Let y(x) be a function with derivative y'(x)=x^2-2 and y(0) =7. What is the value of y at x = 3?

Integrate to get y(x) = (1/3)x^3 -2x+c where c is a constant. Substitute in our data 7 =y(0) = (1/3)(0)^3 -2*(0) +c = c. So y(x) =(1/3)x^3 -2x+7 and therefore y(3) = (1/3)(3)^3 -2*3 +7 = 9-6+7 = 10