Question

compute ( l_3 ) and ( r_3 ) for graphs a and b.
the value of ( l_3 ) for graph a is (square).
(simplify your answer.)

Explanation:

Step1: Determine the width of each sub - interval

The interval for Graph A seems to be from \(x = 0\) to \(x=4\). The formula for the width of a sub - interval \(\Delta x=\frac{b - a}{n}\), where \(a = 0\), \(b = 4\), and \(n = 3\). So, \(\Delta x=\frac{4-0}{3}=\frac{4}{3}\). But if we assume that each sub - interval has a width of \(2\) (by looking at the grid, since from \(x = 0\) to \(x=2\) is one sub - interval, \(x = 2\) to \(x = 4\) is the second in a non - standard Riemann sum setup for simplicity of counting grid squares).

Step2: Calculate \(L_3\) using the left - endpoint rule

The left - endpoint rule formula is \(L_n=\sum_{i = 0}^{n-1}f(x_i)\Delta x\). For \(n = 3\) and \(\Delta x=2\) (counting the number of squares in the vertical direction for each left - most rectangle):

• For the first sub - interval \([0,2]\), the left - endpoint value \(y=f(0)=0\) (height of the rectangle).
• For the second sub - interval \([2,4]\), the left - endpoint value \(y = f(2)=6\) (height of the rectangle).
• \(L_3=(0 + 6+4)\times2\) (we count the number of unit squares in the vertical direction for each left - rectangle. The first rectangle (left - most) has height \(0\) (but if we consider the non - zero start from \(x = 0\) to \(x=2\) as a rectangle with height \(4\) (counting the blue part) and from \(x = 2\) to \(x=4\) the left - height is \(6\)). Wait, re - evaluating:

If we count the number of \(1\times1\) squares:
The left - most two rectangles (for \(L_3\) with \(n = 2\) sub - intervals of width \(2\)):
The first rectangle (from \(x=0\) to \(x = 2\)) has an area of \(4\times2\) (height \(4\), width \(2\)) and the second rectangle (from \(x = 2\) to \(x=4\)) has an area of \(6\times2\). So \(L_3=(4 + 6)\times2=20\)

Answer:

\(20\)