Calculus Integration
- Thread starter rinspd
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Hello, rinspd!
This is a trick question.
I wouldn't call it divergent . . . It's indeterminate.
. . And that may be that's the wrong term, too.
We must evaluate: .\(\displaystyle \displaystyle -\cos x\,\big]^{\infty}_{\text{-}\infty} \;=\;\lim_{t\to\infty}\big[-\cos x\big]^t_{\text{-}t}\)
. . . . . . . \(\displaystyle \displaystyle =\; \lim_{t\to\infty}\big[-\cos t\big] - \lim_{t\to\infty}\big[-\cos(\text{-}t)\big] \)
And neither limit has a determinable value.
On the other hand, we are asked to find the area "under" the sine curve.
Since the area above the x-axis "equals" the area below the x-axis,
. . we could conclude that the total area is zero.
The evaluation can be performed like this:
We have: .\(\displaystyle -\cos x\,\bigg]^t_{\text{-}t} \;=\; \big[-\cos(t)\big] - \big[-\cos(\text{-}t)\big] \)
. . . . . . . . . . . . . . . . \(\displaystyle =\;\big[-\cos t\big] - \big[-\cos t\big] \)
. . . . . . . . . . . . . . . . \(\displaystyle =\;-\cos t + \cos t \)
. . . . . . . . . . . . . . . . \(\displaystyle =\;0\;\;\hdots\text{ regardless of the value of }t\)
This is a trick question.
I wouldn't call it divergent . . . It's indeterminate.
. . And that may be that's the wrong term, too.
We must evaluate: .\(\displaystyle \displaystyle -\cos x\,\big]^{\infty}_{\text{-}\infty} \;=\;\lim_{t\to\infty}\big[-\cos x\big]^t_{\text{-}t}\)
. . . . . . . \(\displaystyle \displaystyle =\; \lim_{t\to\infty}\big[-\cos t\big] - \lim_{t\to\infty}\big[-\cos(\text{-}t)\big] \)
And neither limit has a determinable value.
On the other hand, we are asked to find the area "under" the sine curve.
Since the area above the x-axis "equals" the area below the x-axis,
. . we could conclude that the total area is zero.
The evaluation can be performed like this:
We have: .\(\displaystyle -\cos x\,\bigg]^t_{\text{-}t} \;=\; \big[-\cos(t)\big] - \big[-\cos(\text{-}t)\big] \)
. . . . . . . . . . . . . . . . \(\displaystyle =\;\big[-\cos t\big] - \big[-\cos t\big] \)
. . . . . . . . . . . . . . . . \(\displaystyle =\;-\cos t + \cos t \)
. . . . . . . . . . . . . . . . \(\displaystyle =\;0\;\;\hdots\text{ regardless of the value of }t\)
pka
Elite Member
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- Jan 29, 2005
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In order to evaluate \(\displaystyle \int_{ - \infty }^\infty {\sin (x)dx} \), there are two limits to consider:
\(\displaystyle \displaystyle\lim_{b \to \infty } \int_0^b {\sin (x)dx} \;\& \,\lim _{a \to - \infty } \int_a^0 {\sin (x)dx} \)
Both must converge. But neither one does.
\(\displaystyle \displaystyle\lim _{b \to \infty } \int_0^b {\sin (x)dx} = \lim _{b \to \infty } \left[ { - \cos (x)} \right]_{x = 0}^{x = b} = \lim _{b \to \infty } \left( {1 - \cos (b)} \right)\)
We see that the limit oscillates.