The value of the integral ∫_-3^3 (x^3 - x) dx is :

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

---

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

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

---

Replace $x$x with $-x$−x :

$$f(-x) = (-x)^3 - (-x)$$f(−x)=(−x)3−(−x)

$$= -x^3 + x$$=−x3+x

$$= -(x^3 - x)$$=−(x3−x)

$$= -f(x)$$=−f(x)

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

---

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

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

---

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