Question

what is the average value of y = sin2x over the interval ( pi/4, pi/3)

Answer 1

average value of \(f(x)\) in interval \((a, b)\) = \(\large \mathbb{\frac{1}{b-a}\int_a^b f(x) dx}\)

Answer 2

take the definite integral in given interval,
divide by the difference b-a

Answer 3

I got, 1/pi/12 ( sin 2x)

Answer 4

but then I did the integral for ( sin2x) and got - cos2x

Answer 5

yes evaluate the limits

Answer 6

-1/pi/6?

Answer 7

i got that answer but the correct answer is - 1/6pi...

Answer 8

average value of \(\sin 2x\) in interval \((\pi/4, \pi/3)\) = \(\large \mathbb{\frac{1}{\pi/3-\pi/4}\int_{\pi/4}^{\pi/3} \sin 2x dx}\)

Answer 9

yea that's what i did ..

Answer 10

how do I evaluate without using the calculator tho?

Answer 11

\(\large \mathbb{\frac{1}{\pi/3-\pi/4}\int_{\pi/4}^{\pi/3} \sin 2x dx}\)

\(\large \mathbb{\frac{1}{\pi/3-\pi/4} \frac{-\cos 2x}{2}\Big|_{\pi/4}^{\pi/3} }\)

Answer 12

\(\large \mathbb{\frac{-6}{\pi} \cos 2x\Big|_{\pi/4}^{\pi/3} }\)

Answer 13

\(\large \mathbb{\frac{-6}{\pi} [ \cos(2\pi/3) - \cos (\pi/2)] }\)

Answer 14

\(\large \mathbb{\frac{-6}{\pi} [ \cos(\pi - \pi/3) - 0] }\)


\(\large \mathbb{\frac{-6}{\pi} [ \cos(\pi - \pi/3) - 0] }\)

\(\large \mathbb{\frac{-6}{\pi} [ -\cos( \pi/3) ] }\)

\(\large \mathbb{\frac{-6}{\pi} [ -1/2 ] }\)

\(\large \mathbb{\frac{3}{\pi} }\)

Answer 15

the right answer is suppose to be - 1/ 6pi tho

Answer 16

for \(\sin(2x)\), the average value in interval \((\pi/4, \pi/3)\) is \(3/\pi\)


check ur question again

Answer 17

the question is correct.

Answer 18

http://www.wolframalpha.com/input/?i=average+value+of+sin%282x%29+pi%2F4%3Cx%3Cpi%2F3

Answer 19

if u think a bit, u wil see that average value has to be POSITIVE.
wat u saying is the answer is negative.

Answer 20

it is negative because the integral of sin is - cos