Why Does My Calculated Integral Value Differ from the Expected Result?

[hangainlover]

Homework Statement

integrate R(t) 2+5sin(4πt/25) to calculate F(t) from 0 to 6

Homework Equations

The Attempt at a Solution

the differential equation is 2+5sin(4πt/25)
so after the integration, i get 2t-5cos(4πt/25)(25/4π)
so i want to get the value in the interval from 0 to 6)

so i put 2+5sin(4πt/25) on my calculator and made it do the integration,
It gives me 31.8159
but when i F(6)-F(0) on this function, 2t-5cos(4πt/25)(25/4π), i get something entirely different.
My calculator is on radian mode and π is pie
What am i doing wrong?

 

[mighty2000]

Thank you for your question. It seems like you are on the right track with your solution, but there may be a few things you need to check to ensure accuracy.

Firstly, when you integrate 2+5sin(4πt/25), you should get 2t-5(25/4π)cos(4πt/25)+C. The constant of integration, C, is important and should be included in your final answer.

Secondly, when you are calculating F(6)-F(0), make sure you are using the correct values for t. In this case, t=6 and t=0, not t=6π and t=0π. Also, make sure you are using the correct units for the values of t. If t is in seconds, then your final answer for F(6)-F(0) should be in seconds as well.

Lastly, double check that your calculator is set to the correct mode (radian or degree) and that you are using the correct value for π. Sometimes calculators have a different key for π, so make sure you are using the correct one.

I hope this helps and good luck with your integration! Don't hesitate to ask for clarification if you need it.