+118658 Here is the graph of
\(y=x^3-3x+4\)
You MUST understand this.
Now when you are integrating this function between 2 set values you are finding the area between the x axis the curve and those 2 vertical lines.
So look at this graph of
\(y=x^3-3x+4\)
The integral is the green area under the curve.
The minumum value is 2 and the maximum value is 22
You should be able to see that the area MUST be bigger than 2*(3-0) and smaller than 22*(3-0)
so
\(6<\int_0^3\;\;(x^3-3x+4)\;dx\;\;<66\)
\(6\;<\;\int_0^3\;(x^3-3x+4)\;dx\;<\;66\)
Understand that the integral sign S is exactly that - a stylized S which stands for 'sum of'
You are summing the area of infintesimally narrow rectagles to find the total area under the curve.
