Question

$$ \text { Without integrating, explain why } \int_{-2}^{2} x\left(x^{2}+1\right)^{2} d x=0 $$

Short answer

The integral of the function \(x(x^2 + 1)^2\) from \(-2\) to \(2\) is \(0\) because the function is an odd function

Step-by-step solution

Analyze Function Type

Determine whether the given function \(x(x^2 + 1)^2\) is an odd function, an even function, or neither. We can determine this by replacing \(x\) with \(-x\) and simplify it. For a function to be odd, \(f(-x) = -f(x)\) and for a function to be even, \(f(-x) = f(x)\). So, let's replace \(x\) with \(-x\) in our function: \[(-x)((-x)^2 + 1)^2 = -x(x^2 + 1)^2 = -f(x)\] Our function is an odd function.

Apply Integral Properties

Now, we can apply the property that the integral of an odd function from \(-a\) to \(a\) is zero i.e, \(\int_{-a}^{a} f(x) dx = 0\) if \(f(x)\) is an odd function. So, for our case, as \(f(x) = x(x^2 + 1)^2\) is an odd function, the integral \(\int_{-2}^{2} x(x^2 + 1)^2 dx = 0\)