Question

\[\int\limits_{2\pi}^{3\pi} \left| sinx/2 \right|dx\]

Answer 1

split the limit make function such that sine is always positive in 2pi to 3pi

Answer 2

the best start would be to draw graph first.

Answer 3

split it as in \[\int\limits_{2\pi}^{3\pi}\left| sinx \right|dx + \int\limits_{2\pi}^{3\pi}\left| 1/2 \right|dx\]

Answer 4

so your function was 1/2 * sin(x)

Answer 5

no my function was the absolute value of sinx/2

Answer 6

oh so rather than the + sign it should be a - sign

Answer 7

was it sin(x/2) for sin(x)/2 ??

Answer 8

it's sinx/2

Answer 9

can put plot that function in wolframalpha ??

Answer 10

i don't know how to use wolfram

Answer 11

im not good with graphing :(

Answer 12

http://www.wolframalpha.com/input/?i=plot+ |sinx%2F2|from+x%3D2pi+to+3pi

Answer 13

and make sure if 1/2 in inside sin or outside??
http://www.wolframalpha.com/input/?i=graph+sin%28x%2F2%29+from+3pi+to+4pi

Answer 14

http://www.wolframalpha.com/input/?i=graph+ |sin%28x%2F2%29|+from+3pi+to+4pi

Answer 15

http://www.wolframalpha.com/input/?i=graph+sin%28x%29%2F2+from+3pi+to+4pi

Answer 16

it's negative so ... integrate -sinx instead of |sinx| ... if you know how to integrate it should be all right.

Answer 17

http://www.wolframalpha.com/input/?i=integrate+-sin%28x%29%2F2+from+3pi+to+4pi

Answer 18

im so confused

Answer 19

you know how to integrate right??

Answer 20

can u explain to me step wise on how to solve it. i'm studying for a test and i don't know how to solve problems like these

Answer 21

a little.

Answer 22

http://www.wolframalpha.com/input/?i=integrate+ |sin%28x%29%2F2|+from+2pi+to+3pi

Answer 23

can you integrate 1/2*sin(x)

Answer 24

i don't know how to integrate with |absolute value|

Answer 25

i know that we can take the constant 1/2 out so it can written as \[1/2\int\limits_{2\pi}^{3\pi}|sinx| dx\]

Answer 26

in place of |sin x| you can place - sin x
because sin x will be negative in interval 2pi to 3pi

Answer 27

\[\int\limits_{}^{} sinx dx= cosx \] i know that

Answer 28

i meant -cosx sorry

Answer 29

\(\huge \int_{2\pi}^{3\pi}|\sin x| dx = \int_{2\pi}^{3\pi}-\sin x dx \)

Answer 30

ok. so when i integrate that i'll get positive cosx

Answer 31

|x| = -x or |x| = +x

you can change values according to make x always positive.

YES

Answer 32

so after integrating it's 1/2cos(3pi)-1/2cos(2pi)

Answer 33

yes ...thats your answer.

Sorry about the limit, .... the limit was mistake.

Answer 34

ok thanks you!!!

Answer 35

yw .. np