Simpson’s Rule Formula: Definition, Derivation, Steps & Solved Examples

Jasmine Grover logo

Jasmine Grover

Education Journalist | Study Abroad Lead

Simpson's Rule is a formula that has been used to calculate the estimated value of a definite integral. A defined integral is defined by its lower and upper limits. To evaluate a definite integral, we usually integrate it (using integration techniques) and then apply the fundamental theorem of Calculus to apply the limits. However, there are situations when no integration strategy can be used to produce an integral, and there are times when we do not even have a function to integrate; instead, we have some observed values of the function (in the case of experiments). In such cases, Simpson's Rule can be used to estimate the value of the definite integral.

Key Terms: Simpson’s Rule, Parabolas, Eclipses, Area, Riemann's Left Sum, Riemann’s Right Sum, Integral, Integration, Area, Trapezoidal Rule


What is Simpson’s Rule?

[Click Here for Sample Questions]

Simpson's Rule is used to approximate the area under the graph of the function f in order to determine the value of a definite integral. Simpson's method calculates the area under a curve by splitting it into parabolas, whereas Riemann Sum calculates the area under a curve by dividing it into rectangles (a definite integral).

Also Read:


Simpson's Rule Formula

[Click Here for Sample Questions]

There are several numerical ways for estimating an integral, such as Riemann's left sum, Riemann's right sum, midpoint rule, trapezoidal rule, Simpson's rule, and so on. Simpson's method, on the other hand, gives a more exact approximation of a definite integral. If f(x) = y is distributed equally amongst [a,b], Simpson's rule formula is

Simpson's Rule Formula

In this formula, 

  • Here, n is an even number which is the number of subintervals that the interval [a, b] should be divided into. (n is usually mentioned in the problem)
  • x0 = a and xn = b
  • h = (b−a) / n
  • x0, x1, ...., xn are the ends of the n subintervals.

Read More: List of Integral Formulas


Simpson's Rule Error Bound

[Click Here for Sample Questions]

Simpson's approach produces merely an approximation of the integral value, not the exact number. As a result, there is always a mistake that may be found using the approach outlined below.

Error bound in Simpson's rule is

 \(\frac{M(b-a)^5}{180n^4}\)

where |f(4)(x)| ≤M

Simpson's Rule Formula

Simpson’s Rule Formula

Read More: Definite Integral Formula


Simpson's Rule Derivation

[Click Here for Sample Questions]

By splitting the area under the curve f(x) into parabolas, we may approximate the value of the definite integral ba f(x) dx using Simpson's method. Divide the interval [a, b] into n subintervals [x0, x1], [x1, x2], [x2, x3], ..., [xn-2, xn-1], [xn-1, xn] of width 'h' each, with x0 equalling a and xn equalling b.

Simpsons rule derivation or proof. For this, the curve is divided into multiple intervals each of width h.

To calculate the area under the curve, assume the next three points are on a parabola. Create a parabola between x0, x1, and x2 to determine the area under the curve between x0 and x2. Of course, none of the three are on the same parabola. Now, plot a graph with approximate values of above given three points.

Simpsons rule derivation includes approximating area using parabolas.

Now, let us make this parabola symmetric about the y-axis. Then it becomes something like this

As a part of Simpson's rule derivation the parabola made in the previous step is made to be symmetric about the y axis.

Let us assume that the equation of the parabola be y = ax2 + bx + c. Then the area between x0 and x2 is approximated by the definite integral:

Area between x0 and x2

area

.......(1) 

Let us have another observation from the above figure.

f(x0) = a(-h)2 + b(-h) + c = ah2 - bh + c

f(x1) = a(0)2 + b(0) + c = c

f(x2) = a(h)2 + b(h) + c = ah2 + bh + c

f(x0) + 4f(x1) + f(x2) = (ah2 - bh + c) + 4c + (ah2 + bh + c) = 2ah2 + 6c.

Substitute this in (1):

Area between x0 and x2 ≈ h/3 (f(x0) + 4f(x1) + f(x2))

Similarly, we can see that:

Area between x2 and x4 ≈ h/3 (f(x2) + 4f(x3) + f(x4))

Calculating the other areas in a similar way, we get

\(\int_a^b f(x)dx\)

= h/3 (f(x0) + 4f(x1) + f(x2))

+ h/3 (f(x2) + 4f(x3) + f(x4))

+ ...

