How do you evaluate the definite integral #int (2t-1)^2 # from #[0,1]#?

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Answer 1 · 2016-08-19T07:27:01

#1/3#

#int_0^1(2t-1)^2dt#

Let #u = 2t-1 implies du = 2dt#

#therefore dt = (du)/2#

Transforming the limits:

#t:0rarr1 implies u:-1rarr1#

Integral becomes:

#1/2int_(-1)^1u^2du = 1/2[1/3u^3]_(-1)^1 = 1/6[1 - (-1)] = 1/3#

Answer 2 · 2016-08-19T07:35:44

#1/3#.

#int_0^1 (2t-1)^2dt=int_0^1(4t^2-4t+1)dt#

#=[4t^3/3-4t^2/2+t]_0^1#

#=[4/3t^3-2t^2+t]_0^1#

#=4/3-2+1-0#

#1/3#, as derived by Euan S.!

Enjoy Maths.! .