Question

using integration find the distance travelled by the particle during the interval t=1s to t=6s for the equation v = 5 + 24t = 3t^2

Answer 1

I know I need to use limits, but I'm not certain how to do this...

Answer 2

\[\int\limits_{6}^{1} ( 5+24t -3t ^{2}\]

Answer 3

Simple I think I first integrate it then apply the limits...

Answer 4

12.47

Answer 5

You have your limits applied upside down. \[\int\limits_{1}^{6}(-3t^{2}+24t+5)dt\]

Answer 6

Once you integrate the expression, take that value, plug in your upper limit and subtract that value with your lower limit. \[(-6^{3}+12(6)^2+5(6)) - (-1^{3}+12(1)^2+5(1))\]

Answer 7

I got 234 as my final answer, is that right?

Answer 8

After integrating I got \[5t + 12t ^{2} - t ^{3}\]

Answer 9

I got the same integration, I didn't work out the math so give me a sec and see if I got the same answer you did.

Answer 10

Yes! Looks like my calculus is improving then. Thank you kindly.

Answer 11

I got an answer of 230

Answer 12

I was out by 4, I made a silly mistake, I put 6 where a 2 should of been. Thank you.

Answer 13

You're welcome.

Answer 14

:)

Answer 15

Could someone please help me on this one?

Answer 16

You are going to do the same thing with this one only that your limits are now 0 to pi/4

Answer 17

So would my integrated answer be \[\frac{ dy }{ dx } = -\sin \theta \]

Answer 18

The derivative of sin is cos, so the anti derivative of cos is sin.

Answer 19

So I'm using: \[\sin \theta \] to place the limits on?

Answer 20

Except that you had y=cos(2theta) as your original equation

Answer 21

So is that, y = sin(2theta) ?