Changes

:<math>=\left ( { 1 \over 3 } \right ) \left( {68579\over2860} \right) = {68579\over8580}</math>

:<math>=\left ( { 1 \over 3 } \right ) \left( {68579\over2860} \right) = {68579\over8580}</math>

:<math>\approx 7.99289...</math>

:<math>\approx 7.99289...</math>

<br />This new answer is much closer than the original; it is off by about <math>.05</math>. Note how much less "error" area that there is on the picture. The graph represents the first parabola we use for Example 2 Continued, evaluating from 4 to 6. By reducing the area that each individual parabola approximates, our total approximation became more accurate.

<br />This new answer is much closer than the original; it is off by about <math>.05</math>. In the picture on the right, note how much less "error" area that there is compared to when we only used one parabola. The graph represents the first parabola we use for Example 2 Continued, evaluating from 4 to 6. By reducing the interval that each individual parabola approximates, our total approximation became more accurate.

  If the number of subintervals is increased, it can be shown that the estimated value approaches the real value:

  If the number of subintervals is increased, it can be shown that the estimated value approaches the real value: