Question

How do you use Riemann sums to evaluate the area under the curve of $f(x) = {x^3}$ on the closed interval $[1,3]$, with $n = 4$rectangles using right, left and midpoints?

Answer 1

Hint : Before getting into the problem let us know about Riemann sum. Riemann sum is a kind of approximation of an integration by a finite sum. Riemann sum is mainly used for approximating the area of lines, functions or curves on a graph. The idea behind the Riemann sum is to divide the region into the known shapes such as rectangles, trapezoids, parabolas or Cubic’s, which form the new region that is somewhat similar to the region we needed to measure, and then add all of the regions to find the area. In the question given here, it is given that the function $f(x) = {x^3}$, forms a curve, and the area of the curve between the points $1,3$ is divided into $4$ rectangles. Now we are going to find the area of the regions by right, left and midpoints Riemann sum.
Formula used:
$\Delta x = \dfrac{{b - a}}{n}$
Right Riemann sum $= \Delta x[f(a + \Delta x) + f(a + 2\Delta x) + ... + f(b)]$
Left Riemann sum $= \Delta x[f(a) + f(a + \Delta x) + f(a + 2\Delta x) + ... + f(b - \Delta x)]$
Midpoint rule $= \Delta x\left[ {f(a + \dfrac{{\Delta x}}{2}) + f(a + \dfrac{{3\Delta x}}{3}) + ... + f(b - \dfrac{{\Delta x}}{2})} \right]$

