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#