Calculus

Integration of Functions

Integration of Functions Logo

Riemann Sums and the Definite Integral

  • Riemann Sums

    Let \(f\left( x \right)\) be a continuous and non-negative function defined on the closed interval \(\left[ {a,b} \right].\) How to find the area of the region \(S\) bounded by the curve \(y = f\left( x \right),\) the \(x-\)axis, and the vertical lines \(x = a\) and \(x = b?\)

    Area of the region S bounded by the curve y=f(x), vertical lines x=a, x=b and the x-axis
    Figure 1.

    We can approximate this area by dividing the region into rectangles.

    Suppose that the interval \(\left[ {a,b} \right]\) is divided into \(n\) subintervals defined by the points

    \[{a = {x_0} < {x_1} < {x_2} < \cdots }<{ {x_i} < \cdots }<{ {x_n} = b}\]

    Then the \(n\) subintervals are

    \[{\left[ {{x_0},{x_1}} \right],\left[ {{x_1},{x_2}} \right], \ldots ,}\kern0pt{\left[ {{x_{n – 1}},{x_n}} \right].}\]

    This subdivision is called the partition of \(\left[ {a,b} \right]\) and is denoted by \(P.\)

    The width of the \(i\)th subinterval \(\left[ {{x_{i – 1}},{x_i}} \right]\) is given by

    \[{\Delta {x_i} = {x_i} – {x_{i – 1}},}\]

    so the subintervals generally may have different widths.

    The width of the largest subinterval is called the norm of the partition \(P\) and is denoted \(\left\| P \right\|.\) Thus

    \[{\left\| P \right\| }={ \max \left\{ {\Delta {x_1},\Delta {x_2}, \ldots ,\Delta {x_n}} \right\}.}\]

    We use the partition \(P\) to divide the region \(S\) into strips \({S_1},{S_2}, \ldots ,{S_n}.\) We then approximate the strips \({S_i}\) using rectangles \({R_i}\) and choosing a sample point \({\xi _i}\) in each subinterval \(\left[ {{x_{i – 1}},{x_i}} \right].\)

    Definition of the Riemann Sum
    Figure 2.

    The point \({\xi _i}\) can be anywhere in its subinterval.

    The area of the \(i\)th rectangle \({R_i}\) is given by

    \[{{A_i} = f\left( {{\xi _i}} \right)\Delta {x_i},}\]

    so the area of the region \(S\) is approximated by the sum of the areas of the rectangles \({R_i}:\)

    \[{A \approx \sum\limits_{i = 1}^n {{A_i}} }={ \sum\limits_{i = 1}^n {f\left( {{\xi _i}} \right)\Delta {x_i}} }={ f\left( {{\xi _1}} \right)\Delta {x_1} }+{ f\left( {{\xi _2}} \right)\Delta {x_2} + \cdots }+{ f\left( {{\xi _n}} \right)\Delta {x_n}.}\]

    The sum \(\sum\limits_{i = 1}^n {f\left( {{\xi _i}} \right)\Delta {x_i}} \) is called the Riemann Sum, which was introduced by Bernhard Riemann \(\left( {1826 – 1866} \right),\) a German mathematician.

    There are several types of Riemann Sums. The Left Riemann Sum uses the left endpoints of the subintervals. The Right Riemann Sum uses the right endpoints, and the Midpoint Riemann Sum is calculated using the midpoints of the subintervals.

    The Definite Integral

    If we take the limit of the Riemann Sum as the norm of the partition \(\left\| P \right\|\) approaches zero, we get the exact value of the area \(A:\)

    \[{A = \lim\limits_{\left| P \right| \to 0} \sum\limits_{i = 1}^n {f\left( {{\xi _i}} \right)\Delta {x_i}} .}\]

    This limit is called the definite integral of the function \(f\left( x \right)\) from \(a\) to \(b\) and is denoted by \(\int\limits_a^b {f\left( x \right)dx}.\)

    The notation for the definite integral is very similar to the notation for an indefinite integral. The new elements \(a\) and \(b\) mean, respectively, the lower and the upper limit of integration.

