How do you evaluate the definite integral int (6x^2 - 4e^(2x))dx from [0,2]?

1 Answer
Jul 17, 2018

-2(e^4 - 9)~~-91.1963

Explanation:

Given: int_0^2 (6x^2 - 4e^(2x)) dx

Break the integral into two pieces and solve separately:

int_0^2 6x^2 dx - int_0^2 4e^(2x) dx =

6 int_0^2 x^2 dx -4 int_0^2 e^(2x) dx

Integration rules:
Use the Power rule: int x^n dx = 1/(n+1) x ^(n+1) + C

Use int e^u du = e^u + C

Let u = 2x; " "du = 2 dx; " "dx = (du)/2
x= 0, => u = 0; " " x= 2 => u = 4

Using the rules:
6 int_0^2 x^2 dx -4 int_0^2 e^(2x) dx =

6 * 1/3 x^3|_0^2 - 4 int_0^4 e^u(du)/2 =

2x^3|_0^2 - 2 int_0^4 e^u du =

2(2)^3 - 0 - 2e^(u)|_0^4 =

16 - 2(e^4 - e^0) = 16 - 2(e^4 - 1) =

16 - 2e^4 +2 = 18 - 2e^4 =

-2(e^4 - 9)~~-91.1963