How do you find the antiderivative of #f(x)=-1/2x^3+2x^2-3x-2#?

Answer 1

#-1/8x^4+2/3x^3-3/2x^2-2x+C#

The antiderivative of the function is basically the integral of the function. So here we have: #int-1/2x^3+2x^2-3x-2 \ dx#.

We employ the constant rule and the power rule, which stipulate that,

#inta^n \ dx=(a^(n+1))/(n+1)+C#
#inta*f(x) \ dx=aintf(x) \ dx#,

correspondingly.

#=-1/8x^4+2/3x^3-3/2x^2-2x+C#
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Answer 2

#-1/8x^4+2/3x^3-3/2x^2-2x+c#

#"integrate each term using the "color(blue)"power rule"#
#•color(white)(x)int(ax^n)=a/(n+1)x^(n+1)ton!=-1#
#int(-1/2x^3+2x^2-3x-2)dx#
#=-1/8x^4+2/3x^3-3/2x^2-2x+c#
#"where c is the constant of integration"#
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Answer 3

To find the antiderivative of the function ( f(x) = -\frac{1}{2}x^3 + 2x^2 - 3x - 2 ), integrate each term separately using the power rule for integration:

[ \int -\frac{1}{2}x^3 , dx = -\frac{1}{8}x^4 + C ] [ \int 2x^2 , dx = \frac{2}{3}x^3 + C ] [ \int -3x , dx = -\frac{3}{2}x^2 + C ] [ \int -2 , dx = -2x + C ]

Putting them together, the antiderivative of ( f(x) ) is:

[ F(x) = -\frac{1}{8}x^4 + \frac{2}{3}x^3 - \frac{3}{2}x^2 - 2x + C ]

Where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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