Question

Use symmetry to evaluate the following integral.


`\int\limits^4_(-4) {(x^3+3x)/(x^4+5) } \, dx`

Answer 1

The integrand is odd.

`f(x) = (x^3 + 4x)/(x^4 + 5) \implies f(-x) = ((-x)^3 + 4(-x))/((-x)^4 + 5) = (-x^3-4)/(x^4+5) = -f(x)`

The integral of an odd function over a symmetric interval is zero, so

`\displaystyle \int_(-4)^4 (x^3+3x)/(x^4+5) \, dx = 0`

This is because

`\displaystyle \int_(-a)^a f(x) \, dx = \int_(-a)^0 f(x) \, dx + \int_0^a f(x) \, dx \\\\ ~~~~ = -\int_0^(-a) f(x) \, dx + \int_0^a f(x) \, dx \\\\ ~~~~ = \int_0^(-a) f(-x) \, dx + \int_0^a f(x) \, dx \\\\ ~~~~ = -\int_0^a f(x) \, dx + \int_0^a f(x) \, dx = 0`

where we substitute
`x\mapsto-x` in the second-to-last equality.