Question

\[\int\limits_{0}^{\pi}1/2(\cos x+\left| \cos x \right|)dx\]=?

Answer 1

\[\int\limits_{0}^{\pi}1/2(\cos x+\left| \cos x \right|)dx\]

Answer 2

I know the answer is 1. But how to get the answer?

Answer 3

\[\frac{1}{2} \int_0^{\pi } \cos (x) \, dx\] +\[\frac{1}{2} \int_0^{\frac{\pi }{2}} \cos (x) \, dx\]+\[-\frac{1}{2} \int_{\frac{\pi }{2}}^{\pi } \cos (x) \, dx\]

Answer 4

\[=\frac{1}{2}\int\limits_{0}^{\pi}cosx dx+\frac{1}{2}\int\limits_{0}^{\pi}\left| cosx \right|\]Now, the first integral is easy:\[\frac{1}{2}sinx|_{0}^{\pi}=0\]

Answer 5

\[\int_0^{\pi } |\cos (x)| \, dx=\int_{\frac{\pi }{2}}^{\pi } -\cos (x) \, dx+\int_0^{\frac{\pi }{2}} \cos (x) \, dx\]

Answer 6

Why become ∫ππ2−cos(x)dx+∫π20cos(x)dx ?

Answer 7

because value of cos is negative between pi/2 to pi

Answer 8

but if we integrate we get sin right?

Answer 9

So, we are left with the absolute value integral. We note that \[cosx \ge0\]When,\[0\le x \le \frac{\pi}{2}\]So we can split the integral into:\[\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}cosx dx + \frac{1}{2}\int\limits_{\frac{\pi}{2}}^{\pi}\left| \cos x \right|dx\]Since cosx is negative in the second interval we write:\[\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}cosx dx - \frac{1}{2}\int\limits_{\frac{\pi}{2}}^{\pi}cos x dx\]This gives:\[\frac{1}{2}sinx|_{0}^{?\frac{\pi}{2}}-\frac{1}{2}sinx|_{\frac{\pi}{2}}^{\pi}\]This gives:\[\frac{1}{2}-\frac{1}{2}(0-1)=1\]

Answer 10

we want value of cos to be positive all through out. So when Cos[x] is positive we leave it as it is. When it is negative, we negate it to turn it into positive

Answer 11

do you remember the definition
|x|=x when x>0
=-x when x<0
?

Answer 12

I only know that |-x|=x.
But I think I understand. Thanks all.

Answer 13

so is |-x|=x true for all x?
what about if x=-3
|-(-3)|=-3 <-this is not true
distance can't be negative so |-x|=x is untrue

Answer 14

I remember that cos is positive for 0 to 90 degree and 270 to 360 degree. So I think I understand how to do it.Thanks.

Answer 15

ok well its also good that you know that definition i mention too
everywhere you see an x you can put cosx instead

Answer 16

What I mean is the value inside|| will be all positive.

Answer 17

or neutral lol

Answer 18

Ya. Thanks a lot.

Answer 19

let us know if you have anymore questions steven
we haven't gotten alot of calculus questions
and cal questions are more fun

Answer 20

today anyways