Calculating the area between graphs of $x^2$ and $x + 1$ This is actually a question from Apostol's Calculus book (find it on p. 94). I would like to know if my reasoning is reasonable.
I need to calculate the area between graphs of $f(x) = x^2$ and $g(x) = x + 1$ defined on $\left[-1, \frac{1 + \sqrt{5}}{2}\right] $. I know that for that I need to calculate the difference $ \int_{-1}^{\frac{1 + \sqrt{5}}{2}}\ f(x) dx - \int_{-1}^{\frac{1 + \sqrt{5}}{2}}\ g(x)dx $ when $ g(x) \leq f(x) $ and $ \int_{-1}^{\frac{1 + \sqrt{5}}{2}}\ g(x) dx - \int_{-1}^{\frac{1 + \sqrt{5}}{2}}\ f(x)dx $ when $ f(x) \leq g(x) $.
I know that at some point $x$ between $-1$ and $0$, $g(x)$ becomes larger than $f(x)$. I know that I should partition the interval to $[ -1, x_i]$ and $[x_i, \frac{1 + \sqrt{5}}{2}]$ where $x_i$ is the point where the two graphs intersect. Then I should sum the two integrals defined on those two intervals.
The problem is that I can't compute $x_i$. I see two options here: (1) find a way to compute $x_i$ which I can't do on my own; (2) leave $x_i$ unknown. I am learning on my own thus I'm not exactly sure when to give up.
I'll be grateful for any kind of clarifications and/or suggestions.
 A: make it easier on yourself by splitting it into 2 integrals. F(x) = x^2, G(x)= x+1.
from [-1,0] we know F(x) > G(x) and from [0,(1+sqrt(5))/2] G(x)>F(x). so integrate on the interval [-1,0] where the function is F(x)-G(x) added to the integral on the interval    [0,(1+sqrt(5))/2] G(x)-F(x). hope this helps.
When you solve for x set both equations equal to eachother so x^2 = x+1. move both equations to the same side so you have x^2 -x -1= 0 and when you do the quadratic formula the answer should be that upper bound of the integral. 
A: We have $f(-1) = 1$ and $g(-1) = 0$, so near the left endpoint of the interval we have $g(x) < f(x)$.  The graphs cross when $g(x) = f(x)$, which is equivalent to
$$
x+1 = x^2.
$$
Solve this equation to determine where the graphs meet.
A: It was a simple equation that prevented me from solving this problem.
As stated by Antonio, the graphs intersect when $ x^2 = x + 1 $. Hence we have $ x^2 - x - 1 = 0 $. Completing the square we get $ (x - \frac{1}{2})^2 = \frac{5}{4} $, hence $ x = \frac{\sqrt{5}}{2} + \frac{1}{2} = \frac{1 + \sqrt{5}}{2} $ or $ x = \frac{1 - \sqrt{5}}{2}$. The latter is useful to us.
My inability to solve this problem was not related to integration. I hope this is insightful for someone anyway.
