How do you evaluate the definite integral #int x^3 dx# from #[-1,1]#?
2 Answers
We will need two ideas here:
#intx^ndx=x^(n+1)/(n+1)+C# #int_a^bf'(x)dx=f(b)-f(a)#
So when we integrate
This is written as:
#int_(-1)^1x^3dx=[x^4/4]_(-1)^1=(1^4/4)-((-1)^4/4)#
#color(white)(int_(-1)^1x^3dx)=1/4-1/4#
#color(white)(int_(-1)^1x^3dx)=0#
We can also note that
An odd function is one where it is reflected over the
graph{x^5-16x [-4, 4, -30, 30]}
We can see that the area from the
#int_(-a)^af(x)dx=0#
(Since the area from
We can also see this as:
#int_(-a)^0f(x)dx=-int_0^af(x)dx#
#=>int_(-a)^0f(x)dx+int_0^af(x)dx=0#
#=>int_(-a)^af(x)dx=0#
The same is applicable to the integral we have here! For
#int_(-1)^1x^3dx=0#