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One of the numerical approaches for evaluating the definite integral is Simpson's rule. To get the definite integral, we usually employ the fundamental theorem of calculus, which requires us to use anti-derivative integration techniques. However, in other cases, such as in Scientific Experiments, where the function must be calculated from observed data, finding the anti-derivative of an integral is difficult. In such situations, numerical approaches are utilized to approximate the integral. Trapezoidal rule, midpoint rule, and left or right approximation using Riemann sums are some of the other numerical methods used. We will go through Simpson's rule formula, the 1/3 rule, the 3/8 rule, and some examples in this section.
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Key takeaways: Simpson’s Rule, 1/3 rule, 3/8 rule, Simpson’s Rule Error, definite integrals, trapezoidal rule, integral, calculus, midpoint rule
Also read: Isosceles Triangle Theorems
Simpson’s Rule Formula
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Simpson's rule can be used to approximate integrals, according to multiple sources. Quadratic polynomials are used to do this. In this case, parabolic arcs have taken the place of the trapezoidal rule's straight line segments. When it comes to finding approximate polynomials, Simpson's one-third rule can provide certain results. This is possible up to a cubic degree. It's crucial to note that the trapezoidal formula can help you find the area of objects that are shaped beneath a curve. Simpson's formula, on the other hand, is the way to go if one wishes to make those approximations better and more precise. It's also worth noting that Simpson’s rule parabolas are used to discover curve components. This indicates that the approximate area under the curve can be determined using the following formula, according to the Simpson definition:
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Simpson’s Rule Derivation
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Let us derive Simpson's rule, which involves splitting the area under the curve f(x) into parabolas to approximate the value of the definite integral ba f(x) dx. Divide the interval [a, b] into n subintervals [x0, x1], [x1, x2], [x2, x3],..., [x n−2, x n−1], [x n−1, x n] each of width 'h', where xâ‚€ = a and xâ‚TM= b.
Let us now approximate the area under the curve by assuming that every three consecutive points lie on a parabola. By sketching a parabola between the points x0, x1, and x2, we may approximate the area under the curve between x0 and x2. Naturally, all three may not appear on the same parabola. However, let us attempt to draw a parabola through these three locations.
Let's make the y-axis of this parabola symmetric. After that, it looks like this:
Let's pretend that the parabola's equation is y = ax2+ bx + c. The definite integral is then used to approximate the area between x0 and x2:
Area between xâ‚€ and xâ‚‚ ≈ â‚‹â‚•∫ʰ (ax2 + bx + c) dx
= (ax3/3 + bx2/2 + cx) ₋ₕ|ʰ
= (2ah3/3 + 0 + 2ch)
= h/3 (2ah2 + 6c) ... (1)
Let's look at another point from the diagram above.
- f(xâ‚€) = a(-h)2 + b(-h) + c = ah2 - bh + c
- f(xâ‚) = a(0)2 + b(0) + c = c
- f(xâ‚‚) = a(h)2 + b(h) + c = ah2 + bh + c
Now, f(xâ‚€) + 4f(xâ‚) + f(xâ‚‚) = (ah2 - bh + c) + 4c + (ah2 + bh + c) = 2ah2 + 6c.
Substitute this in (1):
Area between xâ‚€ and xâ‚‚ ≈ h/3 (f(xâ‚€) + 4f(xâ‚) + f(xâ‚‚))
Similarly, we can see that:
Area between xâ‚‚ and xâ‚„ ≈ h/3 (f(xâ‚‚) + 4f(x₃) + f(xâ‚„))
Calculating the other areas in a similar way, we get
b∫â‚ f(x) dx
= h/3 (f(xâ‚€) + 4f(xâ‚) + f(xâ‚‚))
+ h/3 (f(x₂) + 4f(x₃) + f(x₄))
+ ...
+ h/3 (f(xn−2n−2) + 4f(xn−1n−1) + f(xn))
≈h3[f(x0)+4f(x1)+2f(x2)+⋯+2f(xn−2)+4f(xn−1)+f(xn)]h3[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.
