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?
Yes, the upper bound function is $\displaystyle y=4$.
I suggest you to do a graph of the situation to follow my answer.
First, we'll find the intersection between $\displaystyle 2y+3x=7$ and $\displaystyle y=4$. Just replacing, we obtain $\displaystyle 8+3x=7\Rightarrow x=-\frac{1}{3}$, $\displaystyle \therefore$ the functions intersects at $\displaystyle \left(-\frac{1}{3},4\right)$.
Doing the same, we conclude that $\displaystyle 2y+3x=7$ and $\displaystyle y=2\sqrt{x}$ insersects at $\displaystyle \left(1,2\right)$.
Finally, the area is given by $\displaystyle \int_{-\frac{1}{3}}^14-\left(-\frac{3}{2}x+\frac{7}{2}\right)dx+\int_1^44-2\sqrt{x}dx$.