Question

integral(f(x))dx from 0 to pi where f(x)= sinx if 0<=x 14 years ago

Answer 1

-2

Answer 2

not -2

Answer 3

that's what i got too

Answer 4

would you not just add the integral of sin(x) from 0 to pi/2 to the integral of cos(x) from pi/2 to pi? which is 1 + (-1) = 0, this is a piecewise defined function

Answer 5

ok, the answer is 0, can u elaborate please?

Answer 6

note the integral of sinx is -cosx

Answer 7

ok, i can follow that

Answer 8

i also know that the integral of cosx is sinx

Answer 9

it's the evaluation idk how to do

Answer 10

f(x) = sin(x) for 0< x </2 and
f(x) = cos(x) for pi/2< x <pi

so in evaluating the integral of sin(x), which is -cos(x) you get:
-cos(pi/2) - (-cos(0)) which is 0 - (-1) which is 1.

then evaluate the integral of cos(x) which is sin(x):
sin(pi) - sin(pi/2) which is -1 - 0 = -1

add the values and you will get 0

Answer 11

but aren't the 0, pi/2, and pi, restrictions of the graph?

I see your logic but the question was integral of f(x)dx where f(x)=sin(x) if 0< = x < pi/2
and
f(x)=cos(x) if pi/2< = x <= pi

or is that irrelevant?

Answer 12

so this says, f(x) = sin(x) where x is greater than or equal to x and less than pi/2,

and f(x) = cos(x) where x is greater than or equal to pi/2 and less than or equal to pi.

all values you mentioned are defined in the graph (0,pi/2,pi)

you are finding area under curve from [0,pi/2) and [pi/2,pi]

Answer 13

COOL! THANKS MAN (or girl/woman/miss, don't know your gender sorry) YOU ROCK!