ok, i tried to do what you told me but i cant solve it for c because they cancel each others out!
the integral for the first one i got is [sin(c)cos(x)-cos(c)sin(x)+sin(x)+c]
and
the integral for the 2nd one i got is [-sin(c)cos(x)+cos(c)sin(x)-sin(x)+c]
I don't know what to do from here. can u pleses help??? To tell you the truth i don't really get what you are saying. I really need to get the correct answer. Please Help.
Here is your respond for my question from yerster day. Thank You.
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Thank you for reposting.
I must have misinterpreted the question, but a sketch of the curves made it crystal clear:
img406.imageshack.us/img406/1995/1291016654.png
We have two distinct areas enclosed by:
1. cos(x), cos(x-c),and x=0
2. cos(x-c), cos(x) and x=¥ð
The y=0 in the second area is extraneous and misleading for someone (like me) who hasn't taken the time to make a proper sketch.
Now for the limits for each of the areas,
1. from -¥ð+c/2 to x=0
2. from c/2 to ¥ð
We are able to find the limits because of the symmetry of the curves cos(x) and cos(x-c).
Finally, the integrals:
1. ¡ò(cos(x)-cos(x-c))dx
2. ¡ò(cos(x-c)-cos(x))dx
Hope this will get you along the way.
Hint: integrate and find each of the areas, equate the two areas and solve for c [if necessary].
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1. Integral ∫(cos(x)-cos(x-c))dx
(C=integration constant)
I1=∫[cos(x)-cos(x-c)]dx
=sin(x)-sin(x-c)+C
Evaluate I1=definite integral from x=-π+c/2 to x=0
[sin(0)-sin(0-c)]-[sin(-π+c/2)-sin(-π+c/2-c)]
=[0+sin(c)]-[-sin(c/2)-sin(c/2)]
=sin(c)+2sin(c/2)
2. Integral ∫(cos(x-c)-cos(x))dx
(C=integration constant)
I1=∫[cos(x-c)-cos(x)]dx
=sin(x-c)-sin(x)+C
Evaluate I2=definite integral from x=c/2 to x=%pi;
[sin(c/2)-sin(c/2-c)] - [sin(π)-sin(π-c)]
=[sin(c)-sin(-c/2)] - [0-sin(π-c)]
=sin(c)+2sin(c/2)
So to find c, we equate I1 and I2
I1=I2
sin(c)+2sin(c/2) = sin(c)+2sin(c/2)
Since it is an identity, we conclude that all values of c on the interval [0,π/2] will satisfy the condition that Area1 = Area2.
Note that the last sentence in the response was:
...equate the two areas and solve for c [if necessary]
and it turns out that it is not necessary because it is an identity.
nope, you are wrong. I got the correct answer. area1 and area2 are not equal to each other since their limits are DIFFERENT. for area1, the limits should be (from 0 to c/2) and for area2. the limits should be (from pi/2+c to pi). by solving it i got the value for c. Thank you anyways.
I apologize for the confusion in my previous response. It seems like I made an error in explaining the problem to you. Let me try to clarify the steps for solving this integral.
First, let's address the issue you mentioned about the cancellation of terms. In the initial expressions you provided for both integrals, some of the terms cancel each other out. This is expected and is a result of the properties of trigonometric functions. However, even though some terms cancel, there are still remaining terms that need to be accounted for.
Let's focus on the integral expressions you provided:
1. Integral of (sin(c)cos(x) - cos(c)sin(x) + sin(x) + c) dx
2. Integral of (-sin(c)cos(x) + cos(c)sin(x) - sin(x) + c) dx
To solve these integrals, you need to apply the integration rules and techniques for each term separately:
1. For the first integral, let's break it down:
- Integral of sin(c)cos(x) dx: This can be solved using the product-to-sum formula for the sin and cos functions.
- Integral of -cos(c)sin(x) dx: This can also be solved using the product-to-sum formula.
- Integral of sin(x) dx: This is a simple integral. The integral of sin(x) with respect to x is -cos(x).
- Integral of c dx: This is a constant term, so its integral is simply c times x.
2. For the second integral, you can follow the same steps as above for each term.
After integrating each term separately, you can combine them to get the final integral expression. Keep in mind that when integrating with respect to x, you should have a constant of integration, which is typically represented as "+ C" at the end of the integral.
Once you have the final integral expressions for both integrals, you can compare them and see if there are any terms that can be simplified or canceled out. You may need to simplify further or apply some trigonometric identities to reach a final expression.
If you can provide the original problem or the context in which these integrals arise, I may be able to assist you further in solving for the variable c.