What is the area of the regions between and on the interval ?

Question

​What is the area of the regions between  and  on the interval  ? ​​

Quizplus · Accepted Answer

The region between two curves is given by the integral of the difference of the two functions. In this case, the two functions are:
f(x) = sqrt(x+1) + 1
g(x) = x^2
We want to find the area between these functions on the interval [0,3]. First, we need to find the points where the two functions intersect:
sqrt(x+1) + 1 = x^2
x^4 - 2x^2 - x + 2 = 0
(x^2 - 2)(x^2 + x - 1) = 0
The quadratic factor has solutions x = (-1 + sqrt(5))/2 and x = (-1 - sqrt(5))/2. However, the negative solution is not in the interval [0,3], so we only need to consider the positive solution. This gives us one point of intersection at x = (-1 + sqrt(5))/2.
Now we can set up the integral:
integral from 0 to (-1 + sqrt(5))/2 of (sqrt(x+1) + 1 - x^2) dx
We can simplify this by expanding out the square and distributing the integral:
integral from 0 to (-1 + sqrt(5))/2 of sqrt(x+1) dx
+ integral from 0 to (-1 + sqrt(5))/2 of 1 dx
- integral from 0 to (-1 + sqrt(5))/2 of x^2 dx
The first two integrals are easy to evaluate:
(2/3)(x+1)^(3/2) evaluated from 0 to (-1 + sqrt(5))/2 
+ x evaluated from 0 to (-1 + sqrt(5))/2 
Now we just need to evaluate the last integral:
(-1/3)x^3 evaluated from 0 to (-1 + sqrt(5))/2 
= (-1/3)[(-1 + sqrt(5))/2]^3
= (2-sqrt(5))/3
Putting everything together, we get:
(2/3)[(-1 + sqrt(5))/2 + 1]^(3/2) + (-1 + sqrt(5))/2 + 2/3 - (2-sqrt(5))/3
= (2/3)(sqrt(2+sqrt(5))^3 + 3) - (sqrt(5)-1)/3
= 2.405 (approx)

​What Is the Area of the Regions Between and

What Is the Area of the Regions Between and