# Calculate the integral ∫_0^2(-x^2+x ) dx using Riemann Sum and a regular partition with sub- intervals

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### 1 Answer

You should evaluate the definite integral using Riemann's sum such that:

`R_n = sigma_(i=1)^n*f(x_i)*(2 - 0)/n`

Notice that the product f(x_i)*(2 - 0)/n represents the area of the i-the rectangle from all n rectangles that approximate the region under the curve `y = -x^2 + x` , having the length `f(x_i)` and the width `(2 - 0)/n` .

You may consider n as large as you wish, hence, considering n = 4 yields:

`R_n = sigma_(i=1)^4*f(x_i)*2/4 => R_n = sigma_(i=1)^4*f(x_i)*1/2`

You may find each `x_i` , giving values to i from 1 to 4 such that:

`x_1 = 0+1/2 = 1/2`

`x_2 = 1/2 + 1/2 = ` 1

`x_3 = 1 + 1/2 = 3/2`

`x_4 = 3/2 + 1/2 = 4/2=2`

Since the fourth x coordinate yields the right endpoint of the interval [0,2] means that the width is evaluated correctly.

You need to evaluate `f(x_i)` such that:

`f(x_1) = (-x_1)^2 + x_1 = 1/4 + 1/2 = 3/4`

`f(x_2) = (-x_2)^2 + x_2 = 1 + 1 = 2`

`f(x_3) = (x_3)^2 + x_3 = 9/4 + 3/2 = 15/4`

`f(x_4) = (-x_4)^2 + x_4 = 4 + 2 = 6`

`R_n = 1/2(3/4 +2 + 15/4 + 6) => R_n = 50/4`

**Hence, evaluating the definite integral using Riemann's sum yields `R_n = int_0^2((-x)^2+x)dx = 50/4` .**