# Area of a region

## Homework Statement

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

2y=3(x^1/2)
y=4
2y+4x=7

## The Attempt at a Solution

I decided to solve for y, since that would make it easier to graph. This got me y=(3(x^1/2)/2) and y=7/2-2x. I then graphed these, finding that the span on the x-axis would be from 0 to 2.

I integrated the problem, getting 3-2*(2^1/2). This seems wrong. What happened? I'm probably overlooking something small...

(By the way, Latex isn't working, so the x^1/2 things means that that number or variable is square rooted.

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## Answers and Replies

Hang on, what function did you integrate exactly?

Well, when I graphed it, I noticed that 7/2-2x was above 3(x^1/2)/2. So I integrated (7/2-2x) - (3(x^1/2)/2) with an upper limit of 2 and a lower limit of 0, and that got me 3-2*(2^1/2).

$$\int_{0}^{2} (\frac{7}{2} - 2x) - (\frac{3*\sqrt{x}}{2}) dx$$

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Okay, I've sort of figured it out. I have to solve for x in the equation, even though I solved for y in the graph. So here it is, revised...

$$\int_{1.5}^{3} \frac{4*y^2}{9} - \frac{-(2*y - 7)}{4} dy$$

It's still not coming out correct, though (I'm doing an internet-generated problem and it's saying that it's wrong). I'm getting 2.5625, by the way. Am I still doing something wrong?

Well, when I graphed it, I noticed that 7/2-2x was above 3(x^1/2)/2. So I integrated (7/2-2x) - (3(x^1/2)/2) with an upper limit of 2 and a lower limit of 0, and that got me 3-2*(2^1/2).

$$\int_{0}^{2} (\frac{7}{2} - 2x) - (\frac{3*\sqrt{x}}{2}) dx$$

Are you sure about " 3-2*(2^1/2)" as an answer, I can see where the 3 comes from, but I don't see how you got the square root of two there, what is the integral of the squareroot of x?

HallsofIvy
Science Advisor
Homework Helper
Why do you need to solve for x rather than y? That's a crucial part of the problem.