Section 5.5 : Area Problem
2. Estimate the area of the region between \(g\left( x \right) = 4 - \sqrt {{x^2} + 2} \) the \(x\)-axis on \(\left[ { - 1,3} \right]\) using \(n = 6\) and using,
- the right end points of the subintervals for the height of the rectangles,
- the left end points of the subintervals for the height of the rectangles and,
- the midpoints of the subintervals for the height of the rectangles.
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a The right end points of the subintervals for the height of the rectangles. Show SolutionThe widths of each of the subintervals for this problem are,
\[\Delta x = \frac{{3 - \left( { - 1} \right)}}{6} = \frac{2}{3}\]We don’t need to actually graph the function to do this problem. It would probably help to have a number line showing subintervals however. Here is that number line.
In this case we’re going to be using right end points of each of these subintervals to determine the height of each of the rectangles.
The area between the function and the \(x\)-axis is then approximately,
\[\begin{align*}{\mbox{Area}} & \approx \frac{2}{3}f\left( { - \frac{1}{3}} \right) + \frac{2}{3}f\left( {\frac{1}{3}} \right) + \frac{2}{3}f\left( 1 \right) + \frac{2}{3}f\left( {\frac{5}{3}} \right) + \frac{2}{3}f\left( {\frac{7}{3}} \right) + \frac{2}{3}f\left( 3 \right)\\ & = \frac{2}{3}\left( {4 - \frac{{\sqrt {19} }}{3}} \right) + \frac{2}{3}\left( {4 - \frac{{\sqrt {19} }}{3}} \right) + \frac{2}{3}\left( {4 - \sqrt 3 } \right) + \frac{2}{3}\left( {4 - \frac{{\sqrt {43} }}{3}} \right) + \frac{2}{3}\left( {4 - \frac{{\sqrt {67} }}{3}} \right) + \frac{2}{3}\left( {4 - \sqrt {11} } \right)\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{7.420752}}\end{align*}\]b The left end points of the subintervals for the height of the rectangles. Show Solution
As we found in the previous part the widths of each of the subintervals are \(\Delta x = \frac{2}{3}\).
Here is a copy of the number line showing the subintervals to help with the problem.
In this case we’re going to be using left end points of each of these subintervals to determine the height of each of the rectangles.
The area between the function and the \(x\)-axis is then approximately,
\[\begin{align*}{\mbox{Area}} & \approx \frac{2}{3}f\left( { - 1} \right) + \frac{2}{3}f\left( { - \frac{1}{3}} \right) + \frac{2}{3}f\left( {\frac{1}{3}} \right) + \frac{2}{3}f\left( 1 \right) + \frac{2}{3}f\left( {\frac{5}{3}} \right) + \frac{2}{3}f\left( {\frac{7}{3}} \right)\\ & = \frac{2}{3}\left( {4 - \sqrt 3 } \right) + \frac{2}{3}\left( {4 - \frac{{\sqrt {19} }}{3}} \right) + \frac{2}{3}\left( {4 - \frac{{\sqrt {19} }}{3}} \right) + \frac{2}{3}\left( {4 - \sqrt 3 } \right) + \frac{2}{3}\left( {4 - \frac{{\sqrt {43} }}{3}} \right) + \frac{2}{3}\left( {4 - \frac{{\sqrt {67} }}{3}} \right)\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{8.477135}}\end{align*}\]c The midpoints of the subintervals for the height of the rectangles. Show Solution
As we found in the first part the widths of each of the subintervals are \(\Delta x = \frac{2}{3}\).
Here is a copy of the number line showing the subintervals to help with the problem.
In this case we’re going to be using midpoints of each of these subintervals to determine the height of each of the rectangles.
The area between the function and the \(x\)-axis is then approximately,
\[\begin{align*}{\mbox{Area}} & \approx \frac{2}{3}f\left( { - \frac{2}{3}} \right) + \frac{2}{3}f\left( 0 \right) + \frac{2}{3}f\left( {\frac{2}{3}} \right) + \frac{2}{3}f\left( {\frac{4}{3}} \right) + \frac{2}{3}f\left( 2 \right) + \frac{2}{3}f\left( {\frac{8}{3}} \right)\\ & = \frac{2}{3}\left( {4 - \frac{{\sqrt {22} }}{3}} \right) + \frac{2}{3}\left( {4 - \sqrt 2 } \right) + \frac{2}{3}\left( {4 - \frac{{\sqrt {22} }}{3}} \right) + \frac{2}{3}\left( {4 - \frac{{\sqrt {34} }}{3}} \right) + \frac{2}{3}\left( {4 - \sqrt 6 } \right) + \frac{2}{3}\left( {4 - \frac{{\sqrt {82} }}{3}} \right)\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{8.031494}}\end{align*}\]