Question

Evaluate the Definite Integral:

Answer 1

\[\int\limits_{0}^{\pi/2} (\sin 2Pit)/T - \alpha dt\]

Answer 2

Please help! :)

Answer 3

ask wolframalpha ... first

Answer 4

\[\int\limits_{0}^{\pi/2} (\sin 2 \pi t)/T - \alpha dt\]

*replace pi with the symbol

Answer 5

http://www.wolframalpha.com/input/?i= \int\limits_{0}^{\pi%2F2}+%28\sin+2Pit%29%2FT+-+\alpha+dt&dataset=

Answer 6

More modification : \[\large{\int\limits_{0}^{\pi/2} \frac{\sin 2 \pi t}{T} - \alpha dt}\]

Answer 7

\[T=t\]?

Answer 8

yup, the modified equation is correct.

Answer 9

so big T is different than little t, right?

Answer 10

Yup

Answer 11

\[\large{T\ne t \space \textbf{or}\space t = T}\]

ok so it is first one T is not equal to t

Answer 12

Correct

Answer 13

so treat big T as a constant

Answer 14

Okay

Answer 15

Then U = 2PIt/T - Alpha, therefore, du = 2pi/T - alpha?

Answer 16

\[\int_0^{\pi/2}\frac{\sin (2 \pi t)}{T} - \alpha dt=\frac1T\int_0^{\pi/2}\sin(2\pi t)dt-\alpha \int_0^{\pi/2}dt\]

Answer 17

for the first integral\[u=2\pi t\implies du=2\pi dt\]the next should be pretty straight forward

Answer 18

Okay, thank you! :)

Answer 19

welcome!

Answer 20

I love Substitution Rule! :)

Answer 21

Can I solve it and double check what I got with you?

Answer 22

... or do I hate it?

Answer 23

I know, I hate it. lol.

Answer 24

go ahead and try to solve it

Answer 25

@TuringTest is the answer 1?

Answer 26

no, that is not possible... let me see what it is (will be an ugly number)

Answer 27

Oh Okay :(.

Answer 28

yeah, ugly answer...
first do\[\frac1T\int_0^{\pi/2}\sin(2\pi t)dt\]what do you get?

Answer 29

I can't view the equation for some reason. I'll keep refreshing the page.

Answer 30

Nope, still can't view it. I can only see the codes.

Answer 31

@TuringTest I got -0.159 for that one.

Answer 32

Sorry for the late reply. My net is acting up.

Answer 33

Where does 1/T go ?

Answer 34

Oh wait, so it's -1/T 0.159

Answer 35

I'm sorry, I'm so bad at this.

Answer 36

Can you ∫ sin 2π tdt ?

Answer 37

Yeah, I got (-cos2π t)/2π t

Answer 38

(-cos2π(π/2))/2π(π/2) - ((-cos2π)(0))/2π(0)

Answer 39

Is that correct?

Answer 40

u = -cos2πt, du = sin2π, 1/2du = sinπ?

Answer 41

sorry about that.

Answer 42

I'm not sure what to do after, definite integrals make no sense to me, again, I'm sorry.

Answer 43

yes

Answer 44

Yes, I can.

Answer 45

u = 2π correct?

Answer 46

2πt*

Answer 47

so du = 2π = 1/2 du = π

Answer 48

From TuringTest's instruction: u= 2πt -> du=2πdt
Since you want dt: dt = du/2π

Answer 49

Okay

Answer 50

Now plug into the equation
(1/T )∫ sin 2π tdt = ... ?

Answer 51

-1/T (cos 2πt)/2πt?

Answer 52

Actually =
(- 1/2πT ) ( cos 2πt )

Answer 53

Okay, so this is when I plug in π/2 and 0?

Answer 54

I think so

Answer 55

1.418T?

Answer 56

I have class in 10 minutes, so I better go.

Answer 57

I appreciate the help!

Answer 58

Okay. Thanks again! :)