Question

$\underset{-1}{\overset{1}{\int }}\left[x^2+x-\left(x\right)\right]dx$

Answer 1

$\begin{array}{c}\underset{-1}{\overset{1}{\int }}\left[x^2+x-\left(x\right)\right]dx\\ =\underset{-1}{\overset{0}{\int }}\left[x^2+x-\left(-1\right)\right]dx+\underset{0}{\overset{1}{\int }}\left[x^2+x-0\right]dx\\ =\underset{-1}{\overset{0}{\int }}\left[x^2+x+1\right]dx+\underset{0}{\overset{1}{\int }}\left[x^2+x\right]dx\\ =\underset{-1}{\overset{0}{\int }}\left[x^2+x\right]dx+\underset{-1}{\overset{0}{\int }}1.dx+\underset{0}{\overset{\alpha }{\int }}0dx+\underset{{}^{\alpha }}{\overset{2}{\int }}1dx\\ =\underset{-1}{\overset{0}{\int }}-1dx+1\left(1\right)+0+1\left[2-\alpha \right]\\ =-1\left(1\right)+\gamma +2-\alpha =2-\alpha \\ =2-\left(\frac{\sqrt{3}-1}{2}\right)\\ =\frac{4-\sqrt{5}+1}{2}=\frac{5-\sqrt{5}}{2}ans.\end{array}$