Finding Lines a & b through Integration

[Rayquesto]

Finding Lines "a" & "b" through Integration

Homework Statement

Let f be the function given by f(x)=4cosx

a) This one I figured out.
b) Find the numbers a and b such that the lines x=a and x=b divide the region R into three regions with equal area.

Homework Equations

∫4cos(x)dx=4sin(x) + c

The Attempt at a Solution

Typing out the question, I started to realize one thing; a and b will be numbers, not variables. So, these lines where x=a and x=b will be vertical. Knowing this is much more helpful. But now that I have this logic deduced, let me try and see if I could do the problem. Hm...

Oh I know what else would help... a very important yet easy to miss rule...The sum rule of integration.

So,

∫[0,a] 4cos(x)dx + ∫[a,b] 4cos(x)dx + ∫[b,∏/2] 4cos(x)dx= 4 since from letter a question which I didn't state is 4 (∫[0,∏/2]4cos(x)dx=4)

Therefore,

∫4cos(x)dx=4sin(x) + c
& assuming c=0 *is this a safe assumption? I would believe so, but I can't seem to prove it 100%

[4sin(a)] + [(4sin(b)-4sin(a)] + [(4sin(∏/2)-4sin(b))]=4

but also

∫[0,a] 4cos(x)dx = ∫[a,b] 4cos(x)dx = ∫[b,∏/2] 4cos(x)dx= 4/3 (three equal areas)
then I could solve for b and a hopefully. Let me come up with some equations...

so,

4sin(b) - 4sin(a)=4/3

4sin(b) - 4/3 = 4sin(a)

∫[0,a] 4cos(x)dx=4sin(a)=4/3

therefore,

4sin(b) - 4/3 = 4/3

4sin(b)=8/3

sin(b)=2/3

b=sininverse(2/3)

b= .7297=x

so,

∫[a,b] 4cos(x)dx=[(4sin(b)-4sin(a)]=4/3

4sin(.7297) - 4sin(a)=4/3

a=sininverse(.33333333)=.3398

Does this look right everyone? a=.3398, b=.7297 it looks right to me. I guess when I attempted the solution, I wasn't thinking right or something.

 

[MisterX]

What's the "region R" ?

 

[Rayquesto]

4 from a. sorry. I did not post the question. That was first question though and I got the answer correct.

 

[Screwdriver]

If I'm understanding correctly, $R$ is the area under the graph of $f(x) = 4cos(x)$ between 0 and π/2, which you've determined correctly to be 4.

∫[0,a] 4cos(x)dx + ∫[a,b] 4cos(x)dx + ∫[b,∏/2] 4cos(x)dx= 4

This is true, but it doesn't really help you.

∫[0,a] 4cos(x)dx = ∫[a,b] 4cos(x)dx = ∫[b,∏/2] 4cos(x)dx= 4/3 (three equal areas)

This is what you want. Note that you don't actually need $\int_{a}^{b}4cos(x)dx=4/3$, since the other two integrals each only have an $a$ and a $b$.

Does this look right everyone? a=.3398, b=.7297 it looks right to me. I guess when I attempted the solution, I wasn't thinking right or something.

That's what I got. Good job. Maybe leave the answers in the form $a = sin^{-1}(1/3)$ and $b = sin^{-1}(2/3)$, but it's up to you.

 

[SammyS]

Homework Statement

Let f be the function given by f(x)=4cosx

a) This one I figured out.
b) Find the numbers a and b such that the lines x=a and x=b divide the region R into three regions with equal area.

Homework Equations

∫4cos(x)dx=4sin(x) + c
...

What's the "region R" ?

4 from a. sorry. I did not post the question. That was first question though and I got the answer correct.

It's understandable that in your original post you didn't post enough information about the question you're solving. But then, after being asked by MisterX, 'What is the "Region R"?', you still didn't post an intelligible question. You didn't even clearly state what Region R is !

Although Screwdriver may be commended for digging through your whole post to figure out what problem you were solving, as well as checking your solution, you shouldn't expect that very many of us would attempt that.

In the future, please, please, please ... post the entire problem you're attempting to solve -- even if you have already solved parts a, b, c, … and only want help with part x . In fact it will also help if you summarize your results for those parts you have already solved !

Those of us who help on this site are volunteers, offering our time freely.

Have a good day !

 

[Rayquesto]

I'll keep that in mind. Thanks. I need to be a little bit more respectful to the people who use this site and the site itself. I must admit that I have been getting much better at it though.

 

[Rayquesto]

OH! I see what the dilemma is now. I feel so dull. The region R contains the limits of integration from [0, pie/2] under f. I'm truly very sorry for NOT emphasizing this fact. I assumed I wrote it already, since I was thinking about different things at once. As I said before, I will respect the site of those who are using it as well much more. I'll remember to start from a and end at z. Assume no one can read my mind.

 

[SammyS]

Thank you, Rayquesto.

SammyS