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anonymous

  • 4 years ago

can anyone help me with integration

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  1. JamesJ
    • 4 years ago
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    Quite a few people on to night are more than competent to help with your likely question. Post it.

  2. anonymous
    • 4 years ago
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    |dw:1327030921905:dw|

  3. anonymous
    • 4 years ago
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    Separate it into two integrals. Divide the sqrt(2) into the top and then you have 2 trivial integrals.

  4. anonymous
    • 4 years ago
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    can you show how you would do it

  5. anonymous
    • 4 years ago
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    |dw:1327031115564:dw|

  6. anonymous
    • 4 years ago
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    |dw:1327031166888:dw|

  7. anonymous
    • 4 years ago
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    Make sense?

  8. anonymous
    • 4 years ago
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    no

  9. anonymous
    • 4 years ago
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    isnt like this

  10. anonymous
    • 4 years ago
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    |dw:1327031304975:dw|

  11. anonymous
    • 4 years ago
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    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:

  12. anonymous
    • 4 years ago
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    |dw:1327031471323:dw|

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

  14. anonymous
    • 4 years ago
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    thats worng because my book says the answer is 2/3

  15. anonymous
    • 4 years ago
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    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

  16. anonymous
    • 4 years ago
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    no i dont think so because iam using the fundemental thereom of calculus

  17. anonymous
    • 4 years ago
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    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.

  18. anonymous
    • 4 years ago
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    thats exactly how it reads and it asks to evaluate the definite integral

  19. anonymous
    • 4 years ago
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    Mhm, the answer is what I said.

  20. anonymous
    • 4 years ago
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    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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