How do you find the definite integral of int (1-2x-3x^2)dx from [0,2]?

1 Answer
Mar 11, 2016

int_0^2(1-2x-3x^2)d x=-10

Explanation:

int_0^2(1-2x-3x^2)d x=|x-2*1/2*x^2-3*1/3*x^3|_0^2
int_0^2(1-2x-3x^2)d x=|x-x^2-x^3|_0^2
int_0^2(1-2x-3x^2)d x=2-2^2-2^3
int_0^2(1-2x-3x^2)d x=2-4-8
int_0^2(1-2x-3x^2)d x
int_0^2(1-2x-3x^2)d x=-10