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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.