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

Jul 6, 2018

The integral is divergent. See explanation.

#### Explanation:

${\int}_{0}^{+ \infty} \left({x}^{2} + 2 x - 1\right) \mathrm{dx} = {\left[{x}^{3} / 3 + {x}^{2} - x\right]}_{0}^{+ \infty} =$

$= {\lim}_{x \to + \infty} \left({x}^{3} / 3 + {x}^{2} - x\right) - \left[{0}^{3} / 3 + {0}^{2} - 0\right] =$

$= {\lim}_{x \to + \infty} x \cdot \left({x}^{2} / 3 + x - 1\right) = \left(+ \infty\right) \cdot \left(+ \infty\right) = + \infty$

Jul 6, 2018

See process below

#### Explanation:

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

We have a parabola ${x}^{2} + 2 x - 1$ which representation is

$g r a p h \left\{{x}^{2} + 2 x - 1 = y \left[- 10 , 10 , - 5 , 5\right]\right\}$

Now, try to evaluate

${\int}_{0}^{\infty} {x}^{2} + 2 x - 1 \mathrm{dx} = {\lim}_{b \to \infty} {\int}_{o}^{b} {x}^{2} + 2 x - 1 \mathrm{dx} =$

$= {\lim}_{b \to \infty} {\left(\frac{1}{3} {x}^{3} + {x}^{2} - x\right)}_{0}^{b} =$

${\lim}_{b \to \infty} {b}^{3} / 3 + {b}^{2} - b = \infty$