integral of dx/x^2 = integral x^-2 dx = -x^-1 + c = -1/x + c
so -20 [ -1/x + c ] = 20/x + some constant b
20/6 + b = 4
10/3 + b = 4
b = 12/3 - 10/3 = 2/3
so 20/x + 2/3
check my math !!!
(1 point) Find the particular antiderivative that satisfies the following conditions: p'(x) = -20/x^2 p(6) = 4
1 answer