Question

How do you calculate `int_-1^1x^2dx` ?

`int_-1^1x^2dx`

Answer 1

`2/3`

Explanation:

We use the theorem:
If a function f is continous and even on `[-a.a],` then `int_-a^af(x)dx=2int_0^af(x)dx,where,ainR^+`
Here,`f(x)=x^2andf(-x)=(-x)^2=x^2=f(x)=>f` is even function
`I=int_-1^1x^2dx=2int_0^1x^2dx=2[x^3/3]_0^1=2[1/3-0]=2/3`
OR

`I=int_-1^1x^2dx=[x^3/3]_-1^1=[(1)^3/3-(-1)^3/3]=[1/3+1/3]=[1/3-(-1)/3]=1/3+1/3=2/3`