Help

g(x)=5x²+3x⁴ on the interval [-1,3]

The average rate of change is given by :  change in y on the interval /  change in x on the interval =

[g(3)  - g(-1)] / [3 - (-1)]  =

[288 - 8 ] / [ 4 ] =

280 / 4  =

70

Find the average rate of change of g(x)=5x2+3x4 on the interval [-1,3]

CPhill's answer is the best one but I was playing around with this

---------------------------------------------------------------------------------------------------------

The instantaneaous rate of change at any given point is given by the first derivative. It is the gradient of the tangent.

\(g'(x)=10x+12x^3\)

Now, The integral sign  is a sylized S and it really stands for 'sum of'

So the sum of all the gradients from -1 to 3 is given by

\(\displaystyle \int_{-1}^3\;10x+12x^3\;dx\\\\\\ = [5x^2+3x^4]_{-1}^3\\\\\\ =(45+243)-(5+3)\\\\ =280\)

Now the average of this would have to be  

\(\frac{280}{3-\;-1} = 70\)

So the average rate of change is 70

I invite other mathematicians to comment on this.  There is something I don't quite understand but I cannot explain what it is.   

It is this statement that is causing me problems

So the sum of all the gradients from -1 to 3 is given by the integral .....