-e-x + C.
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.
Dividend: x3+4x2-9x-36 Divisor: x+3 Quotient: x2+x-12
The antiderivative of -3 with respect to x is -3x+C. C being any real number.
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.
(2/3)*x^(3/2)
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sqrt(1) + 3*sqrt(x) = 1 + 3*x^1/2So the antiderivative is x + [3*x^(3/2)]/(3/2) + c = x + 2*x^(3/2) + c where c is the constant of integration.
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1/x2
It is x^2 + xy + y^2
X2 (X squared)
Antiderivative of x/-1 = -1(x^2)/2 + C = (-1/2)(x^2) + C Wolfram says antiderivative of x^-1 is log(x) + C
Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.
x^3 - x^3 = 0 Remember , whilst 'x' is an unknown value, that unkonwn is a fixed value. As a numerical example 3^(3) - 3^(3) = 27 - 27 = 0 The '3' is 'x' in this case
By antiderivative do you mean integral? If yes, integral x^1 dx= (x^2)/2
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