- #1
DivGradCurl
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The general idea I have in mind when it comes to analyzing a graph that has a derivative [tex]f[/tex] and its antiderivative [tex]F[/tex] ([tex]\mathrm{C}=0[/tex]) is the following:
- When [tex]f[/tex] is positive, [tex]F[/tex] increases.
- When [tex]f[/tex] is negative, [tex]F[/tex] decreases.
However, I came across a couple of problems in my textbook that don't seem to work exactly that way. That's when either the derivative or the integral is an even function. Here is what I'm talking about:
[tex]\int \frac{x}{\sqrt{x^2 +1}} \: dx = \sqrt{x^2 +1} + \mathrm{C} \qquad (1)[/tex]
[tex]\int \tan ^2 \theta \sec ^2 \theta \: d\theta = \frac{\tan ^3 \theta}{3} + \mathrm{C} \qquad (2)[/tex]
The graphs are located at http://photos.yahoo.com/thiago_j
Note: the blue curves represent [tex]f[/tex], while the red ones represent [tex]F[/tex]. The elements from Eq. (1) are depicted in "calculus-5-5---34" while those from Eq. (2) appear in "calculus-5-5---36".
Is this correct?
Do I need to modify the domain so that I only show the part of the plot that work as expected?
Thank you very much
- When [tex]f[/tex] is positive, [tex]F[/tex] increases.
- When [tex]f[/tex] is negative, [tex]F[/tex] decreases.
However, I came across a couple of problems in my textbook that don't seem to work exactly that way. That's when either the derivative or the integral is an even function. Here is what I'm talking about:
[tex]\int \frac{x}{\sqrt{x^2 +1}} \: dx = \sqrt{x^2 +1} + \mathrm{C} \qquad (1)[/tex]
[tex]\int \tan ^2 \theta \sec ^2 \theta \: d\theta = \frac{\tan ^3 \theta}{3} + \mathrm{C} \qquad (2)[/tex]
The graphs are located at http://photos.yahoo.com/thiago_j
Note: the blue curves represent [tex]f[/tex], while the red ones represent [tex]F[/tex]. The elements from Eq. (1) are depicted in "calculus-5-5---34" while those from Eq. (2) appear in "calculus-5-5---36".
Is this correct?
Do I need to modify the domain so that I only show the part of the plot that work as expected?
Thank you very much