f(x)~d(f(x))-5\int^{6}_{4}d(f(x))$

$\frac{f^4(x)}{4} - \frac{3f^2(x)}{2} - 5f(x) ~\displaystyle{{|}^{6}_{4}}$

Now go on..