Let #P# be a point on the parabola. The coordinates with respect to #x# will be #(x,x^2)#.
We can make a function out of the distance to the origin using the distance formula:
#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#
Applying this to our point #P=(x,x^2)# and the origin #(0,0)#, we get:
#d(x)=sqrt((x-0)^2+(x^2-0)^2)#
#d(x)=sqrt(x^2+x^4)#
To work out the average value of the function, we can use the average value formula. For a function #f(x)# on the interval #[a,b]#, the average value will be:
#(int_a^bf(x)\ dx)/(b-a)#
In our case, we get:
#(int_0^4d(x)\ dx)/(4-0)=1/4int_0^4sqrt(x^2+x^4)\ dx#
I will first work out the antiderivative:
#int\ sqrt(x^2+x^4)\ dx=#
#=int\ sqrt(x^2(1+x^2)\ dx=#
#=int\ xsqrt(x^2+1)\ dx=#
Now we can introduce a u-substitution with #u=x^2+1#. The derivative of #u# is then #2x#, so we divide through by that:
#=int\ (cancel(x)sqrtu)/(2cancel(x))\ du=#
#1/2int\ sqrtu\ du=1/2*2/3u^(3/2)+C=2/6(1+x^2)^(3/2)+C#
Now we can evaluate the original expression:
#1/4int_0^4sqrt(x^4+x^2)\ dx=1/4[1/3(1+x^2)^(3/2)]_0^4=#
#=1/4(1/3(1+16)^(3/2)-1/3(1+0)^(3/2))=#
#=1/4(1/3*17^(3/2)-1/3*1^(3/2))=1/4((17sqrt(17))/3-1/3)=#
#=1/4*(17sqrt17-1)/3=(17sqrt17-1)/12#
So, the average distance from a point to the origin on the parabola on the interval #[0,4]# is:
#(17sqrt17-1)/12~~5.84#
