Sketch the region enclosed by the given curves. finally, find its area.

Question

y=2/x, y=8x, y=(1/2)x, x>0

Answer 1

A = ∫₀^1/2∫(8x - 1/2x)dx + ∫_1/2²(2/x - 1/2x)dx = 15/2·x²/2₀^1/2 + (2lnx - 1/2x²/2)_1/2² = 15/4·1/4 + (2ln2 -1/4·4) -

(2ln(1/2) - 1/4·1/4) = 15/16 + 2ln2 - 1 + 2ln2 + 1/16 = 4ln2 = 2.773

Answer 2

Renumber to keep better track of the equations

y₁ = 2/x

y₂ = 8x

y₃ = 1/2 x

x > 0, only look at the right side of the y-axis.

First graph each line.

y₂ and y₃ are the easiest, so start there and keep the numbers as integers if possible.

y₂ ordered pairs are [0, 0], and [1, 8]

y₃ ordered pairs are [0, 0], and [2, 1]

y₁ is simply a scaled 1/x plot with order pairs being [1/4, 8], [1, 2], [2, 1]

The above should be enough info to determine which graph to choose.

To calculate the area :

Subtract the integral of y₃ from the integral of y₂ using x = 0 to the intersection of y₁ and y₂ as boundaries. This is the area up to that point.

Then

Subtract the integral of y₃ from the integral of y₁ using the intersection of y₁ and y₂ to the intersection of y₁ and y₃ as boundaries. This is the area up to that point.

Add the two areas taken to get the total area.