Chapter 4: Integration
Section 4.1: Area by Riemann Sums
Example 4.1.2
Use a Riemann sum to obtain the area bounded by the graph of fx=x3−2x2−5 x+6 and the x-axis.
Solution
Because of the following simplification, namely,
x3−2x2−5 x+6= factor x−1x−3x+2
the graph in Figure 4.1.2(a) accurately shows the region enclosed by the curve and the x-axis shaded in yellow, and the x-intercepts at x=−2,1,3.
Since fx<0 in 1,3, any Riemann sum for this interval will be negative. Sums have to be taken over two different intervals, −2,1 and 1,3 so that this sign difference can be taken into account.
The calculations in Table 4.1.2(a) show how the desired area can be calculated interactively with left Riemann sums.
Figure 4.1.2(a) Graph of fx=x3−2x2−5 x+6
Define the function f
Control-drag fx=… Context Panel: Assign Function
fx=x3−2x2−5 x+6→assign as functionf
Area of region A
Write the ratio defining the stepsize hA for region A
Context Panel: Assign to a Name≻h[A]
1−−2n = 3n→assign to a namehA
Expression palette: Limit and Summation templates
Context Panel: Evaluate and Display Inline
limn→∞∑k=0n−1f−2+k hA hA = 634
Area of region B
Write the ratio defining the stepsize hB for region B
Context Panel: Assign to a Name≻h[B]
3−1n = 2n→assign to a namehB
Expression palette: Limit and Summation templates Note the minus sign!
−limn→∞∑k=0n−1f1+k hB hB = 163
Total enclosed area
Add the areas of regions A and B Context Panel: Evaluate and Display Inline Context Panel: Approximate≻5 (digits)
634+163 = 25312→at 5 digits21.083
Table 4.1.2(a) Area by left Riemann sum for a region that is both above and below the x-axis
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