Hi, I'm getting really weird answers for the answer of this integral
$\displaystyle \int_{0}^{\pi}\frac{1}{2}+cosx~dx$
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Hi, I'm getting really weird answers for the answer of this integral
$\displaystyle \int_{0}^{\pi}\frac{1}{2}+cosx~dx$
remember what absolute values mean. $\displaystyle x = \left \{ \begin{array}{cc} x & \mbox{ if } x \ge 0 \\ & \\ x & \mbox{ if } x < 0 \end{array} \right.$
thus we need to find where our function (without absolute values) is negative, and negate its integral on that region. we can find that 1/2 + cos(x) is negative for [2pi/3, pi], thus:
$\displaystyle \int_0^{\pi} \left \frac 12 + \cos x \right~dx = \int_0^{\frac {2 \pi}3} \left( \frac 12 + \cos x \right)~dx  \int_{\frac {2 \pi}3}^{\pi} \left( \frac 12 + \cos x \right)~dx$
oh so does the whole 1/2+cosx change to 1/2cosx?
or does just cosx change to cosx
Hello, akhayoon!
Did you make a sketch?
Quote:
$\displaystyle \int_{0}^{\pi}\left\frac{1}{2}+\cos x\right\,dx$
It is the cosine curve raised a halfunit.Code:
*
:::*
:::::*
::::::*
:::::::
:::::::*
:::::::: π
+o+
 ↑*::::::
 ↑ *::::
 2π/3 *

I totally agree with Dan's integrals and his final answer.