0
f'(x) = x^(3) / 3
Hence
f(x) = 3x^(2) / 3 +C
f(x) = x^(2) + C ' The '3' cancels out. The 'C' is a constant that comes with antoderivation.
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What is the antiderivative of e to the -x?
-e-x + C.
What is the answer to x cubed plus 4x squared minus 9x minus 36 divided by x plus 3?
Dividend: x3+4x2-9x-36 Divisor: x+3 Quotient: x2+x-12
Can anyone of can prove that antiderivative is wrong?
The way to disprove an antiderivative is to simply differentiate the function and see if it matches the integral expression. Remember that an antiderivative expression must include a term often coined "C-" an arbitrary constant. For example, ∫(x^3 +14x)dx= (1/4)X^4+ 7X^2 +C. To verify that this is correct, take the derivative. You get x^3 +14x.
What is the Anti-derivative of -3?
The antiderivative of -3 with respect to x is -3x+C. C being any real number.
Can a function have more than one antiderivative?
yes, look at the function f(x)=3x^2 The antiderivative is x^3+C where C is the constant and is more than one value for C. In fact, 3x^2 will have an infinite number of antiderivatives.
