# 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.

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.

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.

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 $$n \to \infty:$$

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