Question

I think this should be a super easy question, because I'm just getting a conflicting answer to what my book has, and I can't find an error in my algebra. Would someone check my answer?

The problem: Evaluate integral[(5-2t+3t^2)dt] from 1 to 4.

Any and all help is greatly appreciated!

Answer 1

What expression did you get for the integral?

Answer 2

Here's what I did:
\[\int\limits_{1}^{4}(5-2t+3t^2)dt\]
\[f(t) = 5 - 2t + 3t^2\]
\[F(t) = 5t - t^2 + t^3\]
\[\int\limits_{1}^{4}(5-2t+3t^2)dt\] =\[F(b) - F(a)\]=
\[(5(4) - (4^2) + (4^3))-(5(1)-(1^2)+(1^3))\]
= \[(20-16+64)-(5-1+1)\]
\[(-4 + 64) - (4+1)\]
\[60 - 5 \]
= 55

Answer 3

I see it... damn it!! My bad... I See my mistake...

Answer 4

sometimes it takes writing it out, and asking a question to make your brain stop and restart...

Answer 5

Simple math error. Sometimes writing out makes it as plain as day. Good job.

Answer 6

algebra is usually the bane of calculus ... and on that rare occasion, arithmetic :)