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# Riemann Sum Calculator

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Riemann sum is an integral that estimates the area under curve on a graph. If f(x) is the function with its interval [a, b], then the area under the curve S is given by,

$\int_{a}^{b} f(dx)=\lim_{x\rightarrow \infty }\sum_{i=1}^{n}f(x_i)$$\frac{b-a}{n}=\lim_{x\rightarrow \infty }\sum_{i=1}^{n}f(x_i)\Delta x Here, \Delta x = \frac{b-a}{n} = width of the intervals x_i = sample points This online Riemann Sum Calculator finds the area bounded under the curve and plot the graph. You just need to plug in the function with its upper and lower bound i.e., intervals. ## Steps Back to Top Step 1 : Note the given function with its intervals. Step 2 : Determine the width of the interval h from the given intervals. Step 3 : Area under the curve is determined by using the above Reimann sum formula given. ## Problems Back to Top Below given are some problems based on riemann sum . ### Solved Examples Question 1: Determine the Reimann Sum of f(x) = x^{2} + 2 on the interval [0,3] into three sub intervals. Solution: Step 1 : Given, f(x) = x^{2} + 2 a = 0, b = 3 Step 2 : Width of the interval \Delta x, \Delta x = \frac{b-a}{n} = \frac{3-0}{4} = \frac{3}{4} = 0.75 Values for the given function are, f(1) = (1)^{2} + 2 = 3 f(2) = (2)^{2} + 2 = 6 f(3) = (3)^{2} + 2 = 11 Step 3 : Area under curve, S is given by \int_{a}^{b} f(dx)=\lim_{x\rightarrow \infty }\sum_{i=1}^{n}f(x_i)$$\frac{b-a}{n}$

$\int_{0}^{3} f(dx)=\lim_{x\rightarrow \infty }\sum_{i=1}^{4}f(x_i) \times 0.75$

$=[f(1)+f(2)+f(3)] \times 0.75$

$=[3+6+11] \times 0.75$

$= 15$

Question 2: Determine the Reimann Sum of $f(x) = 4x^2$ on the interval [1,4] into three sub intervals.
Solution:

Step 1: Given, $f(x) = 4x^{2}$
a = 1, b = 4

Step 2: Width of the interval $\Delta x$,
$\Delta x$ = $\frac{b-a}{n}$

= $\frac{4-1}{4}$

= $\frac{3}{4}$

= 0.75

Values for the given function are,
f(2) = $4(2)^{2}$ = 16
f(3) = $4(3)^{2}$    = 36
f(4) = $4(4)^{2}$ = 64

Step 3 :
Area under curv, S is given by
$\int_{a}^{b} f(dx)=\lim_{x\rightarrow \infty }\sum_{i=1}^{n}f(x_i)$$\frac{b-a}{n}$

$\int_{1}^{4} f(dx)=\lim_{x\rightarrow \infty }\sum_{i=1}^{4}f(x_i) \times 0.75$

$=[f(2)+f(3)+f(4)] \times 0.75$

$=[16+36+64] \times 0.75$

$= 84$