Integral using Riemann sums?

See the expression below.

There are various approaches to Riemann sums. However, at the limit, they would end up giving you the same value.

Imagine the curve drawn on graph paper. It will go from (x, y) = (1, 5) to (3, 13)

[in fact, just do it - the drawing will help you understand]

To do this graphically, you begin with a vertical line at x=1 (from the x axis to the curve) and another vertical line at x=3

The value of the integral should be equal to the surface area enclosed by the figure (x-axis, vertical line at 1, vertical line at 3 and the graph of the function)

Draw a vertical line at x=2 (dividing the surface in two parts).

From (1, 3), draw a horizontal line until x=2. This gives you a rectangle measuring 1*3 = 3

This area is obviously less than the real area of that half.

Draw a horizontal line from (2, 8) until x=3

This gives you a rectangle (1*8 = 8) also obviously smaller than the real area of that half.

Adding the two rectangles will give an area (11) that is obviously less than the real value of the integral.

Go the other way:  from (3, 13), draw a horizontal line until x=2. This rectangle (area = 13) is obviously more than the real area of that half.

From (2, 8), horizontal line to x=1. Area of that rectangle is 8, and is obviously more than that half.

Adding the two rectangles will give an area (21) that is obviously more than the real value of the integral.

The real value of the integral will be somewhere between 11 and 21.

In this first example, the width of each rectangle is 1. However, we can get more accuracy is we make more rectangles.  The area of each tiny rectangle will be

f(x) times h (where h is the width of each rectangle).

Since the length of the interval (on the x axis) is 2 in this case, we could make 20 intervals of 0.1 units each.

If x represents the beginning x value for each rectangle, then we would have the sum of 20 rectangles as f(x) times 0.1.

This sum would still be less than the real value of the integral... but it would be a much closer estimate.

If (x+h) represents the end x-value for each rectangle, then we would have the sum of 20 rectangles as f(x+h) times h.

This sum would still be more than the real value of the integral... but it would be a much closer estimate.

The real value of the integral would be between these two values.

At the limit (as h gets closer and closer to zero... giving you more and more rectangles), both sums would converge to the real value of the integral.

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You will have 2h rectangles, each one having an area of

f(x) times h

where x is the start of each interval.