Consider the region below ​f( x ) = ​( 6 −​ x ),above the​ x-axis, and between x = 0 and x = 6. Let x_{i} be the midpoint of the i th subinterval. Approximate the area of the region using six rectangles. Use the midpoints of each subinterval for the heights of the rectangles. The area is approximately how many square units?

Question
Confidence intervals
asked 2021-01-07
Consider the region below ​\(f( x ) = ​( 6\ −\​ x ),\)above the​ x-axis, and between
\(x = 0\ and\ x = 6.\)
Let \(x_{i}\) be the midpoint of the i th subinterval.
Approximate the area of the region using six rectangles. Use the midpoints of each subinterval for the heights of the rectangles. The area is approximately how many square units?

Answers (1)

2021-01-08
Step 1
We have to estimate \(\int_{0)^{6}(6\ -\ x)dx\) by using the mid−point ruleusing six sub−interval.length of sub−interval
\(\Delta\ x = \frac{6\ -\ 0}{6} = 1.\)
Therefore,the sub−intervals consists of \([0,\ 1],\ [1,\ 2],\ [2,\ 3],\ [3,\ 4],\ [4,\ 5]\ and\ [5,\ 6].\)
The mid−point of these sub−intervals are \(\{\frac{1}{2},\ \frac{3}{2},\ \frac{5}{2},\ \frac{7}{2},\ \frac{9}{2},\ \frac{11}{2}\}.\)
Thus,
\(M_{6}=1.\ f\left(\frac{1}{2}\right)\ +\ 1.\ f\left(\frac{3}{2}\right)\ +\ 1.\ f\left(\frac{5}{2}\right)\ +\ 1.\ f\left(\frac{7}{2}\right)\ +\ 1.\ f\left(\frac{9}{2}\right)\ +\ 1.\ f\left(\frac{11}{2}\right) = 11/2 + 9/2 + 7/2 + 5/2 + 3/2 + 1/2\)
\(= 18\)
hence, the approximated area \(\int_{0}^{6}\ (6\ -\ x)\ \approx\ 18 unit^{2}.\)
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