• May 26th 2006, 04:19 AM
bobby77
what is the area beneath the curve f(x)= x^2sqrt(3-x^3) from x=-2 to x=-1
<a> 38/9
<b> (2.11^(3/2)-16) / 9
<c> this problem cannot be solved
<d> 2/3
• May 26th 2006, 06:44 AM
earboth
Quote:

Originally Posted by bobby77
what is the area beneath the curve f(x)= x^2sqrt(3-x^3) from x=-2 to x=-1
<a> 38/9
<b> (2.11^(3/2)-16) / 9
<c> this problem cannot be solved
<d> 2/3

Hello,

2. Use substitution method:
Let u(x) = 3 - x^3. Then du/dx = -3x^2. Thus x^2 = -1/3*du/dx.

Area: $\displaystyle \int_{-2}^{-1}{-{1\over 3} \cdot u^{{1\over 2}}\cdot {du\over dx} }dx$ = $\displaystyle \left[-{1\over 3} \cdot {2\over 3} \cdot u^{{3\over 2}}\right]_{4\sqrt{11}}^{2}$ = $\displaystyle \left[-{2\over 9}\cdot \sqrt{\left(3-x^3 \right)^3} \right]_{-2}^{-1}$

Plug in the values and you'll get the answer <b>

Greetings

EB
• May 26th 2006, 07:49 AM
ThePerfectHacker
It looks to me you used my method of using the substitution rule, did you?

I prefer this method because splitting the fraction du/dx is inappropirate and informal.
• May 26th 2006, 08:40 AM
earboth
Quote:

Originally Posted by ThePerfectHacker
It looks to me you used my method of using the substitution rule, did you?
...

Hello,

I've seen this method when you answered a few questions about integration. It is very easy to adopt(?) (and adapt) and it is very near the method I had been taught and I teach. See attached image.

I would appreciate it very much if I'm allowed to use your method and teach it to my pupils .

Greetings

EB
• May 26th 2006, 08:53 AM
ThePerfectHacker
Quote:

Originally Posted by earboth
Hello,

I've seen this method when you answered a few questions about integration. It is very easy to adopt(?) (and adapt) and it is very near the method I had been taught and I teach. See attached image.

I would appreciate it very much if I'm allowed to use your method and teach it to my pupils .

Greetings

EB

It is not trademarketed so of course you can.
In fact we can call it "Earboth's Substitution".

In the picture you gave, you use the classical,informal method schools use. Are you asking me if I should show you how to do it with the Earboth Substitution?
------
I think this method once adopted is much easier then the du and dx substitution. Because many people have trouble knowing which function to substitute, in this method you need find such a function whose derivative appears in the integrand and that is what makes the substituion easier. As you do realize this follows directly from the chain rule for derivatives expressed in integration form.
• May 26th 2006, 11:10 AM
earboth
Quote:

Originally Posted by ThePerfectHacker
It is not trademarketed so of course you can.
In fact we can call it "Earboth's Substitution".

...

hello,

please don't do that. Nobody will take this method seriously any more with such a name!

I only noticed that my pupils understand better your way to do the substitution because it is obvious by simple calculation of fractions how the integrand function changes the argument, so the integral is calculable.

And even I did understand this method and, as you've seen by my avatar, it's very hard to get something into my head.

Greetings

EB