    Properties of the Definite Integral

    We assume below that \(f\left( x \right)\) and \(g\left( x \right)\) are continuous functions on the closed interval \(\left[ {a,b} \right].\)

    1. \(\int\limits_a^b {1dx} = b – a\)
    2. \(\int\limits_a^b {kf\left( x \right)dx} \) \(= k \int\limits_a^b {f\left( x \right)dx} ,\) where \(k\) is a constant;
    3. \(\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} \) \(= \int\limits_a^b {f\left( x \right)dx} \) \(+{ \int\limits_a^b {g\left( x \right)dx} }\)
    4. \(\int\limits_a^b {f\left( x \right)dx} \) \(= \int\limits_a^c {f\left( x \right)dx} \) \(+ \int\limits_c^b {f\left( x \right)dx} ,\) where \(a \lt c \lt b;\)
    5. If \(0 \le f\left( x \right) \le g\left( x \right)\) for all \(x \in \left[ {a,b} \right],\) then \(0 \le \int\limits_a^b {f\left( x \right)dx} \) \(\le \int\limits_a^b {g\left( x \right)dx} .\)
    6. \(\int\limits_a^a {f\left( x \right)dx} = 0\)
    7. \(\int\limits_a^b {f\left( x \right)dx} \) \(= – \int\limits_b^a {f\left( x \right)dx}\)
    8. If \(f\left( x \right) \ge 0\) in the interval \(\left[ {a,b} \right],\) then \(\int\limits_a^b {f\left( x \right)dx} \ge 0\)

  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Estimate the area under \(f\left( x \right) = {x^2}\) on the interval \(\left[ {0,10} \right]\) using the midpoint Riemann Sum for \(n = 5.\)

    Example 2

    Find the norm of the partition \[P = \left\{ {1,1.1,1.7,2.4,2.9,3.1,3.8,4} \right\}\]

    Example 3

    Find the norm of the partition \[{P \text{ = }}\kern0pt{\left\{ {-5,-4.3,-3.2,-2.3,-1.8,-1} \right\}}\]

    Example 4

    Find the Riemann Sum for the function \(f\left( x \right) = 5x – 2\) and the partition \(\left\{ {1,3,6,7} \right\}\) using the sample points \({\xi _i} = \left\{ {2,5,7} \right\}.\)

    Example 5

    Find the Riemann Sum for the function \(f\left( x \right) = \large{\frac{1}{{2x + 1}}}\normalsize\) and the partition \(\left\{ {0,2,5,8} \right\}\) using the sample points \({\xi _i} = \left\{ {1,3,6} \right\}.\)

    Example 6

    Use the Right Riemann Sum with \(n = 5\) to approximate the integral \(I = \int\limits_{-1}^4 {\left( {16 – {x^2}} \right)dx}.\)

    Example 7

    Use the Left Riemann Sum with \(n = 5\) to approximate the integral \(I = \int\limits_1^6 {\left( {x + {x^2}} \right)dx}.\)

    Example 8

    Use the Midpoint Riemann Sum with \(n = 4\) to approximate the integral \(I = \int\limits_0^8 {{x^3}dx}.\)

    Example 9

    Use the Midpoint Riemann Sum with \(n = 4\) to approximate the integral \(I = \int\limits_{ – 3}^5 {\left( {1 + 2{x^2}} \right)dx}.\)

    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 \(n \to \infty.\)

    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 \(n \to \infty.\)

    Example 1.

    Estimate the area under \(f\left( x \right) = {x^2}\) on the interval \(\left[ {0,10} \right]\) using the midpoint Riemann Sum for \(n = 5.\)

    Solution.

    We partition the interval \(\left[ {0,10} \right]\) into \(5\) equal subintervals with endpoints

    \[{{x_i} = \left\{ {0,2,4,6,8,10} \right\}.}\]

    Calculating area under the curve f(x)=x^2 on the interval [0,10] using the midpoint Riemann Sum
    Figure 5.

    The midpoints \({\xi _i}\) of the subintervals have the coordinates:

    \[{{\xi _i} = \left\{ {1,3,5,7,9} \right\}.}\]

    Hence, the midpoint Riemann Sum is given by

    \[{M_5} = \sum\limits_{i = 1}^5 {f\left( {{\xi _i}} \right)\Delta x} ,\]

    where \(\Delta x = 2\) is the width of each subinterval.

    Calculating the values of \({f\left( {{\xi _i}} \right)},\) we find the approximate value of the area:

    \[{A \approx {M_5} }={ \sum\limits_{i = 1}^5 {f\left( {{\xi _i}} \right)\Delta x} }={ \Delta x\sum\limits_{i = 1}^5 {f\left( {{\xi _i}} \right)} }={ \Delta x\left[ {f\left( {{\xi _1}} \right) + f\left( {{\xi _2}} \right) + \cdots + f\left( {{\xi _5}} \right)} \right] }={ 2\left( {{1^2} + {3^2} + {5^2} + {7^2} + {9^2}} \right) }={ 2\left( {1 + 9 + 25 + 49 + 81} \right) }={ 330.}\]

    Example 2.

    Find the norm of the partition \[P = \left\{ {1,1.1,1.7,2.4,2.9,3.1,3.8,4} \right\}\]

    Solution.

