Riemann Sums and the Definite Integral
Solved Problems
Example 7.
Use the Left Riemann Sum with \(n = 5\) to approximate the integral \[I = \int\limits_1^6 {\left( {x + {x^2}} \right)dx}.\]
Solution.
Figure 5. The partition points are \(1,\) \(2,\) \(3,\) \(4,\) \(5,\) \(6.\) Hence, the left endpoints of the subintervals are
\[{\xi _i} = \left\{ {1,2,3,4,5} \right\}.\]
The Left Riemann Sum with \(n = 5\) is defined as
\[{L_5} = \sum\limits_{i = 1}^5 {f\left( {{\xi _i}} \right)\Delta x} ,\]
where \(\Delta x = 1.\)
Calculate the values of the function \({f\left( {{\xi _i}} \right)}\) at the left endpoints:
\[f\left( {{\xi _1}} \right) = f\left( 1 \right) = 1 + {1^2} = 2;\]
\[f\left( {{\xi _2}} \right) = f\left( 2 \right) = 2 + {2^2} = 6;\]
\[f\left( {{\xi _3}} \right) = f\left( 3 \right) = 3 + {3^2} = 12;\]
\[f\left( {{\xi _4}} \right) = f\left( 4 \right) = 4 + {4^2} = 20;\]
\[f\left( {{\xi _5}} \right) = f\left( 5 \right) = 5 + {5^2} = 30.\]
This yields
\[I = \int\limits_1^6 {\left( {x + {x^2}} \right)dx} \approx {L_5} = \sum\limits_{i = 1}^5 {f\left( {{\xi _i}} \right)\Delta x} = \Delta x\sum\limits_{i = 1}^5 {f\left( {{\xi _i}} \right)} = 1 \cdot \left( {2 + 6 + 12 + 20 + 30} \right) = 70.\]
Example 8.
Use the Midpoint Riemann Sum with \(n = 4\) to approximate the integral \[I = \int\limits_0^8 {{x^3}dx}.\]
Solution.
Figure 6. The partition points are \(0,\) \(2,\) \(4,\) \(6,\) \(8.\) Hence, the midpoints of the subintervals are
\[{\xi _i} = \left\{ {1,3,5,7} \right\}.\]
The Midpoint Riemann Sum with \(n = 4\) is written in the form
\[{M_4} = \sum\limits_{i = 1}^4 {f\left( {{\xi _i}} \right)\Delta x} ,\]
where \(\Delta x = 2.\)
Determine the values of the function \({f\left( {{\xi _i}} \right)}\) at the midpoints:
\[f\left( {{\xi _1}} \right) = f\left( 1 \right) = {1^3} = 1;\]
\[f\left( {{\xi _2}} \right) = f\left( 3 \right) = {3^3} = 27;\]
\[f\left( {{\xi _3}} \right) = f\left( 5 \right) = {5^3} = 125;\]
\[f\left( {{\xi _4}} \right) = f\left( 7 \right) = {7^3} = 343.\]
Then
\[I = \int\limits_0^8 {{x^3}dx} \approx {M_4} = \sum\limits_{i = 1}^4 {f\left( {{\xi _i}} \right)\Delta x} = \Delta x\sum\limits_{i = 1}^4 {f\left( {{\xi _i}} \right)} = 2 \cdot \left( {1 + 27 + 125 + 343} \right) = 992.\]
Example 9.
Use the Midpoint Riemann Sum with \(n = 4\) to approximate the integral \[I = \int\limits_{ - 3}^5 {\left( {1 + 2{x^2}} \right)dx}.\]
Solution.
Figure 7. The partition points are \(-3,\) \(-1,\) \(1,\) \(3,\) \(5.\) So the midpoints of the subintervals are
\[{\xi _i} = \left\{ {-2,0,2,4} \right\}.\]
The Midpoint Riemann Sum with \(n = 4\) has the form
\[{M_4} = \sum\limits_{i = 1}^4 {f\left( {{\xi _i}} \right)\Delta x} ,\]
where \(\Delta x = 2.\)
Calculate the values of the function \({f\left( {{\xi _i}} \right)}\) at the midpoints:
\[f\left( {{\xi _1}} \right) = f\left( { - 2} \right) = 1 + 2 \cdot {\left( { - 2} \right)^2} = 9;\]
\[f\left( {{\xi _2}} \right) = f\left( 0 \right) = 1 + 2 \cdot {0^2} = 1;\]
\[f\left( {{\xi _3}} \right) = f\left( 2 \right) = 1 + 2 \cdot {2^2} = 9;\]
\[f\left( {{\xi _4}} \right) = f\left( 4 \right) = 1 + 2 \cdot {4^2} = 33.\]
Then
\[I = \int\limits_{ - 3}^5 {\left( {1 + 2{x^2}} \right)dx} = \sum\limits_{i = 1}^4 {f\left( {{\xi _i}} \right)\Delta x} = \Delta x\sum\limits_{i = 1}^4 {f\left( {{\xi _i}} \right)} = 2 \cdot \left( {9 + 1 + 9 + 33} \right) = 104.\]
Example 10.
Find the area \(A\) under the curve \[y = 1 - {x^2}\] on the interval \(\left[{0,1}\right]\) by calculating the right Riemann Sum for \(n\) subintervals and taking a limit of the sum as \(n \to \infty.\)
Solution.
