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wafaa

  • 3 years ago

evaluate s x( (8+2x-x^2) )^1/2 dx p/s: s is for integrate

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  1. Mimi_x3
    • 3 years ago
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    I assume its an integral \[\int\limits x\sqrt{8+2x-x^{2}}dx \] Well, complete the square first; then looks like a sub

  2. wafaa
    • 3 years ago
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    yes. I've done the completing the square but not sure whats the next step. by part or substitution?

  3. Mimi_x3
    • 3 years ago
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    looks like a substitution would be easier

  4. wafaa
    • 3 years ago
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    ok. i"ll try again. tq :)

  5. wafaa
    • 3 years ago
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    still dont get it. help me plez :(

  6. Mimi_x3
    • 3 years ago
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    \[\int\limits x\sqrt{-x^{2}+2x+8} dx => \int\limits x\sqrt{-(x^{2}-2x-8)} dx => \int\limits x\sqrt{-((x^{2}-1)^{2}-(1)^{2}+8))} dx\] \[=> \int\limits x\sqrt{-(x^{2}-1)-9} dx => \int\limits x\sqrt{9-(x-1)^{2}} dx\] let u = x-1 -> x = u+1 \[=> \int\limits\left(u+1\right)\sqrt{9-u^{2}}du \]

  7. Mimi_x3
    • 3 years ago
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    then it's trig sub

  8. Mimi_x3
    • 3 years ago
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    \[ x= sin\theta\]

  9. wafaa
    • 3 years ago
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    ok mimi.i'll try. thanks again.

  10. Mimi_x3
    • 3 years ago
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    np and for this u = 3sin\theta

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