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Evaluate. Then interpret the result in terms of the area above and/or below the X-axis. 1 / 1 f (x3 - 2x) dx 2 2 17 5 (43 - 2x) dx = (Type an integer or a simplified fraction.)

Evaluate. Then interpret the result in terms of the area above and/or below the X-example-1
User Astley
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1 Answer

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The integral can be decomposed and written as


\int ^{(1)/(2)}_{-(1)/(2)}(x^3-2x)dx=\int ^{(1)/(2)}_{-(1)/(2)}x^3dx-2\int ^{(1)/(2)}_{-(1)/(2)}xdx

Now,


\int ^{(1)/(2)}_{-(1)/(2)}x^3dx=(x^4)/(4)|^{(1)/(2)}_{-(1)/(2)}=(((1)/(2))^4)/(4)-((-(1)/(2))^4)/(4)
=0

And


\int ^{(1)/(2)}_{-(1)/(2)}xdx=(x^2)/(2)|^{(1)/(2)}_{-(1)/(2)}=(((1)/(2))^2)/(2)-((-(1)/(2))^2)/(2)=0

Hence,


\begin{gathered} \int ^{(1)/(2)}_{-(1)/(2)}(x^3-2x)dx=\int ^{(1)/(2)}_{-(1)/(2)}x^3dx-2\int ^{(1)/(2)}_{-(1)/(2)}x=0-2(0) \\ \end{gathered}
\boxed{\therefore\int ^{(1)/(2)}_{-(1)/(2)}(x^3-2x)dx=0}

User Yae
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