int_{-1}^1 (a x^3 + b x) d x = 0 for | vedclass

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(A) Let $I = \int_{-1}^1 (a x^3 + b x) d x$.The integrand $f(x) = a x^3 + b x$ is an odd function because $f(-x) = a(-x)^3 + b(-x) = -(a x^3 + b x) =

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any value of $a$ and $b$

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(A) Let $I = \int_{-1}^1 (a x^3 + b x) d x$.The integrand $f(x) = a x^3 + b x$ is an odd function because $f(-x) = a(-x)^3 + b(-x) = -(a x^3 + b x) = -f(x)$.According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^a f(x) d x = 0$.Alternatively,evaluating the integral:$\int_{-1}^1 (a x^3 + b x) d x = \left[ a \frac{x^4}{4} + b \frac{x^2}{2} \right]_{-1}^1$$= \left( \frac{a(1)^4}{4} + \frac{b(1)^2}{2} \right) - \left( \frac{a(-1)^4}{4} + \frac{b(-1)^2}{2} \right)$$= \left( \frac{a}{4} + \frac{b}{2} \right) - \left( \frac{a}{4} + \frac{b}{2} \right) = 0$.This result holds true for any real values of $a$ and $b$.

$\int_{-1}^1 (a x^3 + b x) d x = 0$ for

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