Question

integrate part of sine in interval x=[0, pi/2]

function:
f(x) = sin(x)

Anti-Derivative:
F(x) = -1/x cos(x)

Answer 1

Is the anti derivative correct ?

Answer 2

The antiderivative of sin(x) is not -1/x cos(x). Just take the derivative of that "antiderivative" to see that the product rule will give you something with 1/x^2 *cos(x).

But you're not completely wrong in guessing that. If it was sin(ax) where a is just a constant, like 2, then you would say it's -1/a*cos(ax). Take the derivative of this antiderivative to check and make sure I'm not lying to you!

Answer 3

you are right

Answer 4

x derived is 1......

Answer 5

by x

Answer 6

F(x) = (1/ 1 ) -cos(x)

Answer 7

@Kainui is that the right anti- derivative? :)

Answer 8

f(x) = sin(x)
then
F(x) = -cos(x)

Answer 9

That is the correct antiderivative!

Answer 10

ok thanks :)

so then I can take the end point - start point:
\[\left[ -\cos(x)\right] \frac{ to }{ from }\]-cos(pi/2) - -cos(0)
-cos(pi/2) + cos(0)
-0 + 1
= 1