# Thread: Finding Area by Integration Problem.

1. ## Finding Area by Integration Problem.

Determine the area bound by y=-x^2 + x + 1 and y = |x|?

Firstly i let y=x and y=-x and sub it in y=-x^2 + x + 1 to find the possible intersection points.

For the substitution of y=x to the curve equation, there are 2 answers for x, one is negative while the other is positive. I choose the positive solution while neglecting the negative solution.

For the substitution of y=-x to the curve equation, there are also 2 answers for x, one is negative while the other is positive. I choose the negative solution while neglecting the positive solution.

So for integration to determine the areas, i split into 2 parts. one part is the integration of the negative solution to 0 using y=-x^2 + x + 1 minus y=-x.

The other part is the integration from 0 to the positive solution using y=-x^2 + x + 1 minus y=x.

Finally, adding the 2 integrals give me the answer. May i know is this approach correct? Thanks.

2. Originally Posted by toffeefan
Determine the area bound by y=-x^2 + x + 1 and y = |x|?

Firstly i let y=x and y=-x and sub it in y=-x^2 + x + 1 to find the possible intersection points.

For the substitution of y=x to the curve equation, there are 2 answers for x, one is negative while the other is positive. I choose the positive solution while neglecting the negative solution.

For the substitution of y=-x to the curve equation, there are also 2 answers for x, one is negative while the other is positive. I choose the negative solution while neglecting the positive solution.

So for integration to determine the areas, i split into 2 parts. one part is the integration of the negative solution to 0 using y=-x^2 + x + 1 minus y=-x.

The other part is the integration from 0 to the positive solution using y=-x^2 + x + 1 minus y=x.

Finally, adding the 2 integrals give me the answer. May i know is this approach correct? Thanks.
It always helps to draw a picture.

Anyway, I agree that you will need to split the problem into two parts, one for everything to the left of the $y$ axis, and one for everything to the right of the $y$ axis.

You also need to notice that in the region bounded by the two functions, $-x^2 + x + 1 \geq |x|$.

Left hand intersection:

$|x| = -x$ when $x < 0$.

So $-x^2 + x + 1 = -x$

$0 = x^2 - 2x - 1$

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

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

$2 = (x - 1)^2$

$\pm \sqrt{2} = x - 1$

$1 \pm \sqrt{2} = x$.

It should be clear that this particular intersection is at $x = 1 - \sqrt{2}$.

Right hand intersection:

$|x| = x$ when $x > 0$.

So $-x^2 + x + 1 = x$

$x^2 = 1$

$x = \pm 1$.

It should be clear that this particular intersection is at $x = 1$.

$\int_{1 - \sqrt{2}}^0 {\left(-x^2 + x + 1\right) - \left(-x\right)\,dx} + \int_0^1 {\left(-x^2 + x + 1\right) - \left(x\right) \,dx}$.