Area between curves (integration)

**turillian@gmail.com** · June 8th 2007, 10:10 AM

Find the area between the curves y=(2/x) and 3x-2y=3, from x=0.5 to x=4.

Simple simple, I'm sure there's something small I'm overlooking. I'll try to get my solution up here so someone can point out where I'm going wrong. No need to draw a diagram for this equation, something small along the way is messing me up.

**Jhevon** · June 8th 2007, 10:20 AM

Originally Posted by turillian@gmail.com

No need to draw a diagram for this equation, something small along the way is messing me up.

actually i think drawing the graph would be helpful. you may even see what's been messing you up. the graphs actually intersect between 0.5 and 4, so you would have to split the integral into two parts

**turillian@gmail.com** · June 8th 2007, 06:15 PM

I got my solution done, I'll post it currently to make sure I didn't mess up somewhere.

Something that's giving me trouble right now is the following. I know I have to use the reverse chain rule and/or the substitution rule, but I'm just not getting it.
(integral symbol)4x^3 * (x^4 -1)^5 * dx

**Krizalid** · June 8th 2007, 06:51 PM

$\int4x^3(x^4-1)^5~dx$

Just set $u=x^4-1$

**turillian@gmail.com** · June 9th 2007, 02:10 PM

I can't figure out how to do that substitution question. My textbook gives one substitution example but it's a very different looking question. I think If I saw one more solution I could figure more out.

I got stuck on an additional question. I'm posting my partial solution (along with another one that I'm not sure I did correctly) currently.

**topsquark** · June 9th 2007, 03:02 PM

Originally Posted by turillian@gmail.com

I can't figure out how to do that substitution question. My textbook gives one substitution example but it's a very different looking question. I think If I saw one more solution I could figure more out.

I got stuck on an additional question. I'm posting my partial solution (along with another one that I'm not sure I did correctly) currently.

A simple version of this is if you are given an integral of the form:
$\int f(x) f^{ \prime} (x) dx$

Make the substitution $u = f(x)$ . Then $du = f^{ \prime} (x) dx$ , so when you substitute into your integral you get:
$\int f(x) f^{ \prime} (x) dx = \int f(x) ( f^{ \prime} (x) dx) = \int u \, du$
which you can easily integrate. Many of the substitution questions you will see will be of this form. So look for integrands that can be split up in this manner.

A similar type of problem would be
$\int f(g(x)) g^{ \prime} (x) dx$

Here we set $u = g(x)$ so that $du = g^{ \prime} (x) dx$ and the integral becomes
$\int f(g(x)) g^{ \prime} (x) dx = \int f(u) du$
which is much simpler to do. (This is the case of your last example. f(u) = u^5 and g(x) = x^4 - 1.)

Other substitutions are done in order to express the integral in a more familiar form, or that of a problem you already know how to solve. There is no template for these except to do problem after problem. Fortunately these questions come up fairly often on the forum and you will see many examples done. (Particularly on trigonometric substitutions.)

Hope this has helped.

-Dan

**Jhevon** · June 9th 2007, 03:05 PM

Originally Posted by turillian@gmail.com

I got my solution done, I'll post it currently to make sure I didn't mess up somewhere.

Something that's giving me trouble right now is the following. I know I have to use the reverse chain rule and/or the substitution rule, but I'm just not getting it.
(integral symbol)4x^3 * (x^4 -1)^5 * dx

by the way, you seem to be really off here. i got $A_{Total} \approx 6.946067037$ . Apparently what you did was integrate the function $f(x) = 3x^2 - 3x - 4$ , that's the wrong function to integrate. That function was just to find the intercepts, once we did that, we are done with it

The area you seek will be given by:

$A_{Total} = \int_{0.5}^{ \frac {3 + \sqrt {57}}{6}} \left( \frac {2}{x} - \frac {3}{2}x + \frac {3}{2} \right)dx + \int_{ \frac {3 + \sqrt {57}}{6}}^{4} \left( \frac {3}{2}x - \frac {3}{2} - \frac {2}{x} \right)dx$

**Plato** · June 9th 2007, 03:37 PM

Once again Jhevon is dead-on correct about this problem.
I also know that many of you will disagree with my views on this sort of problem.
I see absolutely no value in assigning such a problem unless we expect the student to use a computer algebra system(CAS).
Consider this: $\int\limits_{0.5}^4 {\left| {\left( {\frac{2}{x}} \right) - \left( {\frac{{3x - 3}}{2}} \right)} \right|dx}$ answers the question correctly.
Don’t we want students to understand why that works?
Or is the point to put the student through algebraic hoops that we ourselves hate?

**Jhevon** · June 9th 2007, 03:50 PM

Originally Posted by Plato

I also know that many of you will disagree with my views on this sort of problem.
I see absolutely no value in assigning such a problem unless we expect the student to use a computer algebra system(CAS).

i agree with you!

Consider this: $\int\limits_{0.5}^4 {\left| {\left( {\frac{2}{x}} \right) - \left( {\frac{{3x - 3}}{2}} \right)} \right|dx}$ answers the question correctly.
Don’t we want students to understand why that works?

in all my years of doing problems like these, i have never seen that approach. thanks for that. i'll have to think about it and make sure that I know why it works. (i think i already do, but just to be safe

). but that's nice.

Or is the point to put the student through algebraic hoops that we ourselves hate?

that's a question that i've always wondered about. why is it that professors do that?!

**qbkr21** · June 9th 2007, 04:52 PM

People make this Volume and Area thing way to complicated. If you happen to get a problem in terms of y just switch "everything" from Y's to X's or vice versa. This makes it super easy, because then you can just stick it directly into your calculator. Here is an sneaky example

**qbkr21** · June 9th 2007, 04:53 PM

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