Simpson’s 1/3 Formula
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The trapezoidal rule is extended by Simpson's 1/3rd rule, in which the integrand is approximated by a second-order polynomial. The Simpson rule can be determined in a variety of ways, including utilizing Newton's divided difference polynomial, Lagrange polynomial, and the coefficients technique. The Simpson's 1/3 rule is defined as follows:
∫ab f(x) dx = h/3 [(y0 + yn) + 4(y1 + y3 + y5 + …. + yn-1) + 2(y2 + y4 + y6 + ….. + yn-2)]
Simpson's One-third Rule is the name of this rule.
Simpson’s 1/3 Rule for Integration
When we divide a tiny interval [a, b] into two halves, we can get a rapid approximation for definite integrals. As a result of splitting the interval, we get:
x0= a, x1= a + b, x2 = b
Hence, the approximation can be written as;
∫ab f(x) dx ≈ S2 = h/3[f(x0) + 4f(x1) + f(x2)]
S2 = h/3 [f(a) + 4 f((a+b)/2) + f(b)]
Where h = (b – a)/2
This is the Simpson’s â…“ rule for integration.
Simpson’s 3/8 Rule
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∫ab f(x) dx = 3h/8 [(y0 + yn) + 3(y1 + y2 + y4 + y5 + …. + yn-1) + 2(y3 + y6 + y9 + ….. + yn-3)]
Because it employs one extra functional value, this rule is more accurate than the normal technique. The composite Simpson's 3/8 rule, which is comparable to the generalized form, also exists for the 3/8 rule. Simpson's second rule of integration is known as the 3/8 rule.
Simpson’s Rule Error
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It's important to note that, while employing Simpson's approach for definite integral calculation yields a more precise approximation, errors still occur. When this occurs, this is specified by n = 2; -(1/ 90) (b - 1 / 2) 5f (4) (ξ)
Here, ξ is some number that exists between a and b.
Things to Remember
- Simpson's rules are numerous approximations for definite integrals in numerical integration, named after Thomas Simpson (1710–1761).
- It is called after Johannes Kepler in German and various other languages, who derived it in 1615 after seeing it used for wine barrels (barrel rule, Keplersche Fassregel). If f is a polynomial of up to a third-degree, the rule's approximation equality becomes precise.
- The composite Simpson's rule is obtained by applying the 1/3 rule to n equal subdivisions of the integration range [a, b]. Weights of 4/3 and 2/3 are alternated for points within the integration range. Simpson's 3/8 rule, sometimes known as Simpson's second rule, necessitates one extra function evaluation within the integration range and results in reduced error limits, but does not improve error order.
- Simpson's rules are substantially less efficient than trapezoidal rules when it comes to estimating the complete area of narrow peak-like functions.
- To put it another way, the composite Simpson's 1/3 rule requires 1.8 times as many points as the trapezoidal rule to obtain the same level of accuracy. Simpson's 3/8 rule for composites is much less precise.
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Sample Questions
Ques. Evaluate ∫01exdx, by Simpson’s â…“ rule. (4 marks)
Ans. Let us divide the range [0, 1] into six equal parts by taking h = 1/6.
If x0 = 0 then y0 = e0 = 1.
If x1 = x0 + h = â…™, then y1 = e1/6 = 1.1813
If x2 = x0 + 2h = 2/6 = 1/3 then, y2 = e1/3 = 1.3956
If x3 = x0 + 3h = 3/6 = ½ then y3 = e1/2= 1.6487
If x4 = x0 + 4h = 4/6 â…” then y4 = e2/3 = 1.9477
If x5 = x0 + 5h = â…š then y5 = e5/6 = 2.3009
If x6 = x0 + 6h = 6/6 = 1 then y6 = e1 = 2.7182
We know by Simpson’s â…“ rule;
∫ab f(x) dx = h/3 [(y0 + yn) + 4(y1 + y3 + y5 + …. + yn-1) + 2(y2 + y4 + y6 + ….. + yn-2)]
Therefore,
∫01exdx = (1/18) [(1 + 2.7182) + 4(1.1813 + 1.6487 + 2.3009) + 2(1.39561 + 1.9477)]
= (1/18)[3.7182 + 20.5236 + 6.68662]
= 1.7182 (approx.)
