Find the area of y=2/x, y=8x, y=1/8x, x>0

Answer 1

The first step in solving this problem would be to sketch the graph of the 3 functions. If you have access to a graphing calculator you can do this quickly.

After sketching you find that the y=8x intersects y = 2/x at (0.5,4)

and y= 2/x intersects y=x/8 at (4,0.5)

We can break the problem into 2 parts. Part 1 is the area between y = 8x and x/8 when 0<= x = 0.5

Part 2 is when 0.5<=x<=4 and the area is between 2/x and x/8

Now for the first part.

Area = integral from 0 to 0.5 of (8x-x/8) dx.

This simplifies to 7.875x. The antiderivative of x is 0.5*x²

Evaluating between 0.5 and 0 yields 7.875*(0.5)^2/2) = 0.984375 or 1-1/64

For the second part

Area = integral from 0.5 to 4.0 of (2/x-x/8) dx.

The antiderivative is 2*ln(x)-x^2/16

Evaluating between the limits yields

(2*ln(4)-4^2/16)-(2*ln(1/2)-1/64) = (2.772 -1)-(-1.386-0.0156)= 1.772 + 1.4016=3.1736

So the final answer is approximately 0.984375 + 3.1736 = 4.1579

Some teachers may not want the answer in decimal form but rather in a form such as

63/64 + ln(16)-1-ln(0.25) +1/64 =ln(16)-ln(.25) = ln(16/0.25) = ln(64) = 6ln(2)

The key point is that the difference between the functions is the integrand for both integrals

Answer 2

Presumably your 3rd expression for x should read (1/8)x; otherwise the question doesn't make geometric sense (try graphing)!

The 3 lines intersect at (1/2,4) and (4,1/2) and the area involved is symmetric about y=x.

You will see that the area is ∫(2 dx/x) from 1/2 to 4.

The areas of the 2 triangles add out, that is you have to add the area of the left hand triangle to the integral and subtract the area of the right hand triangle....but the two triangles have equal area.