Integral of a function

[lethalasian]

V(t) = sin 3pi t s(0) = 1
determine the position function using theorem 6.1 the fundamental theorm of calculus.
I tried doing it on my own but when I looked at the answer online it was very confusing.

Here is what I did

s(t)=s(0) +intergal sin 3pi t
s(t) = 1+ cos 3t / pi |^t ₀
s(t) = 1 + cos 3t/pi - 1/pi

the answer was s(t) = 1+3/pi +cos t

Slader :: Homework Answers and Solutions

Tomorrow's answer's today! Find correct step-by-step solutions for ALL your homework for FREE!

www.slader.com

 

[Steven G]

I have a few question for you based on the work you stated.
1) Where did the limits 0 to t come from?
2) Why are you not using the limits for s(0) =1. When t=t, 1=1. When t=0, 1=1. Then 1-1 =0.
My advice is NOT to put in s(0) until you are done with your integration.

Compute the integral (sin(3pi*t)dt. Please get a C at the end. The compute C knowing that s(0)=1.
Please post back.

 

[lethalasian]

In the question
part, a asks for find the position function with t>0 using the antiderivative method
part b determines the position function for t>0 using theorem 6.1 the fundamental theorem of calculus.

the limits are [0,4] on the problem but the solution is shown used 0 and T. (maybe the solution is wrong?)
The only information given was what I listed above.

For part B (my work in the first post I made)
When I plug in the limits I just leave the s(0) alone in the front because it is a constant right?
so it would be s(0) + cost 3(t)/pi - cos 3(0)/pi

For part A i got
s(t)=intergal of v(t) dt
s(t) = intergal sin 3pi t (dt)
s(t) = (-3/pi) cos t + c
at t =0 c= 1 +3/pi
so s(t) = (-3c/pi) cos t + (1+3/pi)
however the solution is 1 + 3/pi + cos t

 

[Romsek]

[MATH]v(t) = \sin(3\pi t),~s(0) = 1\\ s(t) = \displaystyle \int \limits_0^t v(\tau) d\tau + s(0) =\\ \displaystyle \int \limits_0^t \sin(3\pi \tau) d\tau + 1= \\ \left . \dfrac{1}{3\pi}\cos(3\pi t) \right |_t^0 = \\ \dfrac{1-\cos(3\pi t)}{3\pi}+1 = \\ 1 + \dfrac{1}{3\pi} - \dfrac{\cos(3\pi t)}{3\pi} [/MATH]

The solution given is incorrect. The integral of \(\displaystyle \sin(3\pi t)\) is not \(\displaystyle \cos(t)\)

 

[Steven G]

[MATH]v(t) = \sin(3\pi t),~s(0) = 1\\ s(t) = \displaystyle \int \limits_0^t v(\tau) d\tau + s(0) =\\ \displaystyle \int \limits_0^t \sin(3\pi \tau) d\tau + 1= \\ \left . \dfrac{1}{3\pi}\cos(3\pi t) \right |_t^0 = \\ \dfrac{1-\cos(3\pi t)}{3\pi}+1 = \\ 1 + \dfrac{1}{3\pi} - \dfrac{\cos(3\pi t)}{3\pi} [/MATH]

The solution given is incorrect. The integral of \(\displaystyle \sin(3\pi t)\) is not \(\displaystyle \cos(t)\)

Since we are not requesting anything else from the OP I feel obligated to add one more piece.
So why was Romsek correct when seemingly arbitrary adding s(0) to the result of the integral, [math]\displaystyle \int \limits_0^t v(\tau), d\tau[/math]??

The result of s(0) must be 1. Now the integral [math]\displaystyle \int \limits_0^0 v(\tau) d\tau =0[/math]. Do you see why I put the upper limit at 0? So to get s(0)=1 we must have s(t)=[math]\displaystyle \int \limits_0^\tau v(\tau) d\tau +s(0)[/math]to get 1

 

[lethalasian]

i don't understand why he flipped the limits in the end and put 0 on top and t on the bottom?

 

[Steven G]

i don't understand why he flipped the limits in the end and put 0 on top and t on the bottom?

Given any definite integral you can multiply by (-1) and change the order of the limits and the result will be the same.

Suppose the integral of f(x)dx is F(x) + c.

Now if you have the integral of -f(x)dx from a to b or -integral f(x) dx from a to b = -F(x) from a to b = (-F(b)) - (-F(a)) = F(a)-F(b) = integral of f(x)dx from b to a.

Romsek had a negative in front of the integral so he got rid of that negative sign and switched the limits around.

 

[lethalasian]

ah understood thanks for the help!