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

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Answer 1 · 2017-08-20T15:49:57

The integral has value $1/4$ .

We have by basic integration:

$I = int_0^1 xdx - int_0^1x^3dx$

$I = [1/2x]_0^1 - [1/4x^3]_0^1$

Now find the value using the 2nd fundamental theorem of calculus.

$I = 1/2(1) - 1/2(0) - (1/4(1) - 1/4(0))$

$I = 1/2 - 1/4$

$I = 1/4$

Answer 2 · 2017-08-20T15:50:29

$1/4$

$int_0^1(x-x^3)dx = int_0^1(x)dx - int_0^1(x^3)dx$

$= x^2/2|_0^1$ $- x^4/4|_0^1$

$= (1/2 - 0) - (1/4 - 0)$

$=1/4$

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