# area between the functions

• Nov 12th 2008, 05:20 AM
jvignacio
area between the functions
find the area between $y = x^3 + 4x$ and $y = 5x^2$

i basically drew out the diagram and the diagrams meet when x is 1 and zero.
how do i find the area now? and from what intervals. i would need to calculate a total area then minus it from the total or sumthing ? thank u
• Nov 12th 2008, 07:16 AM
toraj58
the digrams meets in 0, 1 and 4

$5x^2 - x^3 - 4x = 0$ then answers for x are: 0,1 and 4

now we can calculate the integral berween intervals.

$s = \int_{0}^{1} \left(f(x) - g(x)\right) + \int_{1}^{4} \left(f(x) - g(x)\right)$
• Nov 12th 2008, 07:58 AM
jvignacio
Quote:

Originally Posted by toraj58
the digrams meets in 0, 1 and 4

$5x^2 - x^3 - 4x = 0$ then answers for x are: 0,1 and 4

now we can calculate the integral berween intervals.

$s = \int_{0}^{1} \left(f(x) - g(x)\right) + \int_{1}^{4} \left(f(x) - g(x)\right)$

so i integrate $s = \int_{0}^{1} x^3 -5x^2+4x dx + \int_{1}^{4} x^3 -5x^2+4x dx$ yeah ?

• Nov 12th 2008, 08:46 AM
Krizalid
Before setting up those integrals, one has to check that $f(x)-g(x)$ is positive or negative on $[0,1],$ in the same fashion check that $f(x)-g(x)$ is positive or negative on $[1,4].$ After concluding that, set up the integrals.
• Nov 12th 2008, 08:50 AM
jvignacio
Quote:

Originally Posted by Krizalid
Before setting up those integrals, one has to check that $f(x)-g(x)$ is positive or negative on $[0,1],$ in the same fashion check that $f(x)-g(x)$ is positive or negative on $[1,4].$ After concluding that, set up the integrals.

i understand.. is my result correct ?
• Nov 12th 2008, 10:10 AM
Krizalid
It can't be correct, the result should be positive.
• Nov 12th 2008, 12:26 PM
jvignacio
Quote:

Originally Posted by Krizalid
It can't be correct, the result should be positive.

so what happens with one of them is negative? the f(x)-g(x)?

my answer is now 71/6 when i get the magnitute of the 2nd integration
• Nov 12th 2008, 01:22 PM
mr fantastic
Quote:

Originally Posted by toraj58
the digrams meets in 0, 1 and 4

$5x^2 - x^3 - 4x = 0$ then answers for x are: 0,1 and 4

now we can calculate the integral berween intervals.

$s = \int_{0}^{1} \left(f(x) - g(x)\right) \, {\color{red}dx} + \int_{1}^{4} \left({\color{red}g(x) - f(x)}\right) \, {\color{red}dx}$

Corrections in red.

To the OP of the thread (NOT the quoted post): You should realise which curve is f(x) and whch curve is g(x). It's always strongly advised to draw the graphs. Did you?
• Nov 12th 2008, 01:25 PM
toraj58
no, i did not....i gave him the clue....he should do...thanks
• Nov 12th 2008, 01:32 PM
mr fantastic
Quote:

Originally Posted by toraj58
no, i did not....i gave him the clue....he should do...thanks

Sorry, I should have been clearer which OP I meant. I didn't mean you lol!
• Nov 12th 2008, 01:41 PM
toraj58
okay; thanks for making clear for him.