Asked by Anonymous
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?
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?
Answers
Answered by
Steve
∫[-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
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
Answered by
Anonymous
How do you come up with 1 - √y if the line is x = -1?
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