Find the area of the region bounded by y = x^2, y = 0, x = -1, and x = 2.
I tried the integral from -1 to 2 of x^2 and got 3 as the answer.
I tried (integral from 0 to 1 of √y + 1) + (integral from 0 to 4 of 2 - √y) and got 13/3.
What is wrong with the way the integrals are set up?
∫[-1,2] x^2 dx = 3
so far, so good
For the horizontal strips, you need the width on the left side as 1-√y, since that is the distance from x = -1 to the parabola. Just as it is 2-√y on the right side.
∫[0,1] (1-√y)) dy + ∫[0,4] (2-√y) dy
= 1/3 + 8/3 = 3
How do you come up with 1 - √y if the line is x = -1?
To find the area of the region bounded by the given curves, you need to set up the correct integral. Let's break down the different sections of the region and calculate their areas separately.
First, let's consider the region between the curve y = x^2 and the x-axis (y = 0). We need to find the points of intersection between these curves in order to determine the limits of integration.
Setting y = 0 in the equation y = x^2, we get:
0 = x^2
This equation has two solutions, x = 0 and x = 0.
Therefore, when finding the area for the region between y = x^2 and y = 0, the limits of integration for x will be from x = 0 to x = 2.
The integral will look like this:
∫[0 to 2] x^2 dx
Evaluating this integral gives us:
[1/3 * x^3] from 0 to 2
= (1/3 * 2^3) - (1/3 * 0^3)
= 8/3 - 0
= 8/3
So, the area of the region between y = x^2 and y = 0 for x ranging from -1 to 2 is indeed 8/3, not 3 as you calculated in the first attempt.
Regarding your second attempt, it seems like you tried to divide the region into two separate sections and integrate each independently. However, the setup for these integrals is incorrect.
To summarize, the correct integral for finding the area of the region bounded by y = x^2, y = 0, x = -1, and x = 2 is:
∫[0 to 2] x^2 dx
The result is 8/3.