Area Enlcosed By Curves

**My Little Pony** · February 5th 2009, 07:40 PM

Find the area of the region enclosed by the curves:

2y=4*sqrtx, y = 4, 2y + 3x = 7

I'm having a little difficulty with this problem. Is the upper bound function y=4?

**Abu-Khalil** · February 5th 2009, 08:37 PM

Yes, the upper bound function is $y=4$ .

I suggest you to do a graph of the situation to follow my answer.

First, we'll find the intersection between $2y+3x=7$ and $y=4$ . Just replacing, we obtain $8+3x=7\Rightarrow x=-\frac{1}{3}$ , $\therefore$ the functions intersects at $\left(-\frac{1}{3},4\right)$ .

Doing the same, we conclude that $2y+3x=7$ and $y=2\sqrt{x}$ insersects at $\left(1,2\right)$ .

Finally, the area is given by $\int_{-\frac{1}{3}}^14-\left(-\frac{3}{2}x+\frac{7}{2}\right)dx+\int_1^44-2\sqrt{x}dx$ .

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