Consider a real-valued, bounded function f(x) defined on the closed and bounded interval[a, b], a < b.
The function f(x) need not have the same sign on [a,b] that is, f(x) may have positive as well as negative values on [a, b] .
Partition the interval [a, b] into n subintervals
[x_{0}, x_{1}], [x_{1}, x_{2}].................,[x_{n-2}, x_{n-1}], [x_{n-1}, x_{n}] such that a = x_{0 }< x_{1 }< x_{2 }<.......x_{n-1 }< x_{n} = b
The sum in (1) is called a Riemann sum of f(x) corresponding to the partition
[x_{0}, x_{1}], [x_{1}, x_{2}].................,[x_{n-2}, x_{n-1}], [x_{n-1}, x_{n}]
By choosing Σ_{i} = x_{i-1}, i = 1, 2,...........n, we have
The above equation is known as the left-end rule for evaluating the Riemann integral.
By choosing Σ_{i} = x_{i-1}, i = 1, 2,...........n, we have
The above equation is known as the right-end rule for evaluating the Riemann integral.
The above equation is known as the mid-point rule for evaluating the Riemann integral
Problem 1 :
Find an approximate value of integral 1 to 1.5 x dx by applying the left-end rule with the partition
{1.1, 1.2, 1.3, 1.4, 1.5}.
Solution :
Here a = 1, b = 1.5, number of sub interval (n) = 5, f(x) = x
Width of interval (h) = Δx = 0.1
x_{0} = 1, x_{1} = 1.1, x_{2} = 1.2, x_{3} = 1.3, x_{4} = 1.4
S = [f(x_{0}) + f(x_{1}) + f(x_{2}) + f(x_{3}) + f(x_{4})]Δx
S = [f(1) + f(1.1) + f(1.2) + f(1.3) + f(1.4)](0.1)
= (1 + 1.1 + 1.2 + 1.3 + 1.4) (0.1)
= 6(0.1)
= 0.6
So, integral 1 to 1.5 x dx is approximately 0.6.
Problem 2 :
Find an approximate value of integral 1 to 1.5 x^{2} dx by applying the right end rule with the partition
{1.1, 1.2, 1.3, 1.4, 1.5}.
Solution :
Here a = 1, b = 1.5, number of sub interval (n) = 5
f(x) = x^{2}
Width of interval (h) = Δx = 0.1
x_{1} = 1.1, x_{2} = 1.2, x_{3} = 1.3, x_{4} = 1.4, x_{5} = 1.5
S = [f(x_{1}) + f(x_{2}) + f(x_{3}) + f(x_{4}) + f(x_{5})]Δx
S = [f(1.1) + f(1.2) + f(1.3) + f(1.4) + f(1.5)](0.1)
= [(1.1)^{2} + (1.2)^{2} + (1.3)^{2} + (1.4)^{2} + (1.5)^{2}](0.1)
= [1.21 + 1.44 + 1.69 + 1.96 + 2.25](0.1)
= 8.55(0.1)
= 0.855
Problem 3 :
Find an approximate value of integral 1 to 1.5 (2-x) dx by applying the mid point rule with the partition
{1.1, 1.2, 1.3, 1.4, 1.5}
Solution :
Here a = 1, b = 1.5, number of sub interval (n) = 5
f(x) = (2-x)
Width of interval (h) = Δx = 0.1
x_{0} = 1, x_{1} = 1.1, x_{2} = 1.2, x_{3} = 1.3, x_{4} = 1.4, x_{5} = 1.5
S = [f(x_{0 }+ x_{1})/2 + f(x_{1 }+ x_{2})/2 + f(x_{2 }+ x_{3})/2 + f(x_{3 }+ x_{4})/2 + f(x_{4 }+ x_{5})/2]Δx
S = [f(1+1.1)/2 + f(1.1+1.2)/2 + f(1.2+1.3)/2 + f(1.3+1.4)/2 + f(1.4+1.5)/2]Δx
S = [f(1.05) + f(1.15) + f(1.25) + f(1.35) + f(1.45)](0.1)
= [(2-1.05) + (2-1.15) + (2-1.25) + (2-1.35) + (2-1.45)](0.1)
= (0.95+0.85+0.75+0.65+0.55)(0.1)
= 3.75(0.1)
= 0.375
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