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How do you integrate #int (2x-3x^2)dx#?

Answer 1

# :. int(2x-3x^2)dx = x^2 - x^3 + C#

The power rule for integration, which is the opposite of the rule for differentiation, is something you should understand.

# int x^ndx = x^(n+1)/(n+1) AA x in RR, x !=-1 #

So # int(2x-3x^2)dx = 2x^(1+1)/(1+1)-3x^(2+1)/(2+1) + C# # :. int(2x-3x^2)dx = 2x^2/2-3x^3/3 + C# # :. int(2x-3x^2)dx = x^2 - x^3 + C#

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Answer 2

Luke Alvarez

To integrate ( \int (2x - 3x^2) , dx ), follow these steps:

Separate the terms in the integral: [ \int 2x , dx - \int 3x^2 , dx ]
Integrate each term separately: [ \int 2x , dx = x^2 + C_1 ] [ \int 3x^2 , dx = x^3 + C_2 ]
Combine the results: [ \int (2x - 3x^2) , dx = x^2 - x^3 + 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.