# calculus II

posted by
**Shelley** on
.

We're doing areas by integrals now, with 2 eqns. I have a few questions.

1. Sketch the region in the xy-plane defined by the inequalities x-2(y^2)> 0 and 1-x-abs(y)>0. and find its area.

Would it be best to solve for x, then graph it as x= something? Then find the integral with dy?

2. a. Find the number a such that the line x=a bisects the area under the curve y=1/x^2, 1<x<4.

b. Find the number b such that the line y=b bisects the area in part a.

For this, I graphed it then tried to draw a line to vertically bisect it. Is it right to just say that b is the x value that it bisects the graph at?

Thank you for your help!!

Yes, I would do the first one that way.

Yes, but you can do it with the integral, let the integral of one side equal the integral of the other side, then solve for b, the center part.