+ h/3 (f(xn-2) + 4f(xn-1) + f(xn))

≈ (h / 3) [f(x0) + 4f(x1) + 2f(x2)

 + ... + 2f(xn-2) + 4f(xn-1) + f(xn)]

The like terms are combined here.

Hence, we have derived Simpson's rule formula.

Read More: Differentiation and Integration Formula


How to Apply Simpson's Rule?

[Click Here for Sample Questions]

The Trapezoidal rule and Simpson's rule both approximate the same regions, however Simpson's rule is more precise. The following are the ways for determining the intergral ba f(x) dx using Simpson's rule.

  • Step 1: From the interval [a, b], get the values of a and b, as well as the value of 'n,' which reflects the number of subintervals.
  • Step 2: Using the formula h = (b - a)/n, calculate the width of each subinterval.
  • Step 3: Divide the interval [a, b] into 'n' subintervals using the interval width 'h.'
  • Step 4: Substitute all of these values into Simpson's rule formula and simplify.

\(\int_a^b f(x) dx\)

≈ (h / 3) [f(x0) + 4f(x1) + 2f(x2)

 + ... + 2f(xn-2) + 4f(xn-1) + f(xn)]

Read More: Relations and Functions


Things to Remember

  • Simpson's Rule is a numerical method for approximating the value of a definite integral using quadratic functions.
  • When using Simpson's rule, we always divide the interval into an even number of subintervals. This means that 'n' must always be an even integer.
  • Simpson's Rule is based on the premise that if you have three points, you can find the a quadratic equation through those points. All subintervals must have the same width.
  • Simpson's principles are used by surveyors to calculate the capacity of rescue boats and the quantity of gunk in a ship's storage tanks.
  • If a function is particularly oscillatory or has missing derivatives in some areas, the above rule may fail to produce proper conclusions.
  • By Simpson's rule: ba f(x) dx ≈ (h / 3) [f(x0) + 4f(x1) + 2f(x2) + ... + 2f(xn-2) + 4f(xn-1) + f(xn)]

Also Read:


Sample Questions

Ques. Solve the integral \(\int_0^2 sin \sqrt{x} dx\) using Simpson's rule by taking n = 8. (5 Marks)

Ans. \(\int_0^2 sin \sqrt{x} dx = \int_a^b f(x) dx\)

Comparing both integrals,

[a, b] = [0, 2] 

 f(x) = sin√x

h = (b – a) / n

= (2 – 0) / 8

=0.25

The sub-intervals are [0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1], [1, 1.25], [1.25, 1.5], [1.5, 1.75], and [1.75, 2].

By Simpson's rule formula,

\(\int_0^2 f(x) dx\)

≈ (0.25 / 3) x [f(0) + 4f(0.25) + 2f(0.5) + ... + 4f(1.75) + f(2)]

= (0.25 / 3) x (0+1.91770215441681 + 1.29927387816012 + 3.04703992566516 + 1.68294196961579 + 3.59696858641514 + 1.88143866748289 +3.87769904361669 +0.987765945992735)

= 1.52423584761378

Ques. Solve the integral \(\int_0^2 \sqrt{1+e^x} dx\) using Simpson's rule by taking n = 4. (5 Marks)

Ans. \(\int_0^2 \sqrt{1+e^x} dx = \int_a^b f(x) dx\)

When the two integrals are compared,

[a, b] = [0, 2]

 f(x) = √1+ex

h = (b – a) / n

= (2− 0) / 4

=0.5

The subintervals calculated are [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2].

