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

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- May 26th 2006, 04:19 AM #1

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- May 26th 2006, 06:44 AM #2Originally Posted by
**bobby77**

1. answer <b>

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 #3

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- May 26th 2006, 08:40 AM #4Originally Posted by
**ThePerfectHacker**

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 #5

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Originally Posted by**earboth**

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?

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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 #6Originally Posted by
**ThePerfectHacker**

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