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

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

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

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

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

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

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

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

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