By Simpson's rule formula,

\(\int_0^2 \sqrt{1+e^x} dx\)

= (0.5 / 3) x [f(0) + 4 f(0.5) + 2 f(1) + 4 f(1.5) + f(2)]

= (0.5 / 3) x (1.414213562 + 6.509957014 + 3.85656937 + 9.36520288 + 2.896386731)

= 4.0070549278

Ques. Solve the integral \(\int_1^2 e^{x^3} dx\) using Simpson's rule by taking n = 4. (5 Marks)

Ans. \(\int_1^2 e^{x^3} dx = \int_a^b f(x) dx\)

When the two integrals are compared,

[a, b] = [0, 2]

\(f(x) = e^{x^3}\)

h = (b – a) / n

= (2− 1) / 4

=0.25

The subintervals calculated are [1, 1.25], [1.25, 1.5], [1.5, 1.75], and [1.75, 2].

By Simpson's rule formula,

\(\int_1^2 f(x) dx\)

= (0.25 / 3) x [f(1) + 4 f(1.25) + 2 f(1.5) + 4 f(1.75) + f(2)]

= (0.25 / 3) x (2.71828182845905 + 28.2027463392796 + 58.4485675624699 + 850.36813958881 + 2980.95798704173)

= 326.724643530062

Ques. Solve the integral \(\int_0^8 \sqrt{x} dx\) using Simpson's rule by taking n = 4. (5 Marks)

Ans. The width of subinterval is 

The width of subinterval is 

With endpoints xi have coordinates

Xi = {0,2,4,6,8}

The function values at the points xi are

F(x0) = f(0) = √0 = 0

F(x1) = f(2) = √2

F(x2) = f(4) = √4 = 2

F(x3) = f(6) = √6

F(x4) = f(8) = √8 = 2√2

Substituting these values into the Simpson’s rule, we get

Substituting these values into the Simpson’s rule, we get

Substituting these values into the Simpson’s rule, we get

Substituting these values into the Simpson’s rule, we get

The solution of the integral is

The solution of the integral is

The solution of the integral is

The error approximation is

The error approximation is

Ques. Solve the integral \(\int_1^2 \frac{dx}{x}\) using Simpson's rule by taking n = 2. (5 Marks)

Ans. For n=2, the Simpson’s rule is

For n=2, the Simpson’s rule is

The width of the intervals is

The width of the intervals is

The function values at the points xi are

F(x0) = f(1) = 1 / 1 = 1

F(x1) = f(3 / 2) = 2 / 3

F(x2) = f(2) = 1 / 2

Then,

The width of the intervals is

The width of the intervals is

The width of the intervals is

The width of the intervals is

Ques. For the function f(x), a value table is supplied. Calculate the area under the curve y = f(x) between x = -4 and x = 8 using Simpson's Rule and n = 6 subintervals. (3 Marks)
For the function f(x), a value table is supplied. Calculate the area under the curve y = f(x) between x = -4 and x = 8 using Simpson's Rule and n = 6 subintervals.

Ans. For n=6, the Simpson’s rule is

For n=6, the Simpson’s rule is

The width of the intervals is

The width of the intervals is

The approximate value of the area falling under the curve is

The approximate value of the area falling under the curve is

The approximate value of the area falling under the curve is

The approximate value of the area falling under the curve is

Ques. Using Simpson's Rule and n = 4 subintervals, find the area underneath the curve y = 3x from x = -2 and x = 2. (5 Marks)
 Using Simpson's Rule and n = 4 subintervals, find the area underneath the curve y = 3x from x = -2 and x = 2

Ans. For n=6, the Simpson’s rule is

32

The function values at the points xi are

F(x0) = f(-2) = 3-2 = 1 / 9

F(x1) = f(-1) =3-1 = 1 / 3

F(x2) = f(0) = 30 = 1

F(x3) = f(1) = 31 = 3

F(x4) = f(2) = 32 = 9

As Δx = 1, we get

As Δx = 1, we get

As Δx = 1, we get

As Δx = 1, we get

Ques. For the function f(x), a value table is supplied. Calculate the area under the curve y = f(x) between x = 0 and x = 4 using Simpson's Rule and n = 4 subintervals. (3 Marks)
For the function f(x), a value table is supplied. Calculate the area under the curve y = f(x) between x = 0 and x = 4 using Simpson's Rule and n = 4 subintervals.

