Area between curves

Question

y = 8 - 7x and y = (1/x)

Area in units²

Answer 1

We first find the bounds of integration by solving for the x-coordinates of the points of intersection between the two curves as follows:

8 - 7x = 1/x

8x - 7x² = 1

7x² - 8x + 1 = 0

(7x - 1)(x - 1) = 0

x = 1/7 or x = 1

Area = ∫_1/7¹ (8 - 7x) - (1/x) dx

= 8x - 7/2x² - lnx ]_1/7¹

= 9/2 - (15/14 + ln7)

= 24/7 - ln7

Answer 2

3 - ln7 = about 1.054 units^2 = the area between the two lines

8-7x = 1/x, solve for x to find where the two lines intersect. those x values are the limits of integration

8x-7x^2 = 1

7x^2 - 8x - 1 = 0

(7x-1)(x-1) = 0

x = 1/7 or 1

it helps to sketch a rough graph of the 2 lines. One is the 1st quadrant upper branch of a rectangular hyperbola, the other a straight downward sloping line with slope = -7.

they intersect at two points (1/7, 7) and (1,1)

the line is above the hyperbolic curve between those two points

area under the line for 1/7<x< 1 = the area of a triangle with base=1-1/7 = 6/7 and height = 7. A=bh/2 = (6/7)(7)/2 = 6/2 = 3

are under the hyperbolic curve for 1/7<x<1 = the integral of 1/x = lnx evaluated from 1/7 to 1 = -ln(1/7) = about 1.946

3-1.946 = about 1.054

exact area = 3+ln(1/7) = 3-ln7

calculating the areas between y=1/x and y=8-7x for x<1/7 or x>1 gives infinite areas.