Riemann Sum

Tags

Calculus

Cegep/2

Word count

421 words

Reading time

3 minutes

Approximation of a region's area by summing rectangles

The Riemann sum of $y=f(x)$ on $[a,b]$ obtained using $n$ rectangles and right endpoints:

$$R_n=\sum _{i=1}^n{A}_i=\sum _{i=1}^{n}f(a+i\frac{b-a}{n})\frac{b-a}{n}$$

Evaluation

Evaluate the area under $f(x)=5x+2$ on $[0,3]$.

$$\begin{array}{rl}A& =\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}f(x_i)\mathrm{\Delta }x\\ & =\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^n(5\left(\frac{3i}{n}\right)+2)\left(\frac{3}{n}\right)\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{3}{n}(\frac{15}{n}\cdot \frac{n(n+1)}{2}+2n)\\ & =\underset{n\to \mathrm{\infty }}{lim}(\frac{45(n+1)}{2n}+6)\\ & =\frac{45}{2}+6\\ & =\frac{57}{2}\end{array}$$

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Evaluate the area under $f(x)=x^2-2x-5$ on $[1,4]$.

$$\begin{array}{rl}A& =\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}f(x_i)\mathrm{\Delta }x\\ & =\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^n({(1+\frac{(4-1)i}{n})}^2-2(1+\frac{(4-1)i}{n})-5)\left(\frac{4-1}{n}\right)\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{3}{n}\sum _{i=1}^n(1+\frac{6i}{n}+\frac{9i^2}{n^2}-2-\frac{6i}{n}-5)\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{3}{n}\sum _{i=1}^n(\frac{9i^2}{n^2}-6)\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{3}{n}(\frac{9}{n^2}\cdot \frac{n(n+1)(2n+1)}{6}-6n)\\ & =\underset{n\to \mathrm{\infty }}{lim}(\frac{9(n+1)(2n+1)}{2n^2}-18)\\ & =\frac{9\cdot 2}{2}-18\\ & =-9\end{array}$$

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Use a limit of Riemann sums to compute ${\int }_{-3}^1(5-x^2)\mathrm{d}x$.

$$\begin{array}{rl}{\int }_{-3}^1(5-x^2)\mathrm{d}x& =\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^n(5-{(-3+\frac{1+3}{n}i)}^2)\left(\frac{1+3}{n}\right)\\ & =\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^n\frac{-16+\frac{96}{n}i-\frac{64}{n^2}{i}^2}{n}\\ & =\underset{n\to \mathrm{\infty }}{lim}(-16+\frac{96}{n^2}\sum i-\frac{64}{n^3}\sum i^2)\\ & =\underset{n\to \mathrm{\infty }}{lim}(-16+\frac{96}{n^2}\frac{n(n+1)}{2}-\frac{64}{n^3}\frac{n(n+1)(2n+1)}{6})\\ & =-16+\frac{96}{2}-\frac{64\cdot 2}{6}\\ & =\frac{32}{3}\end{array}$$

Contributors

Ajitani Hifumi

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