Question

Find the average value of
f(x) = sin^2x+ cos^2x+x
on the interval [0,pi]

Answer 1

Average value over interval (a,b) is just
\(\Large f_{avg} = \dfrac{1}{b-a}\int_a^b f(x)dx\)

Answer 2

i broke it into 3 integrands and solved
got pi/2,pi/2, and pi^2/2

Answer 3

and then yes put it over pi
but according to wolfram the solution is (pi+2)/2

Answer 4

why not just use the fact that
\(\Large \sin^2x+ \cos^2x =1\)

Answer 5

your f(x) is just 1+x

Answer 6

thanks for making me feel like a jack retricelol. i should really study those trig identities

Answer 7

lol, you're learning, it happens :)

Answer 8

ok still somewhat incorrect. that still gives pi+pi/2

Answer 9

over pi for the avg

Answer 10

integral of x is x^2/2
so you mean

pi+pi^2/2
?

Answer 11

yeah

Answer 12

wolfram has 1/2pi(pi+2)

Answer 13

or (2+pi)

Answer 14

1+pi/2
wolfram and you, both get this

Answer 15

pi+pi^2/2

this divided by pi
is
1+pi/2

or
(2+pi)/2

Answer 16

i don't see any mismatch

Answer 17

no your absolutely correct

Answer 18

ask if anymore doubts :)

Answer 19

40 gold stars for you today. snaps for hartnn

Answer 20

lol, thank you :D