What would you like to ask?how do you determing c for this curve

I have done the A already (hopefully right). Now I need help with b.

Consider the curve y=1/x^2 for 1 A. Graph the region and calculate the area under the curve. (I got 5/6)
B. Determine c so that the line x=c bisects the area of part a. (This is where i need help)

Since the area from 1 to 6 is 5/6, Then half would be 5/12.

INT [ 1/x^2] dx on [1,c] = 5/12

-1/x | (1,c)= -1/c -(-1/1)= 1-1/c= 5/12

7/12= 1/c

C= 12/7

Check by integration.

Hoping this helps!