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Math Help - simple definite integral

  1. #1
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    simple definite integral

    \int_2^{6} \frac {6(x^2+1)}{ \sqrt{x}} dx

    I'm having a bit of trouble. I know I can draw out the 6 and bring the denominator up, but how do I follow that up?
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  2. #2
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    \frac{{6\left( {x^2  + 1} \right)}}<br />
{{\sqrt x }} = 6\left( {x^{\frac{3}<br />
{2}}  + x^{ - \frac{1}<br />
{2}} } \right)
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  3. #3
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    Make substitution

    x=z^2

    dx=2zdz
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  4. #4
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    I think you might be confusing the issue a little by substituting z for x, as x is the original variable in the function. Plato's way makes it easier, as you can pull the 6 out in front of the integral, then evaluate using the exponent rule for integrals.
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  5. #5
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    Quote Originally Posted by Plato View Post
    \frac{{6\left( {x^2  + 1} \right)}}<br />
{{\sqrt x }} = 6\left( {x^{\frac{3}<br />
{2}}  + x^{ - \frac{1}<br />
{2}} } \right)
    Not sure how you got this. The x^-(1/2) is from the denominator, yeah, but what's that first term? What happened to the x^2+1?
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  6. #6
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    <br />
\frac{{6\left( {x^2 + 1} \right)}}{{\sqrt x }} = <br />
6(x^2 + 1) \frac{1}{x^{\frac{1}{2}}} = <br />
6\left( {(x^{4 \over 2} + 1) x^{ {-}{1\over 2}} } \right) = <br />
6\left( {x^{\frac{3}{2}} + x^{ - \frac{1}{2}} } \right)

    Makes sense?
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  7. #7
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    Quote Originally Posted by Archduke01 View Post
    Not sure how you got this. The x^-(1/2) is from the denominator, yeah, but what's that first term? What happened to the x^2+1?
    Very basic “calculus confusion” is almost always the result of someone who is ‘algebra-deficient’.
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  8. #8
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Archduke01 View Post
    Not sure how you got this. The x^-(1/2) is from the denominator, yeah, but what's that first term? What happened to the x^2+1?
    When you divide a variable with an exponent by the same variable with a different exponent you subtract the exponent of the denomintor from the numerator.

    So

     x^2/x = x^(2-1) = x

    These math tags are killing me...how do i properly show that an exponent is ^(2-1), it doesnt display like this :P
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  9. #9
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    Quote Originally Posted by AllanCuz View Post
     x^2/x = x^(2-1) = x
    These math tags are killing me...how do i properly show that an exponent is ^(2-1), it doesnt display like this :P
    [tex]x^{2-1}[/tex] gives  x^{2-1} .
    Put the exponent in {}
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  10. #10
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    Quote Originally Posted by losm1 View Post
    <br />
\frac{{6\left( {x^2 + 1} \right)}}{{\sqrt x }} = <br />
6(x^2 + 1) \frac{1}{x^{\frac{1}{2}}} = <br />
6\left( {(x^{4 \over 2} + 1) x^{ {-}{1\over 2}} } \right) = <br />
6\left( {x^{\frac{3}{2}} + x^{ - \frac{1}{2}} } \right)

    Makes sense?
    EDIT: Nevermind I see my mistake.
    Last edited by Archduke01; April 10th 2010 at 04:13 PM.
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