Sketch the region enclosed by the given curves.y = 2/x, y = 8x, y =1/8x x > 0
Find its area.
I found that this is the correct graph
In order to find its area, do we need to take the integral of each side and add them together? How do we find the limits? Do we need to find in terms of x instead of the y= values they give us?
I thought for graph problems you subtract the two equations but I'm not sure..
You do subtract then integrate; but as topsquark pointed out, you need to find the intersection point of the "curves" (i.e. functions) bounding the region on the top. (See your vertical blue line.) Then you can split the region up into two areas, bounded above and below by only two functions, (one area to the left of that blue line, and one to the right.)
More important than "thinking" that "for graph problem you subtract", you should understand why you subtract. Think back to "Riemann sums"- you want to break the area into many thin rectangles. The area of each rectangle is the "thickness", dx, times the length- that is the difference between the two formulas.
Thanks guys.
Would the limits be .5 and 4? I followed the blue lines down and x= .5 and x=4
so integral 0 to .5 [1/8x] - integral .5 to 4 [2/x] ? Am I on the right track?
If so,
Let us first name the equations:
y = 8x ---- 1.
y = 2/x ---2.
y = (1/8) x
I would suggest that first you find the x coordinates of points of intersection.
Intersection of 1 and 2 x = 1/2
Intersection of 2 and 3 x = 4
Now the required area = (area under the curve 1 from x = 0 to x = 1/2) + (area under the curve 2 from x = 01/2 to x = 4) - (area under the curve 3 from x = 0 to x = 4)
Now use integration to get the result