    The set of partition points is

    \[{{x_0} = 1,\;}\kern0pt{{x_1} = 1.1,\;}\kern0pt{{x_2} = 1.7,\;}\kern0pt{{x_3} = 2.4,\;}{{x_4} = 2.9,\;}\kern0pt{{x_5} = 3.1,\;}\kern0pt{{x_6} = 3.8,\;}\kern0pt{{x_7} = 4.}\]

    Calculate the width of each subinterval:

    \[{\Delta {x_1} = {x_1} – {x_0} }={ 1.1 – 1 }={ 0.1}\]

    \[{\Delta {x_2} = {x_2} – {x_1} }={ 1.7 – 1.1 }={ 0.6}\]

    \[{\Delta {x_3} = {x_3} – {x_2} }={ 2.4 – 1.7 }={ 0.7}\]

    \[{\Delta {x_4} = {x_4} – {x_3} }={ 2.9 – 2.4 }={ 0.5}\]

    \[{\Delta {x_5} = {x_5} – {x_4} }={ 3.1 – 2.9 }={ 0.2}\]

    \[{\Delta {x_6} = {x_6} – {x_5} }={ 3.8 – 3.1 }={ 0.7}\]

    \[{\Delta {x_7} = {x_7} – {x_6} }={ 4 – 3.8 }={ 0.2}\]

    Then the norm of the partition is given by

    \[{\left\| P \right\| \text{ = }}\kern0pt{ \max \left\{ {0.1,0.6,0.7,0.5,0.2,0.7,0.2} \right\} }={ 0.7}\]

    Example 3.

    Find the norm of the partition \[{P \text{ = }}\kern0pt{\left\{ {-5,-4.3,-3.2,-2.3,-1.8,-1} \right\}}\]

    Solution.

    The partition contains the following points:

    \[{{x_0} = -5,\;}\kern0pt{{x_1} = -4.3,\;}\kern0pt{{x_2} = -3.2,\;}\kern0pt{{x_3} = -2.3,\;}{{x_4} = -1.8,\;}\kern0pt{{x_5} = -1.}\]

    Determine the width of each subinterval:

    \[{\Delta {x_1} = {x_1} – {x_0} }={ -4.3 – \left({-5}\right) }={ 0.7}\]

    \[{\Delta {x_2} = {x_2} – {x_1} }={ -3.2 – \left({-4.3}\right) }={ 1.1}\]

    \[{\Delta {x_3} = {x_3} – {x_2} }={ -2.3 – \left({-3.2}\right) }={ 0.9}\]

    \[{\Delta {x_4} = {x_4} – {x_3} }={ -1.8 – \left({-2.3}\right) }={ 0.5}\]

    \[{\Delta {x_5} = {x_5} – {x_4} }={ -1 – \left({-1.8}\right) }={ 0.8}\]

    Hence, the norm of the partition is equal to

    \[{\left\| P \right\| \text{ = }}\kern0pt{ \max \left\{ {0.7,1.1,0.9,0.5,0.8} \right\} }={ 1.1}\]

    Example 4.

    Find the Riemann Sum for the function \(f\left( x \right) = 5x – 2\) and the partition \(\left\{ {1,3,6,7} \right\}\) using the sample points \({\xi _i} = \left\{ {2,5,7} \right\}.\)

    Solution.

    The widths of the subintervals are

    \[\Delta {x_i} = \left\{ {2,3,1} \right\}.\]

    Calculate the function values at the sample points:

    \[f\left( {{\xi _1}} \right) = f\left( 2 \right) = 5 \cdot 2 – 2 = 8\]

    \[f\left( {{\xi _2}} \right) = f\left( 5 \right) = 5 \cdot 5 – 2 = 23\]

    \[f\left( {{\xi _3}} \right) = f\left( 7 \right) = 5 \cdot 7 – 2 = 33\]

    Then the Riemann Sun is given by

    \[{\sum\limits_{i = 1}^3 {f\left( {{\xi _i}} \right)\Delta {x_i}} }={ f\left( {{\xi _1}} \right)\Delta {x_1} }+{ f\left( {{\xi _2}} \right)\Delta {x_2} }+{ f\left( {{\xi _3}} \right)\Delta {x_3} }={ 8 \cdot 2 + 23 \cdot 3 + 33 \cdot 1 }={ 16 + 69 + 33 }={ 118.}\]

    Example 5.

    Find the Riemann Sum for the function \(f\left( x \right) = \large{\frac{1}{{2x + 1}}}\normalsize\) and the partition \(\left\{ {0,2,5,8} \right\}\) using the sample points \({\xi _i} = \left\{ {1,3,6} \right\}.\)

    Solution.