Assuming that the interval \(\left[ {0,1} \right]\) is divided into \(n\) equal subintervals, we find the width of each subinterval:
\[\Delta x = \frac{1}{n}.\]
The coordinates of partition points are
\[{x_i} = a + i\Delta x = 0 + i \cdot \frac{1}{n} = \frac{i}{n}.\]
For the right Riemann Sum, the arbitrary points \({\xi _i}\) are chosen to be
\[{\xi _i} = {x_i} = \frac{i}{n}.\]
Hence
\[f\left( {{\xi _i}} \right) = f\left( {{x_i}} \right) = 1 - x_i^2 = 1 - {\left( {\frac{i}{n}} \right)^2} = \frac{{{n^2} - {i^2}}}{{{n^2}}},\]
and the right Riemann Sum \({R_n}\) is written in the form
\[{R_n} = \sum\limits_{i = 1}^n {f\left( {{x_i}} \right)\Delta x} = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{n^2} - {i^2}}}{{{n^2}}}} = \frac{1}{{{n^3}}}\sum\limits_{i = 1}^n {\left( {{n^2} - {i^2}} \right)} = \frac{1}{{{n^3}}}\left( {\sum\limits_{i = 1}^n {{n^2}} - \sum\limits_{i = 1}^n {{i^2}} } \right).\]
Calculate the sums in the last expression:
\[\sum\limits_{i = 1}^n {{n^2}} = {n^2}\sum\limits_{i = 1}^n 1 = {n^2} \cdot n = {n^3};\]
\[\sum\limits_{i = 1}^n {{i^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} = \frac{{2{n^3} + 3{n^2} + n}}{6}.\]
Then
\[{R_n} = \frac{1}{{{n^3}}}\left( {\sum\limits_{i = 1}^n {{n^2}} - \sum\limits_{i = 1}^n {{i^2}} } \right) = \frac{1}{{{n^3}}}\left( {{n^3} - \frac{{2{n^3} + 3{n^2} + n}}{6}} \right) = \frac{1}{{{n^3}}} \cdot \frac{{4{n^3} - 3{n^2} - n}}{6} = \frac{{4{n^2} - 3n - 1}}{{6{n^2}}}.\]
Now, to find the area \(A\) under the curve, we take a limit of the Riemann Sum as \(n \to \infty:\)
\[A = \lim\limits_{n \to \infty } {R_n} = \lim\limits_{n \to \infty } \frac{{4{n^2} - 3n - 1}}{{6{n^2}}} = \lim\limits_{n \to \infty } \frac{{4 - \frac{3}{n} - \frac{1}{{{n^2}}}}}{6} = \frac{4}{6} = \frac{2}{3}.\]
Example 11.
Find the area \(A\) under the curve \[y = {x^2}\] on the interval \(\left[{0,3}\right]\) by calculating the right Riemann Sum for \(n\) subintervals and taking a limit of the sum as \(n \to \infty.\)
Solution.
We divide the interval \(\left[ {0,3} \right]\) into \(n\) equal subintervals. The width of each subinterval is
\[\Delta x = \frac{3}{n}.\]
The partition points have the following coordinates:
\[{x_i} = a + i\Delta x = 0 + i \cdot \frac{3}{n} = \frac{3i}{n}.\]
For the right Riemann Sum, we choose the arbitrary points \({\xi _i}\) to be
\[{\xi _i} = {x_i} = \frac{3i}{n}.\]
Then
\[f\left( {{\xi _i}} \right) = f\left( {{x_i}} \right) = {\left( {\frac{{3i}}{n}} \right)^2} = \frac{{9{i^2}}}{{{n^2}}},\]
so the right Riemann Sum \({R_n}\) is written in the form
\[{R_n} = \sum\limits_{i = 1}^n {f\left( {{\xi _i}} \right)\Delta x} = \sum\limits_{i = 1}^n {f\left( {{x_i}} \right)\Delta x} = \sum\limits_{i = 1}^n {f\left( {{x_i}} \right)\Delta x} = \Delta x\sum\limits_{i = 1}^n {\frac{{9{i^2}}}{{{n^2}}}} = \frac{3}{n} \cdot \frac{9}{{{n^2}}}\sum\limits_{i = 1}^n {{i^2}} = \frac{{27}}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2}} .\]
Given that \(\sum\limits_{i = 1}^n {{i^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6},\) we have
\[{R_n} = \frac{{27}}{{{n^3}}} \cdot \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} = \frac{{27}}{{{n^3}}} \cdot \frac{{2{n^3} + 3{n^2} + n}}{6} = \frac{{27\left( {2{n^3} + 3{n^2} + n} \right)}}{{6{n^3}}} = \frac{{9\left( {2{n^2} + 3n + 1} \right)}}{{2{n^2}}} = \frac{{18{n^2} + 27n + 9}}{{2{n^2}}}.\]
To determine the area \(A\) under the curve, we take a limit of the Riemann Sum \({R_n}\) as \(n \to \infty:\)
\[A = \lim\limits_{n \to \infty } {R_n} = \lim\limits_{n \to \infty } \frac{{18{n^2} + 27n + 9}}{{2{n^2}}} = \lim\limits_{n \to \infty } \frac{{18 + \frac{{27}}{n} + \frac{9}{{{n^2}}}}}{2} = \frac{{18}}{2} = 9.\]
