Question

a)Find the number a such that the line x = a bisects the area under the curve y= 1/x^2, 1<x<3.

a=3/2

b)Find the number b such that the line y = b bisects the area in part (a)

b=?

Answer 1

int dx/x^2 = -1/x + c

from 1 to 3 = -1/3 - (-1/1) = 2/3
half of that is 1/3

from 1 to a = -1/a - (-1/1) = -1/a + 1
so
-1/a + 3/3 = 1/3
-1/a = -2/3
2 a = 3
a = 3/2