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# 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.

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

### 6 Answers

- llafferLv 72 months agoFavorite 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 HLv 72 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

- jeffdanielkLv 42 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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- 2 months ago
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

- az_lenderLv 72 months ago
The antiderivatives are (2/9)x^(9/2) + (9/2)x^2 + C, where C is any constant.