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

Question

2788 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-17T03:30:15

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

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#