How do you test the improper integral $\int (x^{2} + 2 x - 1) d x$ $int (x^2+2x-1)dx$ from $[0, \infty)$ $[0,oo)$ and evaluate if possible?

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Answer 1 · 2018-07-06T08:17:40

The integral is divergent. See explanation.

$\int_{0}^{+ \infty} (x^{2} + 2 x - 1) d x = {[\frac{x^{3}}{3} + x^{2} - x]}_{0}^{2} =$ $int_0^{+oo}(x^2+2x-1)dx=[x^3/3+x^2-x]_0^{+oo}=$

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

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

Answer 2 · 2018-07-06T08:18:24

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

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

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

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

How do you test the improper integral int (x^2+2x-1)dx∫(x2+2x−1)dxint (x^2+2x-1)dx from [0,oo)[0,∞)[0,oo) and evaluate if possible?