Find area bounded by graph

[Tiaan]

Find the area bounded by the graph of

y=(3sinx-1)(sinx+3)e^x(2/3)sinx (0,x,2pi)

This is not my actual question it is a similar example, could you please go through the steps of how to intergrate to find the area bound by the graph.

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ \frac{2}{3}e^{x}sin(x)[3sin(x)-1][sin(x)+3]$

Is this the function you want solved?

 

[Tiaan]

Yes it is....

 

[BigGlenntheHeavy]

You sound hesitant. Are you sure?

 

[Tiaan]

Yep I am sure

 

[Deleted member 4993]

I'll do a simple problem - then you could tell me where you are stuck with your actual problem.

Find the area bounded by the graph of

y = x (0,x,2?)

$\displaystyle Area \, bounded \, between (a\le x\le b)\, by \, a \, given \, graph \, [y \, = \, f(x)] \, is \, = \, \int_a ^ b f(x) dx$

for the problem above:

$\displaystyle Area \, = \, \int_0 ^ {2 \pi} x dx \, = \left [ \frac{x^2}{2}\right]^{2\pi}_0 \, = \, \frac{4\pi^2}{2} \, - \, 0 \, = \, 2\pi^2$

Now tell us how you are stuck applying the same principle to your problem.....

 

[Tiaan]

I am stuck where I need to find the definite intergrals.

I intergrated the three separate parts so e^x(2/3)sinx,(sinx+3) and(3sinx-1), then I have three parts rather than two parts (which I have had previously on all other examples) and so don't know where to put my o and 2pi.
Or do I intergrate ,(sinx+3) and(3sinx-1), together ??

 

[galactus]

You have, if I understand correctly,

$\displaystyle \int_{0}^{2\pi}(3sin(x)-1)(sin(x)+3)e^{x}(\frac{2}{3})sin(x)dx$

If this is right, then we can expand it to:

$\displaystyle \frac{-1}{2}\int_{0}^{2\pi}e^{x}sin(3x)dx-\frac{8}{3}\int_{0}^{2\pi}e^{x}cos(2x)dx-\frac{1}{2}\int_{0}^{2\pi}e^{x}sin(x)dx+\frac{8}{3}\int_{0}^{2\pi}e^{x}dx$

and integrate term by term.

The last one is easy, but you could try using parts on the other three. Frankly, I would just run them through a calculator.

 

[BigGlenntheHeavy]

Forsooth, galactus, I concur wholeheartedly, as indeed to do this one manually, other than the last one, one is dealt a protracted exercise in integration by parts, not to mention the trees that will die an untimely death.

 

[Deleted member 4993]

I sort of agree - but

for the first integral substitute y = 3x

for the second integral substitute y = 2x

Those all become similar to the third one. Lots of trees can be saved....

 

[Tiaan]

Im really sorry if I am being stupid, but i really struggle with intergrals to be honest I dont understand them. Could you explain how you expanded them please.
Sorry for being an idiot

 

[BigGlenntheHeavy]

OK tiaan, as galactus set them up, I'll do the first one for you; then, I expect for you to follow suit.

$\displaystyle \frac{-1}{2}\int_{0}^{2\pi}e^{x}sin(3x)dx, \ I \ by \ P: \ \int udv \ = \ uv \ -\int vdu.$

$\displaystyle Do \ \int_{0}^{2\pi}e^{x}sin(3x)dx \ = \ e^{x}sin(3x)\bigg]_{0}^{2\pi}-3\int_{0}^{2\pi}e^{x}cos(3x)dx$

For the above, we let $\displaystyle dv \ = \ e^{x}dx, \ then \ v \ = \ e^{x} \ and \ u \ = \ sin(3x), \ du \ = \ 3cos(3x)dx.$

$\displaystyle Now \ \int_{0}^{2\pi}e^{x}cos(3x)dx \ = \ e^{x}cos(3x)\bigg]_{0}^{2\pi}+3\int_{0}^{2\pi}e^{x}sin(3x)dx.$

For the above, same routine as before.

