The value of int_{-2012}^{2012} (sin(x^3) + x | vedclass

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Question

(D) Let $I = \int_{-2012}^{2012} (\sin(x^3) + x^5 + 1) dx$. Using the property of linearity of integrals,we can write: $I = \int_{-2012}^{2012} \sin(x

Vedclass · Accepted Answer

$4024$

Vedclass · Answer

(D) Let $I = \int_{-2012}^{2012} (\sin(x^3) + x^5 + 1) dx$. Using the property of linearity of integrals,we can write: $I = \int_{-2012}^{2012} \sin(x^3) dx + \int_{-2012}^{2012} x^5 dx + \int_{-2012}^{2012} 1 dx$. Since $f(x) = \sin(x^3)$ and $g(x) = x^5$ are odd functions (i.e.,$f(-x) = -f(x)$),the integral of these functions over a symmetric interval $[-a, a]$ is $0$. Thus,$\int_{-2012}^{2012} \sin(x^3) dx = 0$ and $\int_{-2012}^{2012} x^5 dx = 0$. Therefore,$I = 0 + 0 + \int_{-2012}^{2012} 1 dx$. $I = [x]_{-2012}^{2012} = 2012 - (-2012) = 2012 + 2012 = 4024$.

The value of $\int_{-2012}^{2012} (\sin(x^3) + x^5 + 1) dx$ is

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