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How do you test the improper integral #int (x^2+2x-1)dx# from #[0,oo)# and evaluate if possible?

Answer 1

Aurora Alvarado

The integral is divergent. See explanation.

#int_0^{+oo}(x^2+2x-1)dx=[x^3/3+x^2-x]_0^{+oo}=#

#=lim_{x->+oo}(x^3/3+x^2-x)-[0^3/3+0^2-0]=#

#=lim_{x->+oo}x*(x^2/3+x-1)=(+oo)*(+oo)=+oo#

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Answer 2

Paisley Johnson

See process below

Think about geometrical meaning of integral: "is the area under the curve".

We have a parabola #x^2+2x-1# which representation is

# graph{x^2+2x-1=y [-10, 10, -5, 5]} #

Now, try to evaluate

#int_0^(oo)x^2+2x-1dx=lim_(btooo)int_o^bx^2+2x-1dx=#

#=lim_(btooo)(1/3x^3+x^2-x)_0^b=#

#lim_(btooo)b^3/3+b^2-b=oo#

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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.