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. <br /> y = x^2 ,\,\,x = 1,\,\,y = 0;\,<br />about the x-axis.

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

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 <br /> \pi \frac{{r^3 }}{3}<br />? 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 <br /> \pi /5<br />?

 

[FrogPad]

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

example)

\int x^4 \,\, dx = \frac{1}{5}\,x^5 + C

Now differentiate it.

\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

... hmmm I just noticed you put
<br /> v = \pi \int\limits_0^1 {r^2 ,\,dx} <br />

Did you mean for the r^2?

 

[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.
v = \pi \int\limits_0^1 {x^2 ,\,dx}

 

[FrogPad]

Ok, well:

\pi \int x^2 \, dx = \pi \frac{x^3}{3} [/itex]<br /> (setting the constant equal to 0)<br /> <br /> \pi \int_a^b x^2 \, dx = \pi \left( \frac{b^3}{3} - \frac{a^3}{3} \right)

 

[tony873004]

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

 

[tony873004]

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

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) f(x)

Let the antiderivative equal F(x).

Thus,

F(x) = \int f(x) \, dx [/itex]<br /> <br /> F(x) is called an indefinite integral.<br /> <br /> If you were to evaluate this integral (as a definite integral), you would have. <br /> <br /> \int_a^b f(x) \, dx =F(b) - F(a)<br /> <br /> This is the first fundamental http://mathworld.wolfram.com/FirstFundamentalTheoremofCalculus.html&quot; of calculus.<br /> <br /> This answers your question in general.<br /> <br /> But to answer your question specifically.<br /> <br /> You asked, what if I integrate from 2 to 3 instead.<br /> <br /> \pi \int_2^3 x^4 \, dx<br /> <br /> From the first fundamental theorem.<br /> \int_a^b f(x) \, dx =F(b) - F(a)<br /> a = 2<br /> b = 3<br /> f(x) = x^4<br /> <br /> Now we have to find F(x)<br /> F(x) = \int x^4 \, dx = \frac{x^5}{5} + C<br /> <br /> Now plugging in a, b<br /> F(b) = F(3) = \frac{3^5}{5} + C<br /> F(a) = F(2) = \frac{2^5}{5} + C<br /> <br /> Finally F(b) - F(a) is equal to:<br /> \left(\frac{3^5}{5} + C \right) - \left( \frac{2^5}{5} + C \right)<br /> <br /> Notice that the constant is dropped.<br /> Also note that this was multiplied by \pi !

 

[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:

V=\pi\int_0^1f(x)^2dx

 

[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.