Find the exact area of the region enclosed by the square root of (x) + the square root of (y) is = 1; x = 0 and y = 0.
I moved the first equation around to get:
y = [1- the square root of (x) ] ^2
Unfortunately, this gives me bounds from 1 to 1.
So I'm stuck with:
integral from 1-1 of [1- the square root of (x) ] ^2
I used FnInt on this equation and it returns an answer of 0.
Any help or suggestions would be greatly appreciated!
Thanks alot,
Mike
The limits of dx integration are from x=0 (the x=0 line) to x=1 (where the curve crosses the y=0 line). What you want to calculate in the integral od y dx from x=0 to x=1. That can be written
(Integral of) (1 - 2 sqrt x + x) dx
To find the exact area of the region enclosed by the curve, you need to integrate the expression (1 - 2 sqrt(x) + x) with respect to x from x=0 to x=1.
Let's break it down step by step:
1. Start with the integral:
∫ (1 - 2 √x + x) dx
2. Integrate each term separately:
∫ 1 dx - ∫ 2 √x dx + ∫ x dx
3. Evaluate each integral:
x - 2/3 * x^(3/2) + 1/2 * x^2
4. Now, substitute the limits of integration, which are x=0 and x=1:
[(1) - 2/3 * (1)^(3/2) + 1/2 * (1)^2] - [(0) - 2/3 * (0)^(3/2) + 1/2 * (0)^2]
Simplifying that gives:
1 - 2/3 + 1/2 - 0 + 0 + 0
= 1 - 2/3 + 1/2
5. To simplify it further, find a common denominator:
= (6 - 4 + 3) / 6
= 5 / 6
So, the exact area of the region enclosed by the curve is 5/6.
Note: It seems that there was a misunderstanding in your calculation using FnInt, which resulted in an incorrect answer of 0. However, by following the steps above, you should arrive at the correct answer of 5/6.