What would be all the antiderivatives of the function f(x) = x^(7/2) + 9. Can't seem to crack it... would appreciate any help.?

What would be all the antiderivatives of the function f(x) = x^(7/2) + 9. Can't seem to crack it... would appreciate any help.

Update:

If you could show your work how you got the answer that would be great.

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  • 2 months ago
    Favorite Answer

    Getting the derivative and anti-derivative of expressions like this is the same as if they were polynomials.  The fact the exponent is a fraction doesn't affect the steps.

    To get the anti-derivative, each term has the exponent increased by 1 then divide the old coefficient by the new exponent to get the new coefficient.  Any constant terms get an "x" added to it, then you add a constant term at the end, so:

    f(x) = x^(7/2) + 9

    ∫ f(x) = (2/9)x^(9/2) + 9x + C

  • Ian H
    Lv 7
    2 months ago

    Just use the power rule

    The integral of x^n is [x^(n + 1)]/(n + 1) + a constant

    The integral of x^(7/2) is [x^(9/2)]/(9/2) + c

    The integral of x^(7/2) + 9 is (2/9)x^(9/2) + 9x + c

  • Vaman
    Lv 7
    2 months ago

    Please use the word integral for anti derivative.

  • 2 months ago

    Antiderivative is F(x) = (2/9)x^(9/2) + 9x + c

    Just follow the usual rules of anti derivatives. Nothing goofy happens in the problem. 

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  • Power rule:  integral of x^n is (1/(n + 1)) * x^(n + 1) + C, as long as n does not equal -1.

    (1/(7/2 + 1)) * x^(7/2 + 1) + (1/(0 + 1)) * 9 * x^(0 + 1) + C =>

    (1/(9/2)) * x^(9/2) + (1/1) * 9 * x^(1) + C =>

    (2/9) * x^(9/2) + 9x + C

  • 2 months ago

    The antiderivatives are (2/9)x^(9/2) + (9/2)x^2 + C, where C is any constant.

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