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How do you evaluate the indefinite integral #int (2x^2-4x+3)dx#?

Answer 1

#int(2x^2-4x+3)dx=2/3x^3 -2x^2+3x+C, C in RR#

#int(2x^2-4x+3)dx=int2x^2dx-int4xdx+int3dx# #=2intx^2dx-4intxdx+3intdx# Also, #int(x^n)dx=(x^(n+1))/(n+1)# So :#int(2x^2-4x+3)dx=2*x^3/3-4*x^2/2+3x+C=2/3x^3-2x^2+3x+C, C in RR# \0/ here's our answer !

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Answer 2

Cameron Alexander

To evaluate the indefinite integral of ( \int (2x^2 - 4x + 3) , dx ), we apply the power rule for integration. This rule states that for any term ( ax^n ), the indefinite integral is ( \frac{a}{n+1}x^{n+1} + C ), where ( C ) is the constant of integration.

So, integrating each term separately: [ \int (2x^2 - 4x + 3) , dx = \int 2x^2 , dx - \int 4x , dx + \int 3 , dx ]

Using the power rule: [ = \frac{2}{3}x^3 - 2x^2 + 3x + 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.