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

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(A) Let $f(x) = \sqrt{1 + x + x^2} - \sqrt{1 - x + x^2}$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \sqrt{1 + (-x) + (-x)^

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(A) Let $f(x) = \sqrt{1 + x + x^2} - \sqrt{1 - x + x^2}$.We check if the function is even or odd by evaluating $f(-x)$:$f(-x) = \sqrt{1 + (-x) + (-x)^2} - \sqrt{1 - (-x) + (-x)^2}$$f(-x) = \sqrt{1 - x + x^2} - \sqrt{1 + x + x^2}$$f(-x) = -(\sqrt{1 + x + x^2} - \sqrt{1 - x + x^2}) = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.For an odd function,the property of definite integrals states that $\int_{-a}^{a} f(x) \, dx = 0$.Therefore,$\int_{-1}^{1} (\sqrt{1 + x + x^2} - \sqrt{1 - x + x^2}) \, dx = 0$.

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

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