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How do you integrate #int (4x^3+6x^2-1)dx#?

Answer 1

Scarlett Allison

We can use the power rule for integration all the way across that integrand.

#int(4x^3+6x^2-1)dx#

Keep in mind that:

#intx^ndx=x^(n+1)/(n+1)+c#

So, unsimplified:

#=4*(x^(3+1))/(3+1)+6*(x^(2+1))/(2+1)-1(x)+c#

#=x^4+2x^3-x+c#

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Answer 2

Adam Adkins

To integrate ( \int (4x^3 + 6x^2 - 1) , dx ), you can use the power rule of integration.

[ \int (ax^n) , dx = \frac{a}{n+1}x^{n+1} + C ]

Apply this rule to each term of the given function:

[ \int (4x^3 + 6x^2 - 1) , dx = \frac{4}{4}x^{4} + \frac{6}{3}x^{3} - x + C ]

[ = x^4 + 2x^3 - x + 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.