How do you find the definite integral of $int x^3-xdx$ from $[0,2]$ ?

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Answer 1 · 2016-03-08T18:37:12

$int_0^2(x^3-x)dx=2$

$int_0^2(x^3-x)dx=[x^4/4-x^2/2]_0^2$

$=2^4/4-2^2/2-0$

$=2$

Answer 2 · 2016-03-08T18:38:55

2

use power integration formula:

$int x^ndx = x^(n+1) / (n+1) + c$

Hence:

$int (x^3-x) dx = x^4 / 4 - x^2 / 2 |_0^2 = 2^4 / 4 - 2^2 / 2 =2$

How do you find the definite integral of int x^3-xdxint x^3-xdx from [0,2][0,2]?