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The value of the integral ∫−33(x3−x)dx\int_{-3}^{3} (x^3 - x) dx is :

Solution

✅ Correct Option: 1

We need to find the value of ∫−33(x3−x) dx\int_{-3}^{3} (x^3 - x)\, dx


Before integrating, let's check if f(x)=x3−xf(x) = x^3 - x is an odd function.

A function is odd if f(−x)=−f(x)f(-x) = -f(x). If it is, then ∫−aaf(x) dx=0\int_{-a}^{a} f(x)\, dx = 0 because the areas on either side of the y-axis cancel out.


Replace xx with −x-x :

f(−x)=(−x)3−(−x)f(-x) = (-x)^3 - (-x)

=−x3+x= -x^3 + x

=−(x3−x)= -(x^3 - x)

=−f(x)= -f(x)

Since f(−x)=−f(x)f(-x) = -f(x), the function is odd.


Since f(x)=x3−xf(x) = x^3 - x is an odd function and we are integrating over the symmetric interval [−3,3][-3, 3] :

∫−33(x3−x) dx=0\int_{-3}^{3} (x^3 - x)\, dx = 0


The value of the integral is 0\boxed{0}

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