Question

A bicyclist is riding with speed v(t) 2t + 5 feet/second. What is the distance traveled from t=1 to t=5

Answer 1

Plug in t=1 and t=1 then do t=5-t=1

That's your answer.

Answer 2

\[v=2t+5\]
Velocity is the rate of change of displacement
\[\frac{dx}{dt}=2t+5\]
\[dx=2t+5dt\]
\[\int\limits_{}^{} dx=\int\limits_{}^{}2t+5dt\]
\[\large x=[t^2+5t]_1^5 \]
\[\large x=50-6=44ft\]