Question

Kuis
thn baru (xvi.)

[tex]\rm\int_0^{\pi}sin(x)dx =[/tex]

(A.)⠀0
(B.)⠀2
(C.)⠀1
(D.)⠀-2
(E.)⠀-1

Answer 1

--

[tex] \\ \\ \\ \\ \\ [/tex]

[tex]\rm\int_0^{\pi}sin(x)dx [/tex]

= [tex] \int sin(x) = dx[/tex]

= [tex] - cos(x)[/tex]

= [tex] - cos (\pi)+ cos(0)[/tex]

= [tex]2[/tex] B.

[tex] \\ \\ [/tex]

[tex] \\ \\ \\ \\ \\ [/tex]

--

Answer 2

Verified answer

Integral

∫d(x) = x + C

∫sin x dx [π 0]

= ∫d(-cos x)

= - cos x

= - (cos π - cos 0)

= - (-1 - 1)

= 2