Question

F(x)=integral from 0 to x of sin(2t)dt
Evaluate F(pi)

Answer 1

\[F(x)=\int\limits_{0}^{x}\sin(2t)dt\]
Evaluate \[F(\pi )\]

Answer 2

the idea is we do the indefinate integral with respect to t
then evaluate from 0 to pi

Answer 3

so t=pi

\[\int\limits_{0}^{\pi} \sin (2t) dt\]

\[=\left[ -\cos(2\pi)-(-\cos (0)) \right]\]

\[=\left[ -\cos(2\pi)+ 1 \right]\]

Answer 4

sorry forgot something

Answer 5

you best use the rule of sin (2t)= 2 sin t cos t

Answer 6

then set u=sin t; du= cos t dt -> dt = du/cos t, substitute in the integral, and eliminate what you can. Then solve the integral using u.du. Then you've only got sin² (t) left. Apply the substraction rule for limited integrals. [sin²(t2)-sin²(t1)] Plug in the limit values pi and 0. Calculate.