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3. Midpoint Rule of Riemann Sum example ( Enter your problem )
  1. Formula & Example-1 (table data)
  2. Example-2 (table data)
  3. Example-3 (table data)
  4. Example-4 `(f(x)=1/x)`
  5. Example-5 `(f(x)=1/(x+1))`
  6. Example-6 `(f(x)=x^3-2x+1)`
  7. Example-7 `(f(x)=2x^3-4x+1)`
Other related methods
  1. Left Riemann Sum
  2. Right Riemann Sum
  3. Midpoint Rule of Riemann Sum
  4. Left endpoint approximation
  5. Right endpoint approximation
  6. Trapezoidal rule
  7. Simpson's 1/3 rule
  8. Simpson's 3/8 rule
  9. Boole's rule
  10. Weddle's rule

5. Example-5 `(f(x)=1/(x+1))`
(Previous example)
7. Example-7 `(f(x)=2x^3-4x+1)`
(Next example)

6. Example-6 `(f(x)=x^3-2x+1)`





Find the approximated integral value of an equation x^3-2x+1 using Midpoint Rule
a = 2 and b = 4
Step value (h) = 0.5


Solution:
Equation is `f(x)=x^3-2x+1`

`a=2`

`b=4`

The value of table for `x`

`x`
`x_0=2`
`x_1=2.5`
`x_2=3`
`x_3=3.5`
`x_4=4`


Method-1:
Using Midpoint Rule of Riemann Sum
`int f(x) dx=Delta x xx(f((x_0+x_1)/2)+f((x_1+x_2)/2)+f((x_2+x_3)/2)+...+f((x_(n-1)+x_n)/2))`


`int f(x) dx=Delta x xx(f((x_0+x_1)/2)+f((x_1+x_2)/2)+f((x_2+x_3)/2)+f((x_3+x_4)/2))`

`=Delta x xx(f((2+2.5)/2)+f((2.5+3)/2)+f((3+3.5)/2)+f((3.5+4)/2))`

`=Delta x xx(f(2.25)+f(2.75)+f(3.25)+f(3.75))`

`=0.5xx(7.8906+16.2969+28.8281+46.2344)`

`=0.5xx(99.25)`

`=49.625`

Solution by Midpoint Rule of Riemann Sum is `49.625`




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5. Example-5 `(f(x)=1/(x+1))`
(Previous example)
7. Example-7 `(f(x)=2x^3-4x+1)`
(Next example)





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