Question

integrals (fundamental theorem of calculus)?

Answer 1

you might have to refresh the screen .. it seems like your stuck

Answer 2

\[\int\limits_{0}^{1} (2t-1)^2dt\]
\[\int\limits_{0}^{1} (2t^2-4t+1)dt\]
\[F(x) = \frac{ 2t^3 }{ 3 } -2t^2 + t\]
\[F(1)-F(0)\]
\[\frac{ 2(1)^3 }{ 3 } -2(1)^2 + (1) - [\frac{ 2(0)^3 }{ 3 } - 2(0) + 0]\]
\[-\frac{ 1 }{ 3 }\]
But, that's not the correct answer. It's supposed to be \[\frac{ 1 }{ 3 }\]
What am I doing wrong?

Answer 3

2t*2t = 4t^2, not 2t^2

Answer 4

the calculus looks fine, but the algebra is a stickler

Answer 5

oh wow....that's just....sad
thanks.

Answer 6

:) good luck