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- anonymous

can anyone help me with integration

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- anonymous

can anyone help me with integration

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- JamesJ

Quite a few people on to night are more than competent to help with your likely question. Post it.

- anonymous

|dw:1327030921905:dw|

- anonymous

Separate it into two integrals. Divide the sqrt(2) into the top and then you have 2 trivial integrals.

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- anonymous

can you show how you would do it

- anonymous

|dw:1327031115564:dw|

- anonymous

|dw:1327031166888:dw|

- anonymous

Make sense?

- anonymous

no

- anonymous

isnt like this

- anonymous

|dw:1327031304975:dw|

- anonymous

I'm not sure what you did there. How did you get a square root 2? That would just make it harder. Remember that since constants don't change with the variables, you can always pull them out of the integral. You can even do the following:

- anonymous

|dw:1327031471323:dw|

- anonymous

\[\frac{1}{\sqrt{2}}\int_1^4 (u-2)du = \frac{1}{\sqrt{2}}\left(\frac12u^2-2u\right|_1^4=\frac{3}{2\sqrt{2}}\]

- anonymous

thats worng because my book says the answer is 2/3

- anonymous

Your book is incorrect. This answer is correct, with wolfram as additional evidence to boot:
http://www.wolframalpha.com/input/?i=integrate+%28x-2%29%2Fsqrt%282%29+from+1+to+4

- anonymous

no i dont think so because iam using the fundemental thereom of calculus

- anonymous

Does your question specifically ask to solve \[\int_1^4\frac{u-2}{\sqrt{2}}du\] Because the answer to that is what I stated. If the problem in the text is different, let me know and we'll go from there.

- anonymous

thats exactly how it reads and it asks to evaluate the definite integral

- anonymous

Mhm, the answer is what I said.

- anonymous

Think of it this way, since u-2 is a linear equation, you are just calculating the areas of a couple of triangles with rational side lengths. However, dividing by the square root of 2 (an irrational number) will necessitate that your integral is irrational. 2/3 is not irrational, so that cannot be the answer.

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