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2. Right 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 Right Riemann Sum
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` and `f(x)`

`x``f(x)`
`x_0=2``f(x_(0))=f(2)=5`
`x_1=2.5``f(x_(1))=f(2.5)=11.625`
`x_2=3``f(x_(2))=f(3)=22`
`x_3=3.5``f(x_(3))=f(3.5)=36.875`
`x_4=4``f(x_(4))=f(4)=57`


Method-1:
Using Right Riemann Sum
`int f(x) dx=Delta x xx(f(x_(1))+f(x_(2))+...+f(x_(n)))`


`int f(x) dx=Delta x xx(f(x_(1))+f(x_(2))+f(x_(3))+f(x_(4)))`

`=0.5xx(11.625+22+36.875+57)`

`=0.5xx(127.5)`

`=63.75`

Solution by Right Riemann Sum is `63.75`




This material is intended as a summary. Use your textbook for detail explanation.
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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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