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anonymous

  • 5 years ago

help with calclulus

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    can you explain to me what was done? I undertsnad what they did with the infinity but I dont get how they added a 2 to the integrand

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  3. campbell_st
    • 5 years ago
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    if you look at the intergrand they have put 1/2 out the front and the numerator is 2y...... remember 1/2 x 2y = y.....

  4. anonymous
    • 5 years ago
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    LOL but why didthey do that?

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  5. anonymous
    • 5 years ago
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    sorry abt the attachment but there seems to be a bug

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  6. campbell_st
    • 5 years ago
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    its to make the numerator look like the derivative of the denominator.... which would suggest the solution would be a ln function

  7. campbell_st
    • 5 years ago
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    the solution might be... let u = y^2 du = 2y \[1/2\int\limits_{0}^{b} du/(u^2 + 1)\] which is tan^(-1)

  8. campbell_st
    • 5 years ago
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    oops should read 1/2 tan^(-1) (u) then re substitute 1/2 tan^(-1) (y^2)

  9. anonymous
    • 5 years ago
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    Thanks that was clear campbell :D That is exactly what I wanted to know

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