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ksaimouli

  • 2 years ago

f(x) is even and g(x) is odd

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  1. ksaimouli
    • 2 years ago
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    \[\int\limits_{0}^{5}f(x)dx=8\]

  2. ksaimouli
    • 2 years ago
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    \[\int\limits_{0}^{5}g(x)=4\]

  3. ksaimouli
    • 2 years ago
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    find\[\int\limits_{-5}^{5}[f(x)+g(x)]dx\]

  4. ksaimouli
    • 2 years ago
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    @wio

  5. ksaimouli
    • 2 years ago
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    \[\int\limits_{-5}^{0}[f(x)+g(x)] +\int\limits_{0}^{5}[f(x)+g(x)] dx\]

  6. ksaimouli
    • 2 years ago
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    @abb0t

  7. ksaimouli
    • 2 years ago
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    after that i have no idea

  8. ksaimouli
    • 2 years ago
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    i know the 2nd part 12

  9. ksaimouli
    • 2 years ago
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    -+12

  10. ksaimouli
    • 2 years ago
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    well even mean f(-x)=f(x) will that mean\[\int\limits_{-5}^{0}f(x)=-8\]

  11. ksaimouli
    • 2 years ago
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    because\[-\int\limits_{5}^{0}f(x)=-8 \]

  12. ksaimouli
    • 2 years ago
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    even aplies if negative in f(x)? f(-x)=f(x)

  13. KingGeorge
    • 2 years ago
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    Why should \[-\int\limits_{5}^{0}f(x)=-8\]be true? I think this should be 8 and not -8.

  14. ksaimouli
    • 2 years ago
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    because that is the rule right to flip flop - should be their

  15. KingGeorge
    • 2 years ago
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    \[\int_0^5 f(x)=8\]\[\int_5^0 f(x)=-8\]\[-\int_5^0 f(x)=8\]

  16. ksaimouli
    • 2 years ago
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    \[-\int\limits_{-5}^{0}f(x)=\]

  17. KingGeorge
    • 2 years ago
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    But you're certainly on the right path to solving this. You know\[\int\limits_{0}^{5}[f(x)+g(x)] dx=12\]So you just need to find \[\int\limits_{-5}^{0}[f(x)+g(x)]dx =\int\limits_{-5}^0f(x)\;dx+\int\limits_{-5}^0g(x)\;dx\]First, lets start with the f(x) part.

  18. zepp
    • 2 years ago
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    I drove by to say hello to @KingGeorge!

  19. KingGeorge
    • 2 years ago
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    Since \(f(x)=f(-x)\), we have that \[\begin{aligned} \int\limits_{-5}^0f(x)\;dx&=\int\limits_{-5}^0f(x)\;dx \\ &=\int\limits_{-5}^0f(-x)\;dx\\ &=\int\limits_{5}^0f(-u)\;(-du)\qquad\text{this is a u-sub for }u=-x.\\ &=\int\limits_0^5f(-u)\;du\\ &=\int\limits_0^5f(u)\;du\\ &=8 \end{aligned}\]

  20. KingGeorge
    • 2 years ago
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    This is basically the same thing you do for \(g(x)\). Instead we have \(-g(x)=g(-x)\).If we repeat the above argument, we get to the point \[\int\limits_0^5g(-u)\;du\]Substitute for \(-g(u)\), and we get \[-\int\limits_0^5g(u)\;du=-4\]

  21. KingGeorge
    • 2 years ago
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    Did this all make sense? And hello to you too @zepp!

  22. ksaimouli
    • 2 years ago
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    i i got it i did little bit different i took u=-x and then -du=dx

  23. KingGeorge
    • 2 years ago
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    There is a typo in my work two posts above. The second line should not be there. (The line that reads \(=\int_{-5}^0f(-x)\;dx\)).

  24. KingGeorge
    • 2 years ago
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    That works as well.

  25. ksaimouli
    • 2 years ago
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    |dw:1356649511965:dw|

  26. ksaimouli
    • 2 years ago
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    ohk thx a lot

  27. KingGeorge
    • 2 years ago
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    You're welcome.

  28. ksaimouli
    • 2 years ago
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    i will work on this if i have nay questions can i post on this wall

  29. KingGeorge
    • 2 years ago
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    sure.

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