Teaching Myself Anti-Derivatives: A Calc II Homework Challenge

[tony873004]

I’ve dropped my Calc II class because my Calc I class, which I took at the community college, never covered anti-derivatives, and my Calc II class started off assuming you already knew anti-derivatives and integration.

So now I have the fun task of teaching myself anti-derivatives and integration so I can take Calc II again next semester. So I might as well attempt the problems given as homework to the Calc II class. But unlike the homework, I’m going to choose the odd numbered problems so I can check my answers.

Q. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. [tex]
y = x^2 ,\,\,x = 1,\,\,y = 0;\,
[/tex]about the x-axis.

After drawing it, I came up with
[tex]
\begin{array}{l}
v = \sum\limits_0^1 {\pi r^2 } \Delta x = \pi \sum\limits_0^1 {r^2 } \Delta x \\
\\
v = \pi \int\limits_0^1 {r^2 ,\,dx} \\
\end{array}
[/tex]

Now here’s where my lack of anti-derivative skills hurt me. What do I do next? To get the anti-derivative of r squared, do I add 1 to the exponent and divide the whole thing by the new exponent? Should I get [tex]
\pi \frac{{r^3 }}{3}
[/tex]? If so, how do I apply this new formula to get an answer? My integral goes from 0 to 1. How do I get from here to the final answer of [tex]
\pi /5
[/tex]?

 

[FrogPad]

Remember what an anti derivative is. This will help you check yourself.

example)

[tex] \int x^4 \,\, dx = \frac{1}{5}\,x^5 + C [/tex]

Now differentiate it.

[tex] \frac{d(\int x^4 \,\, dx)}{dx} = \frac{d(\frac{1}{5}\,x^5 + C)}{dx} = \frac{1}{5} \, \left( \frac{d(x^5)}{dx} + \frac{d(C)}{dx}\right) = \frac{1}{5} \, 5x^4 + 0 [/tex]

... hmmm I just noticed you put
[tex]
v = \pi \int\limits_0^1 {r^2 ,\,dx}
[/tex]

Did you mean for the [itex] r^2 [/itex]?

 

[tony873004]

To get the volume, I'm adding up all the areas of the slices of the shape. They'll form disks with a thickness of delta x when rotated about the x-axis, so for the area I want to do pi r squared. Since radius of each slice would be y(x), I should have put x² there instead of r², especially since I put dx.
[tex] v = \pi \int\limits_0^1 {x^2 ,\,dx} [/tex]

 

[FrogPad]

Ok, well:

[tex] \pi \int x^2 \, dx = \pi \frac{x^3}{3} [/itex]
(setting the constant equal to 0)

[tex] \pi \int_a^b x^2 \, dx = \pi \left( \frac{b^3}{3} - \frac{a^3}{3} \right) [/tex]

 

[tony873004]

Or maybe I meant
[tex] v = \pi \int\limits_0^1 {(r^2) ,\,dx} [/tex]
give me a few minutes to think about this.
edit ^^ my tex didn't do what I wanted it to do

 

[tony873004]

[tex]
\begin{array}{l}
v = \pi \int\limits_0^1 {\left( {x^2 } \right)^2 \,,\,dx} \\
v = \pi \int\limits_0^1 {x^4 \,,\,dx} \\
v = \pi \frac{{x^5 }}{5} \\
v = \frac{{\pi x^5 }}{5} \\
\end{array}
[/tex]

So if I plug in 1 for x, I get the same answer as the back of the book. What would I have done if the integration went from 2 to 3 instead of 0 to 1? Do I subtract these numbers and plug it in the formula?

 

[FrogPad]

Let's say you have a general function (that's integrable) [itex] f(x) [/itex]

Let the antiderivative equal [itex] F(x) [/itex].

Thus,

[tex] F(x) = \int f(x) \, dx [/itex]

[itex] F(x) [/itex] is called an indefinite integral.

If you were to evaluate this integral (as a definite integral), you would have.

[tex] \int_a^b f(x) \, dx =F(b) - F(a)[/tex]

This is the first fundamental http://mathworld.wolfram.com/FirstFundamentalTheoremofCalculus.html" of calculus.

This answers your question in general.

But to answer your question specifically.

You asked, what if I integrate from 2 to 3 instead.

[tex] \pi \int_2^3 x^4 \, dx [/tex]

From the first fundamental theorem.
[tex] \int_a^b f(x) \, dx =F(b) - F(a)[/tex]
[tex] a = 2 [/tex]
[tex] b = 3 [/tex]
[tex] f(x) = x^4 [/tex]

Now we have to find [itex] F(x) [/itex]
[tex] F(x) = \int x^4 \, dx = \frac{x^5}{5} + C[/tex]

Now plugging in a, b
[tex] F(b) = F(3) = \frac{3^5}{5} + C [/tex]
[tex] F(a) = F(2) = \frac{2^5}{5} + C [/tex]

Finally [itex] F(b) - F(a) [/itex] is equal to:
[tex] \left(\frac{3^5}{5} + C \right) - \left( \frac{2^5}{5} + C \right) [/tex]

Notice that the constant is dropped.
Also note that this was multiplied by [itex] \pi [/itex] !

 

[FrogPad]

Just practice some.

You can check your work with,
http://integrals.wolfram.com/index.jsp

Here is a good free calculus book.
http://ocw.mit.edu/ans7870/resources/Strang/strangtext.htm

You can read up on integrals there.

 

[J77]

tony - you're going a bit wrong bringing the r into play.

For solids of revolution, you just need the curves y=f(x).

Sketch these curves first.

You can then apply the formula:

[tex]V=\pi\int_0^1f(x)^2dx[/tex]

 

[tony873004]

Thank you FrogPad for your help and for the links.

And Thanks J77. I figured that out that I needed to replace r with the function that gives me r, and it must be in terms of x.