Ques. Find Out the Integral of the Function f(x) = 2x in the Interval (0, 2). (5 marks)
Ans. It is given that A = 0 and B = 2. Let’s assume that n = 6
Hence, it can be said that
H = b - an = 2 - 0 x 6 = 13
X = 0 = a = 0
X1 = x0 + h = 0 = 13 = 13
X1 + h = 13 + 13 = 23
X2 + h = 23 + 13 = 33 = 1
X3 + h = 33 + 13 = 43
X4 + h = 43 + 13 = 53
X5 + h = 53 + 13 = 63 = 2
X6 + h = 63 + 13 = 1
X7 = b = 1
Y0 = f(0) = 2(0) = 0
Y1 = 2(13) = 23
Y2 = 2(23) = 43
Y3 = 2(33) = 2
Y4 = 2(43) = 83
Y5 = 2(53) = 103
Y6 = 2(63) = 4
Hence, according to the formula,
∫ba f(x) dx = h3 (y0+yn)+4(y1+y3+…+yn−1)+2(y2+y4+…+yn−2)(y0+yn)+4(y1+y3+…+yn−1)+2(y2+y4+…+yn−2)
f(x) dx = 133 (0+4)+4(23+2+103)+2(43+83+4)(0+4)+4(23+2+103)+2(43+83+4)
= 1 / 9 4+24+164+24+16
= 44 / 9
= 4.89
Ques. Evaluate the integral ∫21ex3dx∫12ex3dx using Simpson's rule by taking n = 4. (4 marks)
Ans. ∫21ex3dx∫12ex3dx = ∫baf(x)dx∫abf(x)dx
Comparing both integrals,
[a, b] = [1, 2] and f(x) = ex3ex3
h=b−an=2−14=0.25h=b−an=2−14=0.25
So the 4 subintervals are [1, 1.25], [1.25, 1.5], [1.5, 1.75], and [1.75, 2].
By Simpson's rule formula,
∫21f(x)dx
≈0.25/3[f(1)+4f(1.25)+2f(1.5)+4f(1.75)+f(2)]
=0.25/3(2.71828182845905+28.2027463392796+58.4485675624699+850.36813958881+2980.95798704173)
=326.724643530062
∫21ex3dx≈∫12ex3dx≈ 326.724643530062.
Ques. Evaluate the integral ∫20sin√xdx∫02sinâ¡xdx using Simpson's rule by taking n = 8. (4 marks)
Ans. ∫20sin√xdx∫02sinâ¡xdx = ∫baf(x)dx∫abf(x)dx
Comparing both integrals,
[a, b] = [0, 2] and f(x) = sin√xsinâ¡x
h=b−an=2−08=0.25h=b−an=2−08=0.25
So the 4 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,
∫20f(x)dx
≈0.25/3[f(0)+4f(0.25)+2f(0.5)+...+4f(1.75)+f(2)]
=0.25/3(0+1.91770215441681+1.29927387816012+3.04703992566516+1.68294196961579+3.59696858641514+1.88143866748289+3.87769904361669+0.987765945992735)
=1.52423584761378
∫20sin√xdx≈∫02sinâ¡xdx≈ 1.52423584761378.
Ques. Evaluate the integral ∫021+exdx" id="MathJax-Element-40-Frame" role="presentation" style="box-sizing:inherit; max-width:none; min-width:0px; -webkit-text-stroke-width:0px" tabindex="-1">∫20√1+exdx∫021+exdx using Simpson's rule by taking n = 4. (4 marks)
Ans. ∫20√1+exdx∫021+exdx = ∫baf(x)dx∫abf(x)dx
Comparing both integrals,
[a, b] = [0, 2] and f(x) = √1+ex1+ex
h=b−an=2−04=0.5h=b−an=2−04=0.5
So the 4 subintervals are [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2].
By Simpson's rule formula,
∫20√1+exdx
≈0.5/3[f(0)+4f(0.5)+2f(1)+4f(1.5)+f(2)]
=0.5/3(1.414213562+6.509957014+3.85656937++9.36520288+2.896386731)
=4.0070549278
∫20√1+exdx≈∫021+exdx≈ 4.0070549278.
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