    Calculate the widths of the subintervals:

    \[\Delta {x_1} = 2 – 0 = 2\]

    \[\Delta {x_2} = 5 – 2 = 3\]

    \[\Delta {x_3} = 8 – 5 = 3\]

    The function has the following values at the sample points:

    \[{f\left( {{\xi _1}} \right) = f\left( 1 \right) }={ \frac{1}{{2 \cdot 1 + 1}} }={ \frac{1}{3}}\]

    \[{f\left( {{\xi _2}} \right) = f\left( 3 \right) }={ \frac{1}{{2 \cdot 3 + 1}} }={ \frac{1}{7}}\]

    \[{f\left( {{\xi _3}} \right) = f\left( 6 \right) }={ \frac{1}{{2 \cdot 6 + 1}} }={ \frac{1}{13}}\]

    Hence, the Riemann Sun is given by

    \[{\sum\limits_{i = 1}^3 {f\left( {{\xi _i}} \right)\Delta {x_i}} }={ f\left( {{\xi _1}} \right)\Delta {x_1} }+{ f\left( {{\xi _2}} \right)\Delta {x_2} }+{ f\left( {{\xi _3}} \right)\Delta {x_3} }={ \frac{1}{3} \cdot 2 + \frac{1}{7} \cdot 3 + \frac{1}{{13}} \cdot 3 }={ \frac{2}{3} + \frac{3}{7} + \frac{3}{{13}} }={ \frac{{182 + 117 + 63}}{{273}} }={ \frac{{362}}{{273}}.}\]

    Example 6.

    Use the Right Riemann Sum with \(n = 5\) to approximate the integral \(I = \int\limits_{-1}^4 {\left( {16 – {x^2}} \right)dx}.\)

    Solution.

    Right Riemann Sum with n=5
    Figure 6.

    The partition points are \(-1,\) \(0,\) \(1,\) \(2,\) \(3,\) \(4,\) so the right endpoints of the subintervals are

    \[{\xi _i} = \left\{ {0,1,2,3,4} \right\}.\]

    The Right Riemann Sum is given by

    \[{R_5} = \sum\limits_{i = 1}^5 {f\left( {{\xi _i}} \right)\Delta x} ,\]

    where \(\Delta x = 1.\)

    Calculate the values of the function \({f\left( {{\xi _i}} \right)}\) at the right endpoints:

    \[f\left( {{\xi _1}} \right) = f\left( 0 \right) = 16 – {0^2} = 16\]

    \[f\left( {{\xi _2}} \right) = f\left( 1 \right) = 16 – {1^2} = 15\]

    \[f\left( {{\xi _3}} \right) = f\left( 2 \right) = 16 – {2^2} = 12\]

    \[f\left( {{\xi _4}} \right) = f\left( 3 \right) = 16 – {3^2} = 7\]

    \[f\left( {{\xi _5}} \right) = f\left( 4 \right) = 16 – {4^2} = 0\]

    Hence

    \[{I = \int\limits_{ – 1}^4 {\left( {16 – {x^2}} \right)dx} }\approx{ {R_5} }={ \sum\limits_{i = 1}^5 {f\left( {{\xi _i}} \right)\Delta x} }={ \Delta x\sum\limits_{i = 1}^5 {f\left( {{\xi _i}} \right)} }={ 1 \cdot \left( {16 + 15 + 12 + 7 + 0} \right) }={ 50.}\]

    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.

    Left Riemann Sum with n=5
    Figure 7.

    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.

    Midpoint Riemann Sum with n=4
    Figure 8.

    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.

    Midpoint Riemann Sum with n=4
    Figure 9.

    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

    \[{I = \int\limits_{ – 3}^5 {\left( {1 + 2{x^2}} \right)dx} }\approx{ {M_4} }={ \sum\limits_{i = 1}^4 {f\left( {{\xi _i}} \right)\Delta x} }={ \Delta x\sum\limits_{i = 1}^4 {f\left( {{\xi _i}} \right)} }={ 2 \cdot \left( {9 + 1 + 9 + 33} \right) }={ 104.}\]

    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 \(\sum\limits_{i = 1}^n {{i^2}} = \large{\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}\normalsize,\) 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 \(n \to \infty:\)

    \[{A = \lim\limits_{n \to \infty } {R_n} }={ \lim\limits_{n \to \infty } \frac{{18{n^2} + 27n + 9}}{{2{n^2}}} }={ \lim\limits_{n \to \infty } \frac{{18 + \frac{{27}}{n} + \frac{9}{{{n^2}}}}}{2} }={ \frac{{18}}{2} }={ 9.}\]