Question

Help Please!

Answer 1

My Professor has been gone for the week and we are suppose to learn this new section on our own. I am a little confused as to how it works. Could someone explain it to me?

Answer 2

I'd be glad to answer YOUR specific questions, but not to "explain it to you" without that first effort on your part.

Answer 3

Here is a question from the homework.

A variable force of 9x^−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1ft to x = 12ft. (Round your answer to two decimal places.)

The answer box says _______ft-lb

Answer 4

Would it just be...
\[\int\limits_{1}^{12} 9x^{-2} dx\]

Answer 5

Good start. Yes. But be certain to include the units of measurement!

Answer 6

Oh okay.... sooo....\[\int\limits_{1ft}^{1ft}9x^{-2}lbs (dx)\]

Answer 7

ooops. I meant 12 for the upper integrand.

Answer 8

Yes: integrate from 1 ft to 12 ft.

Answer 9

\[9\int\limits_{1ft}^{12ft}x^{-2}lbs(dx)\]
\[9\left[ -\frac{ 1 }{ x } \right]_{1ft}^{12ft}\]
\[9\left[ -\frac{ 1 }{ 12 }+\frac{ 12 }{ 12 } \right]\]\[9\left[ \frac{ 11 }{ 12 } \right] = \frac{ 99 }{ 12 }\]

Answer 10

So the answer is 8.25
I got it right! :D

Answer 11

Hm... maybe it was really simple because it was the first problem

Answer 12

I'm so glad. But I'd be even happier if you'd include the units of measurement.

Answer 13

Oh right right sorry.

\[9\left[ -\frac{ 1 }{ 12ft } lbs - \frac{ 12 }{ 12ft }lb\right]\]
\[9\left[ \frac{ 11 }{ 12ft } lbs\right] = \frac{ 99 }{ 12ft }lbs\]

Answer 14

How would I go about this one then? Eh...

Answer 15

it's the area under the graph :-)

Answer 16

Oh I got it. Thanks