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

2 Answers
Aug 19, 2016

1/3

Explanation:

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

Aug 19, 2016

1/3.

Explanation:

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.! .