How do you bisect the area of definate integral 1 to 6 of 1 divided by x sqared?

Answer 1

First, find the area under the curve y = 1/x2, with boundary lines x = 1 and x = 6, by calculating the integral of 1/x2 with lower limit 1, and upper limit 6. Then divide it by 2.

(6)integral(1) of (1/x2) dx = (6)integral(1) of (x--2) dx = -x-1|(6),(1) = -1/x|(6)(1) = -1/6 +1 = 5/6.

Thus, the half of the area under the curve is 5/12.