Ans. For n = 4, the Simpson’s rule is

For n = 4, the Simpson’s rule is

The width of the intervals is

The width of the intervals is

The approximate value of the area falling under the curve is

The approximate value of the area falling under the curve is

The approximate value of the area falling under the curve is

The approximate value of the area falling under the curve is

Ques. Using Simpson's Rule and n = 6 subintervals, find the area underneath the curve y = f(x) from x = -1 and x = 5. (3 Marks)
Using Simpson's Rule and n = 6 subintervals, find the area underneath the curve y = f(x) from x = -1 and x = 5

Ans. For n = 4, the Simpson’s rule is

For n = 4, the Simpson’s rule is

The function values at the points xi are

F(x0) = f(-1) = 4

F(x1) = f(0) = 3

F(x2) = f(1) = 2

F(x3) = f(2) = 3

F(x4) = f(3) = 6

F(x5) = f(4) = 6

F(x6) = f(5) = 4

As Δx = 1, we get

A = S6 ≈ \(\frac{1}{3} [ 4 + 4.3 + 2.2 + 4.3 + 2.6 + 4.6 +4]\)

\(= \frac{1}{3} [4 + 12 + 4 + 12 + 12 + 24 + 4]\)

\(=\frac{1}{3} . 72 = 24\)

Ques. Evaluate \(\int_0^1 e^x dx\), by Simpson’s ? rule. (5 Marks)

Ans. Divide the range [0, 1] into 6 parts by assuming h = 1/6.

If x0 = 0, then, y0 = e0 = 1.

If x1 = x0 + h = 1/6, then,y1 = e1/6 = 1.18

If x2 = x0 + 2h = 2/6 = 1/3, then, y2 = e1/3 = 1.39

If x3 = x0 + 3h = 3/6 = 1/2, then, y3 = e1/2= 1.64

If x4 = x0 + 4h = 4/6 = 2/3, then, y4 = e2/3 = 1.94

If x5 = x0 + 5h = 5/6 , then, y5 = e5/6 = 2.30

If x6 = x0 + 6h = 6/6 = 1, then, y6 = e1 = 2.71

According to Simpson’s rule;

According to Simpson’s rule;

Then,

\(\int_0^1 e^x dx\) = (1/18) [(1 + 2.71) + 4(1.18 + 1.64 + 2.30) + 2(1.39 + 1.94)]

= (1/18) [3.71 + 20.52 + 6.68]

= 1.71

Ques. Find the answer using Simpson’s rule (3 Marks)
Find the answer using Simpson’s rule

Ans. Using Simpson’s Rule

\(\int y dx\) = (h / 3) x [(y0 + y4) + 4(y1 + y3) + 2(y2)]

 = (0.1 / 3) x [(1+0.8604) + 4x(0.9975 + 0.9776) + 2x(0.99)]

= (0.1 / 3) x [(1+0.8604) + 4x(1.9751) + 2x(0.99)]

= 0.39136

Ques. Find the answer of 1/x where x1 = 1 and x2 = 2, h = 0.25, using Simpson’s rule. (3 Marks)

Ans. F(x) = 1 / x

x 1 1.25 1.5 1.75 2
y 1 0.8 0.6667 0.5714 0.5

Using Simpson’s rule

\(\int y dx\) = (h / 3) x [(y0 + y4) + 4(y1 + y3) + 2(y2)]

 = (0.25 / 3) x [(1+0.5) + 4x(0.8 + 0.5716) + 2x(0.6667)]

= (0.25 / 3) x [(1+0.5) + 4x(1.3714) + 2x(0.6667)]

= 0.6933

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates


Check-Out: 

CBSE CLASS XII Related Questions

  • 1.

    At a birthday party, children are being served orange juice in conical cups, as shown in the figure. 


    Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
    On the basis of the above information, answer the following questions :


      • 2.
        Find:

        The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

          • \(-\frac{\pi}{2}\)
          • \(-\frac{\pi}{4}\)
          • \(\frac{\pi}{4}\)
          • \(\frac{\pi}{2}\)

        • 3.
          Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]


            • 4.
              Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).


                • 5.
                  Find:

                  If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

                    • \(0\)
                    • \(-2\)
                    • \(-1\)
                    • \(2\)

                  • 6.
                    Find:

                    The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

                      CBSE CLASS XII Previous Year Papers

                      Comments


                      No Comments To Show