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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

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2. Example-2 (table data)





Find the approximated integral value using Right Riemann Sum
xf(x)
0.01.0000
0.10.9975
0.20.9900
0.30.9776
0.40.8604


Solution:
The value of table for `x` and `f(x)`

`x``f(x)`
`x_0=0``f(x_(0))=1`
`x_1=0.1``f(x_(1))=0.9975`
`x_2=0.2``f(x_(2))=0.99`
`x_3=0.3``f(x_(3))=0.9776`
`x_4=0.4``f(x_(4))=0.8604`


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)))`

`=0.1xx(0.9975+0.99+0.9776+0.8604)`

`=0.1xx(3.8255)`

`=0.3826`

Solution by Right Riemann Sum is `0.3826`




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