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

Answer 1

Paisley Johnson

#"A"(x)=1/3x^3-x^2+4x+"c"#

#intx^2-2x+4# #dx#

Use the reverse power rule on each term independently to calculate the integral.

Rule of reverse power:

#intx^n# #dx=1/(n+1)x^(n+1)+ "constant"#

#intx^2-2x+4# #dx=1/3x^3-2/2x^2+4/1x+"c"# #=1/3x^3-x^2+4x+"c"#

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Answer 2

Paisley Johnson

To evaluate the indefinite integral ∫(x^2 - 2x + 4)dx, you need to integrate each term separately.

∫x^2 dx = (1/3)x^3 + C, where C is the constant of integration. ∫(-2x) dx = -x^2 + C. ∫4 dx = 4x + C.

Putting these together, the integral of x^2 - 2x + 4 dx is:

(1/3)x^3 - x^2 + 4x + 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.