Question

${\int }_{-1}^1{x}^{3}dx$ = ___

Answer 1

We can definitely solve this problem by integrating the $x^3$ But we’ll use one more approach, a very useful approach to solve such problems. We’ll use this property of definite integral -
$\left(i\right){\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$,if f is even,i.e.,$iff(-x)=f\left(x\right)$
$\left(ii\right){\int }_{-a}^{a}f\left(x\right)dx=0,$ if f is odd,i.e.,$iff(-x)=-f\left(x\right)$.
We can see that the limits given in the question are similar to the formula.
Let a be = 1
Also let f(x) be $=x^3$
We know $f\left(x\right)=x^3$ is an odd function . {since , f(-x) = - f(x)}
$So,{\int }_{-1}^1{x}^{3}dx=0$