$\displaystyle Ergo, \ \int_{0}^{2\pi}e^{x}sin(3x)dx \ = \ e^{x}sin(3x)\bigg]_{0}^{2\pi}-3e^{x}cos(3x)\bigg]_{0}^{2\pi}-9\int_{0}^{2\pi}e^{x}sin(3x)dx$

$\displaystyle Therefore, \ \int_{0}^{2\pi}e^{x}sin(3x)dx \ = \ \frac{e^{x}sin(3x)}{10}\bigg]_{0}^{2\pi} \ - \ \frac{3}{10}e^{x}cos(3x)\bigg]_{0}^{2\pi} \ = \ \frac{-3}{10}(e^{2\pi}-1).$

$\displaystyle And \ \frac{-1}{2}\int_{0}^{2\pi}e^{x}sin(3x)dx \ = \ (\frac{-1}{2})( \frac{-3}{10})(e^{2\pi}-1) \ = \ \frac{3}{20}(e^{2\pi}-1). QED$

Now, can you do the other three? If not, review integration by parts.

 

[mmm4444bot]

… i really struggle with intergrals … explain how you expanded them …

Are you using the pronoun "them" to refer to the integrals used by Galactus?

The expansion of your original expression doesn't involve integration. It involves some algebra (eg: the Distributive Property, combining like-terms) and some trigonometic identities (eg: power-reducing formulas, double-angle formulas).

In math, the verb "to expand" means "to multiply".

Let's first look at the two factors [3 sin(x) - 1] and [sin(x) + 3].

The expression 3 sin(x) - 1 is a binomial expression (i.e., two terms).

The expression sin(x) + 3 is also a binomial expression.

We can multiply two binomials using an algorithm known as FOIL.

Here's how the Distributive Property looks symbolically, when applied twice in the FOIL algorithm.

(A + B)(C + D) = AC + AD + BC + BD

I hope that FOIL looks familiar, to you. I think it's important for students of calculus to know this!

So, if we use the FOIL algorithm to multiply the two binomials in your exercise, we could make the following assignments (although, most calculus students do FOIL mentally).

A = 3 sin(x)

B = -1

C = sin(x)

D = 3

Next: AC + AD + BC + BD

[3 sin(x)] [sin(x)] + [3 sin(x)] [3] + [-1] [sin(x)] + [-1] [3]

3 sin(x)^2 + 9 sin(x) - sin(x) - 3

3 sin(x)^2 + 8 sin(x) - 3

Of course, your exercise came with three other factors: 2/3, e^x, and sin(x), so there's another wave of expansion to complete.

[2/3 e^x sin(x)] [3 sin(x)^2 + 8 sin(x) - 3]

Use the Distributive Property again, to expand this.

In other words, multiply 3 sin(x)^2 by 2/3 e^x sin(x).

Multiply 8 sin(x) by 2/3 e^x sin(x).

Multiply -3 by 2/3 e^x sin(x).

Combine like-terms.

(Rats, I'm running out of time, here. Gotta go, soon.)

If my algebra is good, then you should wind up with the following.

2 e^x sin(x)^3 + 16/3 e^x sin(x)^2 - 2 e^x sin(x)

Apply some trigonometric identities, to simplify the result and arrive at the expanded form used by Galactus.

If your use of the pronoun "them" was intended to ask how to do integration, then I've misunderstood your question.

 

[BigGlenntheHeavy]

galactus, good show in setting up the integral.

I couldn't figure out the first and third one until I resorted to the "Product to Sum Formulas".

Again, good show.

 

[Tiaan]

Thank you very much you really helped my understanding

 

[BigGlenntheHeavy]

For What Its Worth Dept.;

\(\displaystyle tExpand \ (F2-9-1) \ [(3sin(x)-1)(sin(x)+3)e^{x}(\frac{2}{3})sin(x)] \ on \ trusty \ TI-89 \ gives: \\)

$\displaystyle \frac{6e^{x}sin^{3}(x)+16e^{x}sin^{2}(x)-6e^{x}sin(x)}{3}$

$\displaystyle Then \ tCollect \ (F2-9-2) \ \frac{6e^{x}sin^{3}(x)+16e^{x}sin^{2}(x)-6e^{x}sin(x)}{3} \ gives:$

