The value of int_{-1}^1 left(sqrt{1+x+x^2} - | vedclass

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(B) Let $I = \int_{-1}^1 \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $a+b =

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$0$

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(B) Let $I = \int_{-1}^1 \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $a+b = -1+1 = 0$. So,$I = \int_{-1}^1 \left(\sqrt{1+(-x)+(-x)^2} - \sqrt{1-(-x)+(-x)^2}\right) dx$. $I = \int_{-1}^1 \left(\sqrt{1-x+x^2} - \sqrt{1+x+x^2}\right) dx$. $I = - \int_{-1}^1 \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx$. $I = -I$. $2I = 0$,which implies $I = 0$.

The value of $\int_{-1}^1 \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx$ is

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