# Integration and differential

## Homework Statement

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

## 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 wanna 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?