$\displaystyle \frac{-e^{x}[3sin(3x)+16cos(2x)+3sin(x)-16]}{6}$

$\displaystyle which \ equals: \ \frac{-1}{2}e^{x}sin(3x)-\frac{8}{3}e^{x}cos(2x)-\frac{1}{2}e^{x}sin(x)+\frac{8}{3}e^{x}.$

$\displaystyle Taking \ the \ integral \ from \ 0 \ to \ 2\pi \ yields: \ \int_{0}^{2\pi}\bigg[\frac{-1}{2}e^{x}sin(3x)-\frac{8}{3}e^{x}cos(2x)-\frac{1}{2}e^{x}sin(x)+\frac{8}{3}e^{x}\bigg]dx$

$\displaystyle Hence, \ \int_{0}^{2\pi}(3sin(x)-1)(sin(x)+3)e^{x}(\frac{2}{3})sin(x)dx$

$\displaystyle = \ \frac{-1}{2}\int_{0}^{2\pi}e^{x}sin(3x)dx-\frac{8}{3}\int_{0}^{2\pi}e^{x}cos(2x)dx-\frac{1}{2}\int_{0}^{2\pi}e^{x}sin(x)dx+\frac{8}{3}\int_{0}^{2\pi}e^{x}dx$

Note: One can expand the above statement manually (which I did, a lot of grunt work, see my thread below) or if one has a trusty TI-89, plug in the statement (without the integral sign) on F2-9-1 (expand) and then F2-9-2 (collect) and you will get the above and avoid a lot of grunt work. Isn't modern technology wonderful?

 

[BigGlenntheHeavy]

Just for kicks, let me show you the grunt work involved in expanding the following trig. statment.

$\displaystyle \frac{2}{3}e^{x}sin(x)[3sin(x)-1][sin(x)+3]$

$\displaystyle = \ \frac{2}{3}e^{x}sin(x)[3sin^{2}(x)+9sin(x)-sin(x)-3]$

$\displaystyle = \ \frac{2}{3}e^{x}sin(x)[3sin^{2}(x)+8sin(x)-3]$

$\displaystyle = \ \frac{2}{3}e^{x}sin(x)[3(1-cos^{2}(x))+8sin(x)-3], \ sin^{2}(x) \ + \ cos^{2}(x) \ =1$

$\displaystyle = \ \frac{2}{3} e^{x}sin(x)[3-3cos^{2}(x)+8sin(x)-3]$

$\displaystyle = \ \frac{2}{3}e^{x}sin(x)[8sin(x)-3cos^{2}(x)]$

$\displaystyle = \ \frac{2}{3}e^{x}[8sin^{2}(x)-3sin(x)cos^{2}(x)]$

$\displaystyle = \ \frac{2}{3}e^{x}\bigg[8\bigg(\frac{1-cos(2x)}{2}\bigg)-3sin(x)\bigg(\frac{1+cos(2x)}{2}\bigg)\bigg], \ Power \ Reducing \ Formulas$

$\displaystyle = \ \frac{2}{3}e^{x}[4-4cos(2x)-\frac{3}{2}(sin(x)+sin(x)cos(2x))]$

$\displaystyle = \ \frac{8}{3}e^{x}-\frac{8}{3}e^{x}cos(2x)-e^{x}[sin(x)+sin(x)cos(2x)]$

$\displaystyle = \ \frac{8}{3}e^{x}-\frac{8}{3}e^{x}cos(2x)-e^{x}[sin(x)+\frac{sin(3x)+sin(-x)}{2}], \ Product-to-Sum \ Formulas$

$\displaystyle = \ \frac{8}{3}e^{x}-\frac{8}{3}e^{x}cos(2x)-e^{x}[sin(x)+\frac{sin(3x)-sin(x)}{2}], \ sin(-x) \ = \ -sin(x ), \ odd \ function$

$\displaystyle = \ \frac{8}{3}e^{x}-\frac{8}{3}e^{x}cos(2x)-e^{x}[\frac{sin(x)}{2}+\frac{sin(3x)}{2}]$

$\displaystyle = \ \frac{8}{3}e^{x}-\frac{8}{3}e^{x}cos(2x)-\frac{1}{2}e^{x}sin(x)-\frac{1}{2}e^{x}sin(3x)$

Done. TI-89 is better (see my above thread).