calculate area of shaded region

Tweety · #1 $Old$ April 30th, 2009, 12:13 PM

firstly, if I want to find the points on intersection on this graph between the curve and line, I'd set thier y values equal to each other.

$y = x^3$

$y = -x+2$

$x^3 = -x +2$

$x^3 +x -2 = 0$

how do I solve that?

or do I not need these points when working out the area ?

derfleurer · #2 $Old$ April 30th, 2009, 12:20 PM

x^3 = 2 - x @ x = 1

So from 0 to 1, we have the area under x^3. And from 1 to 2, we have area under 2 - x.

$\int_{0}^{1} x^3dx + \int_{1}^{2} (2 - x)dx$

Tweety · #3 $Old$ April 30th, 2009, 12:22 PM

Quote:

Originally Posted by derfleurer $View Post$

x^3 = 2 - x @ x = 1

$\int_{0}^{1} x^3dx + \int_{1}^{2} (2 - x)dx$

How did you work out x=1?

derfleurer · #4 $Old$ April 30th, 2009, 12:26 PM

Just a straight forward solution. Couldn't really tell you a method for finding it (maybe someone else can).

(1)^3 = 2 - (1)

Spec · #5 $Old$ April 30th, 2009, 12:30 PM

Rational root theorem - Wikipedia, the free encyclopedia

In this case, we can easily see that x=1 is the root we're looking for.

Tweety · #6 $Old$ April 30th, 2009, 12:31 PM

Can you factor it like this;

$x^3 +x -2 = 0$

$x(x^2+1)-2= 0$

$x-2 = 0$

$(x+1)(x-1) = 0$

$x = 2 , x =-1 , x = 1$

? Is that a valid proccess?

Spec · #7 $Old$ April 30th, 2009, 12:38 PM

The typical process for finding all roots is to first guess one root using the theorem I linked to above, and then use polynomial long division to factor the polynomial.

For example, if one root is x=1, then you divide the original polynomial with x-1.

And no, your process isn't correct. x=2 and x=-1 aren't roots to the posted polynomial.