# Maths help for integration

• Jan 19th 2010, 01:09 AM
maay
Maths help for integration
hey i do not know how to do this question. I don't know how to simplify the power of 3.

Find the area enclosed between the curve y = x^3 , the x - axis and the line y = -3 x + 4.

thank you.
• Jan 19th 2010, 01:28 AM
pickslides
You need to find

$
\int_0^ax^3~dx+\int_a^{\frac{4}{3}}-3x+4~dx
$

$a$ is where $x^3 = -3x+4$
• Jan 19th 2010, 02:51 AM
maay
hey i don't get what you mean.

sorry
• Jan 19th 2010, 03:17 AM
maay
Quote:

Originally Posted by pickslides
You need to find

$
\int_0^ax^3~dx+\int_a^{\frac{4}{3}}-3x+4~dx
$

$a$ is where $x^3 = -3x+4$

hey i know how to do it but the thing it i dont know how to find the x^3 + 3x -4 = 0 how do you simplify that to find the x - values ?
• Oct 19th 2011, 04:27 AM
mr fantastic
Re: Maths help for integration
Quote:

Originally Posted by maay
hey i know how to do it but the thing it i dont know how to find the x^3 + 3x -4 = 0 how do you simplify that to find the x - values ?

It should be clear by inspection that x = 1 satisfies the equation. Therefore x - 1 is a factor of x^3 + 3x - 4 and anyone studying calculus should be able to factorise a cubic if a linear factor is known and therefore establish that x = 1 is the only real solution.