Complete step-by-step answer :
The given question can be mathematically represented as $\int\limits_1^3 {{x^3}} dx$, where $f(x) = {x^3}$, $a = 1,b = 3$ and $n = 4$.
$\Delta x$ is the subinterval, in each subinterval we will get a rectangle.
$\Delta x = \dfrac{{b - a}}{n} = \dfrac{{3 - 1}}{4} = \dfrac{2}{4} = \dfrac{1}{2}$
Now we have the subinterval as $\dfrac{1}{2}$.
Add the subintervals with the existing points in $[1,3]$ to get new points.
$1 + \dfrac{1}{2} = \dfrac{3}{2}$ and $2 + \dfrac{1}{2} = \dfrac{5}{2}$
Divide the interval $[1,3]$ into $n = 4$ subintervals of the length $\Delta x = \dfrac{1}{2}$ with the following endpoints: $1,\dfrac{3}{2},2\dfrac{5}{2},3$
Now we have four rectangles with points: $\left( {1,\dfrac{3}{2}} \right),\left( {\dfrac{3}{2},2} \right),\left( {2,\dfrac{5}{2}} \right)$ and $\left( {\dfrac{5}{2},3} \right)$
Now the Riemann sum is the sum of the area of the $4$ rectangles.
Right sum
For the right sum, we have to use the right end points of the subintervals. Therefore points are $\dfrac{3}{2},2\dfrac{5}{2},3$.
Here we do not consider the interval $1$ because it is the left end point.
We have the formula Right Riemann sum $= \Delta x[f(a + \Delta x) + f(a + 2\Delta x) + ... + f(b)]$. Here each term inside the bracket represents the points of the rectangles. Already we have found the points of the rectangle on the right.
Therefore the formula of area can be written as: $= \Delta x[f({x_1}) + f({x_2}) + f({x_3}) + f({x_4})]$
Substitute the values: $\Delta x = \dfrac{1}{2},{x_1} = \dfrac{3}{2},{x_2} = 2,{x_3} = \dfrac{5}{2}$ and ${x_4} = 3$.
$= \dfrac{1}{2}\left[ {f\left( {\dfrac{3}{2}} \right) + f(2) + f\left( {\dfrac{5}{2}} \right) + f(3)} \right]$
Now evaluate the functions of the right endpoints of the subintervals with function $f(x) = {x^3}$,
$f({x_1}) = f\left( {\dfrac{3}{2}} \right)$, now applying $f(x) = {x^3}$, $= {\left( {\dfrac{3}{2}} \right)^3} = \dfrac{{27}}{8}$ on further division we will get, $\dfrac{{27}}{8} = 3.375$, that is,
$f({x_1}) = f\left( {\dfrac{3}{2}} \right) = \dfrac{{27}}{8} = 3.375$
Similarly, $f({x_2}) = f\left( 2 \right) = 8$
$f({x_3}) = f\left( {\dfrac{5}{2}} \right) = \dfrac{{125}}{8} = 15.625$
$f({x_4}) = f\left( 3 \right) = 27$
Substitute the values in the formula,
$= \dfrac{1}{2}\left[ {3.375 + 8 + 15.625 + 27} \right]$
Adding the values,
$= \dfrac{1}{2}\left[ {54} \right]$
Dividing $54$ by 2 we will get,
$= 27$
Hence by Right Riemann sum, $\int\limits_1^3 {{x^3}} dx \approx 27$.
Left sum
For the left sum, we have to use the left endpoints of the subintervals. Therefore points are $1,\dfrac{3}{2},2\dfrac{5}{2}$.
Here we do not consider the interval $3$ because it is the right end point.
We have the formula Left Riemann sum $= \Delta x[f(a) + f(a + \Delta x) + f(a + 2\Delta x) + ... + f(b - \Delta x)]$. Here each term inside the bracket represents the points of the rectangles. Already we have found the points of the rectangle on the left.
Therefore the formula of area can be written as: $= \Delta x[f({x_1}) + f({x_2}) + f({x_3}) + f({x_4})]$
Substitute the values: $\Delta x = \dfrac{1}{2},{x_1} = 1,{x_2} = \dfrac{3}{2},{x_3} = 2$ and ${x_4} = \dfrac{5}{2}$.
$= \dfrac{1}{2}\left[ {f(1) + f\left( {\dfrac{3}{2}} \right) + f(2) + f\left( {\dfrac{5}{2}} \right)} \right]$
Now evaluate the functions of the right endpoints of the subintervals with function $f(x) = {x^3}$,
$f({x_1}) = f\left( 1 \right)$, now applying $f(x) = {x^3}$, we will get, ${1^3} = 1$, that is,
$f({x_1}) = f\left( 1 \right) = 1$
Similarly, $f({x_2}) = f\left( {\dfrac{3}{2}} \right) = \dfrac{{27}}{8} = 3.375$
$f({x_3}) = f\left( 2 \right) = 8$
$f({x_4}) = f\left( {\dfrac{5}{2}} \right) = \dfrac{{125}}{8} = 15.625$
Substitute the values in the formula,
$= \dfrac{1}{2}\left[ {1 + 3.375 + 8 + 15.625} \right]$
Adding the values,
$= \dfrac{1}{2}\left[ {28} \right]$
Dividing $28$ by 2 we will get,
$= 14$
Hence by Left Riemann sum, $\int\limits_1^3 {{x^3}} dx \approx 14$.
Midpoint rule
The midpoints use the midpoints of the subintervals $\left( {1,\dfrac{3}{2}} \right),\left( {\dfrac{3}{2},2} \right),\left( {2,\dfrac{5}{2}} \right)$ and $\left( {\dfrac{5}{2},3} \right)$.
We have the formula Left Riemann sum $= \Delta x\left[ {f(a + \dfrac{{\Delta x}}{2}) + f(a + \dfrac{{3\Delta x}}{3}) + ... + f(b - \dfrac{{\Delta x}}{2})} \right]$. Here each term inside the bracket represents the midpoint of an interval.
Midpoint of an interval is the average of its end points.
Therefore the formula of area can be written as: $= \Delta x[f({x_1}) + f({x_2}) + f({x_3}) + f({x_4})]$
Midpoint of the point $\left( {1,\dfrac{3}{2}} \right)$, ${x_1} = \dfrac{{1 + \dfrac{3}{2}}}{2} = \dfrac{{\dfrac{5}{2}}}{2} = \dfrac{5}{2} \times \dfrac{1}{2} = \dfrac{5}{4}$
Midpoint of the point $\left( {\dfrac{3}{2},2} \right)$, ${x_2} = \dfrac{{\dfrac{3}{2} + 2}}{2} = \dfrac{{\dfrac{7}{2}}}{2} = \dfrac{7}{2} \times \dfrac{1}{2} = \dfrac{7}{4}$
Midpoint of the point $\left( {2,\dfrac{5}{2}} \right)$, ${x_3} = \dfrac{{2 + \dfrac{5}{2}}}{2} = \dfrac{{\dfrac{9}{2}}}{2} = \dfrac{9}{2} \times \dfrac{1}{2} = \dfrac{9}{4}$
Midpoint of the point $\left( {\dfrac{5}{2},3} \right)$, ${x_4} = \dfrac{{\dfrac{5}{2} + 3}}{2} = \dfrac{{\dfrac{{11}}{2}}}{2} = \dfrac{{11}}{2} \times \dfrac{1}{2} = \dfrac{{11}}{4}$
Substitute the values: $\Delta x = \dfrac{1}{2},{x_1} = \dfrac{5}{4},{x_2} = \dfrac{7}{4},{x_3} = \dfrac{9}{4}$ and ${x_4} = \dfrac{{11}}{4}$.
$= \dfrac{1}{2}\left[ {f\left( {\dfrac{5}{4}} \right) + f\left( {\dfrac{7}{4}} \right) + f\left( {\dfrac{9}{4}} \right) + f\left( {\dfrac{{11}}{4}} \right)} \right]$
Now evaluate the functions of the right endpoints of the subintervals with function $f(x) = {x^3}$,
$f({x_1}) = f\left( {\dfrac{5}{4}} \right)$, now applying $f(x) = {x^3}$, $= {\left( {\dfrac{5}{4}} \right)^3} = \dfrac{{125}}{{64}}$ we will get, $\dfrac{{125}}{{64}} = 1.953$, that is,
$f({x_1}) = f\left( {\dfrac{5}{4}} \right) = \dfrac{{125}}{{64}} = 1.953$
Similarly, $f({x_2}) = f\left( {\dfrac{7}{4}} \right) = \dfrac{{343}}{{64}} = 5.359$
$f({x_3}) = f\left( {\dfrac{9}{4}} \right) = \dfrac{{729}}{{64}} = 11.390$
$f({x_4}) = f\left( {\dfrac{{11}}{4}} \right) = \dfrac{{1331}}{{64}} = 20.796$
Substitute the values in the formula,
$= \dfrac{1}{2}\left[ {1.953 + 5.359 + 11.390 + 20.796} \right]$
Adding the values,
$= \dfrac{1}{2}\left[ {39.498} \right]$
Dividing $39.498$ by 2 we will get,
$= 19.749$
By rounding the decimal we will get,
$= 19.75$
Hence by midpoint rule, $\int\limits_1^3 {{x^3}} dx \approx 19.75$ Sq units.

Note : In the left sum the right endpoint will be excluded and in the right sum the left end point will be excluded. The values we got from right, left and midpoint are not the exact values because the region filled by the small shapes is usually not exactly the same shape as the region being measured. Thus Riemann sum is an approximation method the area we get will be approximately equal to function.