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Math Help - Antiderivative check

  1. #1
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    Antiderivative check

    integral from 0 to 1 (3+xsqrtx)dx

    Soo i just need to see if im on the right track
    I rewrote it to (3+x.x^-1/2) soo then would the antiderivative be (3x+ (2/5)x^(5/2) )??
    obviously then subbing in the x values after but i just want to make sure i got the antiderivative correct.
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  2. #2
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    3+x\sqrt{x}= 3+x\times x^{\frac{1}{2}} = 3+ x^{\frac{3}{2}}
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  3. #3
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    Quote Originally Posted by pickslides View Post
    3+x\sqrt{x}= 3+x\times x^{\frac{1}{2}} = 3+ x^{\frac{3}{2}}
    So 3x+(2/5)x^(5/2) would be correct since taking the derivative of it is 3+x^(3/2). I relized i made a mistake in writing it and put -(1/2).

    I finishe the substitutions and got:
    [3(1)+(2/5)(1)^(5/2)]-[3(0)+(2/5)(0)^(5/2)]
    which is 17/5 if im not mistaken?
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  4. #4
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    \frac{17}{5} I agree
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