Question

Calculus Q..(below)

Answer 1

\[g(x)=\int\limits_{0}^{x} \cos4tdt \] then \[g(x+\pi)\] equals?

Answer 2

[write in terms of g(x) and g(pi)]

Answer 3

running on ups so i will see as much as i can now and the rest as soon as the connection returns...

Answer 4

any idea anyone?

Answer 5

@freckles @ParthKohli @zepdrix

Answer 6

\[\Large\rm g(\color{orangered}{x})=\int\limits\limits_{0}^{\color{orangered}{x}} \cos4tdt\]So then,\[\Large\rm g(\color{orangered}{x+\pi})=\int\limits\limits_{0}^{\color{orangered}{x+\pi}} \cos4tdt\]

Answer 7

Do the integration, plug in the new limit

Answer 8

write in terms of g(x) and g(pi) ?
Hmm interesting

Answer 9

@zepdrix ?

Answer 10

sup? :D

Answer 11

First get a solution for g(x).
Integrate, plug in the stuff.

Answer 12

hey i did that already...

Answer 13

You did..?

Answer 14

\[g(x)=\frac{ \sin4x }{ 4x }\]

Answer 15

Woops,\[\large\rm g(\color{orangered}{x})=\frac{\sin4\color{orangered}{x}}{4}\]right?
I'm not sure where the bottom x is coming from :D

Answer 16

oh typo

Answer 17

and \[g(x+\pi) =\frac{ \sin(4\pi +4x )}{ 4 }\]

Answer 18

which is..\[g(x+\pi) = \sin4x/4\]

Answer 19

Pull the 4 out front, it's so ugly down there >.<\[\large\rm g(\color{orangered}{x})=\frac{1}{4}\sin4\color{orangered}{x}\]
k looks good D\[\large\rm g(\color{orangered}{x+\pi})=\frac{1}{4}\sin4\color{orangered}{(x+\pi)}\]

Answer 20

Because adding 4pi just spins us around twice,
landing in the same place?
Good good good.

So we've discovered that g(x+pi) = g(x).

Answer 21

yup!

Answer 22

but the problem is... g(pi)=0

Answer 23

what? :U

Answer 24

\[g(\pi) = 1/4 (\sin4\pi) =0\] right?

Answer 25

yes.

Answer 26

yeah so \[g(x+\pi)=g(x) \pm d(\pi)\] right?

Answer 27

thats not d! thats g

Answer 28

Why would \(\large\rm g(x+\pi)=g(x)+g(\pi)\)

? :o hmmm
Oh because g(pi)=0.
Ya ya ya I suppose that's a clever observation :)
You could include that in your result.

Answer 29

Thanks!

Answer 30

but the problem isit has both options..

Answer 31

both + and - separately...

Answer 32

\[\large\rm g(x+\pi)=\frac14\sin(4x+4\pi)\]\[\large\rm g(x+\pi)=\frac14\left[\sin4x \cos4\pi+\sin4\pi \cos4x\right]\]\[\large\rm g(x+\pi)=g(x)\cos4\pi+g(\pi) \cos4x\]If you use your angle sum formula for sine, you get some crazy business like this.
So maybe they want you to pick the one with +.
Sorry I'm not sure :(
Weird problem... hmm

Answer 33

well i still think its plus or minus..maybe its a printing mistake..Thanks

Answer 34

ya plus/minus makes more sense :D