Question

Evaluate the integral pi on top, 0 on the bottom f(x) dx where f(x) =
2 sin x if 0 ≤ x < π/2
3 cos x if π/2 ≤ x ≤ π
Please help, I got it wrong and I only have one more try!

Answer 1

\[\int\limits_0^{\pi} f(x) dx=\int\limits_0^\frac{\pi}{2} 2 \sin(x)dx+\int\limits_\frac{\pi}{2}^\pi 3 \cos(x) dx\]

Answer 2

so i would get 2cos(x)+(-3sin(x))?

Answer 3

@freckles

Answer 4

derivative of cos(x) is -sin(x) so no on the first one
and derivative of -sin(x) is -cos(x) and so no on the second one

Answer 5

-2sin(pi/2)+(-cos(pi/2))?

Answer 6

would my answer be -2?

Answer 7

did you actually find the antiderivative yet of each integrand ?

Answer 8

yes? I'm guessing thats not the right answer

Answer 9

then what did you get as the antiderivative for each integrand?

Answer 10

\[\frac{d}{dx} \text{ what ? }=\sin(x) \\ \text{ and } \\ \frac{d}{dx } \text{ what ? } =\cos(x)\]

Answer 11

-cos(x)
sin(x)

Answer 12

ok great so
\[\int\limits \sin(x) dx=-\cos(x)+C \\ \text{ and } \int\limits \cos(x) dx=\sin(x)+C \\ \text{ though we don't really care about the constant of integration since we have } \]
\[\text{ since we have definite integral }\]

Answer 13

\[\int\limits_0^\frac{\pi}{2} 2 \sin(x) dx=-2 \cos(x)|_0^\frac{\pi}{2}=?\]

Answer 14

and
\[\int\limits_\frac{\pi}{2}^\pi 3 \cos(x) dx=3 \sin(x)|_\frac{\pi}{2}^\pi=? \]

Answer 15

-2cos(pi/2)-2cos(0)+3sin(pi)-3sin(pi/2)

Answer 16

not exactly but almost you have one incorrect sign

Answer 17

\[-2 \cos(\frac{\pi}{2})--2 \cos(0) \\ \text{ or } -2 \cos(\frac{\pi}{2})+2 \cos(0)\]

Answer 18

oh okay, so it would give me -1?

Answer 19

yes

Answer 20

okay! thank you!

Answer 21

np