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.
For the first question, you can indeed solve for x and graph the resulting equation in order to sketch the region. Here's a step-by-step process:
1. Solve the first inequality for x: x - 2(y^2) > 0.
To do this, rearrange the equation: x > 2(y^2).
2. Solve the second inequality for x: 1 - x - |y| > 0.
Rewrite the equation by solving the absolute value: 1 - x - |y| > 0
Divide it into two cases: (1 - x - y > 0) and (1 - x + y > 0)
Solve each case separately.
3. Now, you have two equalities:
x = 2(y^2) and x = 1 - y and x = 1 + y.
To graph these equations:
a. Sketch the graph of y = x/2 by either plotting points or finding the intercepts, and draw a smooth curve.
b. For x = 1 - y, rearrange it: y = 1 - x. Plot this line.
c. For x = 1 + y, rearrange it: y = -1 + x. Plot this line.
4. The resulting graph will show the region defined by these inequalities. Adjust your window settings to clearly see the intersection points of the curves.
To find the area of this region, you can integrate with respect to y. Here's how:
1. Determine the limits of integration by finding the intersection points of the curves.
a. Find the y-values where the curves intersect and call them y_1 and y_2.
b. The limits of integration for the area will be from y_1 to y_2.
2. Set up the integral to find the area:
∫(y_1 to y_2) [x(y) - 2(y^2)] dy.
3. Integrate the expression, evaluating it from y_1 to y_2, to find the area.
For the second question, finding the number a such that the line x = a bisects the area under the curve y = 1/x^2 involves finding the area on each side of the line and setting them equal to each other:
1. Determine the limits of integration for the area on each side of the line.
a. The area to the left of x = a will go from 1 to a.
b. The area to the right of x = a will go from a to 4.
2. Set up the integrals to find the areas on each side:
Area on the left: ∫(1 to a) (1/x^2) dx.
Area on the right: ∫(a to 4) (1/x^2) dx.
3. Set these two areas equal to each other:
∫(1 to a) (1/x^2) dx = ∫(a to 4) (1/x^2) dx.
4. Solve the equation to find the value of a that satisfies the equality.
For part (b) of the second question, to find the number b such that the line y = b bisects the area found in part (a), you are correct in saying that b is the x-value that bisects the graph. Draw a vertical line and find the x-coordinate where it intersects the curve. That x-coordinate will be the value of b.
I hope this helps, and feel free to ask more if you have any further questions!