Calculus II

John Marsh

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Hello Everyone please help me with this question :confused:

Find` int_(pi/2)^(-pi/2)` x sinx dx

Thanks in advance :p
 
Hello Everyone please help me with this question :confused:

Find` int_(pi/2)^(-pi/2)` x sinx dx

Thanks in advance :p

Hint: use integration by parts.

Please share your work with us.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (e.g. "are these correct?")
 
\(\displaystyle \int(\pi/2)^{-\pi/2} x sinx dx\) - Let's make it more readable :cool:
 
\(\displaystyle \int(\pi/2)^{-\pi/2} x sinx dx\) - Let's make it more readable :cool:

no it is:

\(\displaystyle \displaystyle \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}x \ \ sin(x) \ \ dx\)
 
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Do you not know the "integration by parts" formula?
\(\displaystyle \int u dv= uv- \int v du\)

Yes, if you let u= x, dv= sin(x)dx then du= dx and v= -cos(x).
Put those into the formula.
 
Do you not know the "integration by parts" formula?
\(\displaystyle \int u dv= uv- \int v du\)

Yes, if you let u= x, dv= sin(x)dx then du= dx and v= -cos(x).
Put those into the formula.

So when your looking at the original problem, then your actually looking at \(\displaystyle u(dv)\). What would be a clue (from just looking at the original problem) that "Integration by Parts" is needed?
 
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\(\displaystyle u = x\)

\(\displaystyle dv = \sin(x)\)

\(\displaystyle v = -\cos(x)\)

\(\displaystyle du = 1\)

\(\displaystyle \int x \sin(x) = x -\cos(x) - \int-\cos(x)(1)\)

\(\displaystyle \int x \sin(x) = x -\cos(x) - \sin(x)x\) - :confused: Is this line right?
 
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\(\displaystyle u = x\)

\(\displaystyle dv = \sin(x)\)

\(\displaystyle v = -\cos(x)\)

\(\displaystyle du = 1\)

\(\displaystyle \int x \sin(x) = x -\cos(x) - \int-\cos(x)(1)\)

\(\displaystyle \int x \sin(x) = x -\cos(x) - \sin(x)x\) - :confused: Is this line right?

No...

\(\displaystyle \int x \sin(x) dx = x* [-\cos(x)] - \int-\cos(x)(1)dx \ = \ -x * cos(x) + sin(x) + C\)
 
\(\displaystyle u = x\)

\(\displaystyle dv = \sin(x)\)

\(\displaystyle v = -\cos(x)\)

\(\displaystyle du = 1\)

\(\displaystyle \int x \sin(x) = x -\cos(x) - \int-\cos(x)(1)\)

\(\displaystyle \int x \sin(x) = x -\cos(x) - \sin(x)x\) - :confused: Is this line right?
Do you not know the difference between multiplying and adding?
You want x times -cos(x), not x plus -cos(x).
 
Do you not know the difference between multiplying and adding?
You want x times -cos(x), not x plus -cos(x).

\(\displaystyle \displaystyle \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}x \ \ sin(x) \ \ dx\)

First solve the integral.

Use the "Integration by Parts" formula (because we are dealing with a product in the original problem):

\(\displaystyle \int u dv= uv- \int v du\)

\(\displaystyle u = x\)

\(\displaystyle du = 1\)

\(\displaystyle v = -\cos(x)\)

\(\displaystyle dv = \sin(x) \)

\(\displaystyle \int x \sin(x) = [x (-\cos(x))] - \int [- cos(x)(1)]\)

\(\displaystyle \int x \sin(x) = -x \cos(x) - \int - cos(x)\)

\(\displaystyle \int x \sin(x) = -x \cos(x) - [- \sin(x)] + C\) - Maybe better understood with this missing step.

\(\displaystyle \int x \sin(x) = -x \cos(x) + \sin(x) + C \)

\(\displaystyle \int x \sin(x) = -x \cos(x) + \sin(x) + C \)

Now moving on to evaluating the integral at two different values (definite integral).
 
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Now finishing the whole thing:

From previous post:
\(\displaystyle \displaystyle \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}x \ \ sin(x) \ \ dx\)

First solve the integral.

Use the "Integration by Parts" formula (because we are dealing with a product in the original problem):

\(\displaystyle \int u dv= uv- \int v du\)

\(\displaystyle u = x\)

\(\displaystyle du = 1\)

\(\displaystyle v = -\cos(x)\)

\(\displaystyle dv = \sin(x) \)

\(\displaystyle \int x \sin(x) = [x (-\cos(x))] - \int [- cos(x)(1)]\)

\(\displaystyle \int x \sin(x) = -x \cos(x) - \int - cos(x)\)

\(\displaystyle \int x \sin(x) = -x \cos(x) - [- \sin(x)] + C\) - Maybe better understood with this missing step.

\(\displaystyle \int x \sin(x) = -x \cos(x) + \sin(x) + C \)

\(\displaystyle \int x \sin(x) = -x \cos(x) + \sin(x) + C \)

Now moving on to evaluating the integral at two different values (definite integral).

Evaluating the integral by plugging in lower and upper bound values.

\(\displaystyle [-(-\dfrac{\pi}{2}) \cos(-\dfrac{\pi}{2}) + \sin(-\dfrac{\pi}{2})] - [ -(\dfrac{\pi}{2}) \cos(\dfrac{\pi}{2}) + \sin(\dfrac{\pi}{2})]\) Upper minus lower bound.

\(\displaystyle (\dfrac{\pi}{2}) \cos(-\dfrac{\pi}{2}) + \sin(-\dfrac{\pi}{2}) + (\dfrac{\pi}{2}) \cos(\dfrac{\pi}{2}) + \sin(\dfrac{\pi}{2})\) - fixing negative and positive signs

\(\displaystyle (\dfrac{\pi}{2}) (0) + (-1) + (\dfrac{\pi}{2}) (0) + (1)\) - :?: - Anything wrong with this line? - evaluating the plugged in values further.

\(\displaystyle -1 + 1 = 0\) The area under the curve is \(\displaystyle 0\) - Answer (I think) from arithmetic.
 
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