1.2 The definite integral  (Page 7/16)

$\underset{n\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^n\left(x_i^{*}\right)\text{Δ}x}$ over $\left[1,3\right]$

$\underset{n\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^n\left(5{\left(x_i^{*}\right)}^2-3{\left(x_i^{*}\right)}^3\right)\text{Δ}x}$ over $\left[0,2\right]$

$\underset{n\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^n{\text{sin}}^2\left(2\pi x_i^{*}\right)\text{Δ}x}$ over $\left[0,1\right]$

$\underset{n\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^n{\text{cos}}^2\left(2\pi x_i^{*}\right)\text{Δ}x}$ over $\left[0,1\right]$

$L_n=\frac{1}{n}{\displaystyle \sum _{i=1}^n\frac{i-1}{n}}$

$R_n=\frac{1}{n}{\displaystyle \sum _{i=1}^n\frac{i}{n}}$

$L_n=\frac{2}{n}{\displaystyle \sum _{i=1}^n\left(1+2\frac{i-1}{n}\right)}$

$R_n=\frac{3}{n}{\displaystyle \sum _{i=1}^n\left(3+3\frac{i}{n}\right)}$

$L_n=\frac{2\pi }{n}{\displaystyle \sum _{i=1}^{n}2\pi \frac{i-1}{n}\text{cos}\left(2\pi \frac{i-1}{n}\right)}$

$R_n=\frac{1}{n}{\displaystyle \sum _{i=1}^n\left(1+\frac{i}{n}\right)\text{log}\left({\left(1+\frac{i}{n}\right)}^2\right)}$

${\displaystyle {\int }_0^3\left(3-x\right)dx}$

${\displaystyle {\int }_2^3\left(3-x\right)dx}$

${\displaystyle {\int }_{\mathrm{-3}}^3\left(3-\left|x\right|\right)dx}$

${\displaystyle {\int }_0^6\left(3-\left|x-3\right|\right)dx}$

${\displaystyle {\int }_{\mathrm{-2}}^2\sqrt{4-x^2}dx}$

${\displaystyle {\int }_1^5\sqrt{4-{\left(x-3\right)}^2}dx}$

${\displaystyle {\int }_0^{12}\sqrt{36-{\left(x-6\right)}^2}dx}$

${\displaystyle {\int }_{\mathrm{-2}}^3\left(3-\left|x\right|\right)dx}$

$\left\{\left(0,0\right),\left(2,1\right),\left(4,3\right),\left(5,0\right),\left(6,0\right),\left(8,3\right)\right\}$ over $\left[0,8\right]$

$\left\{\left(0,2\right),\left(1,0\right),\left(3,5\right),\left(5,5\right),\left(6,2\right),\left(8,0\right)\right\}$ over $\left[0,8\right]$

$\left\{\left(\mathrm{-4},\mathrm{-4}\right),\left(\mathrm{-2},0\right),\left(0,\mathrm{-2}\right),\left(3,3\right),\left(4,3\right)\right\}$ over $\left[\mathrm{-4},4\right]$

$\left\{\left(\mathrm{-4},0\right),\left(\mathrm{-2},2\right),\left(0,0\right),\left(1,2\right),\left(3,2\right),\left(4,0\right)\right\}$ over $\left[\mathrm{-4},4\right]$

${\displaystyle {\int }_0^4\left(f\left(x\right)+g\left(x\right)\right)dx}$

${\displaystyle {\int }_2^4\left(f\left(x\right)+g\left(x\right)\right)dx}$

${\displaystyle {\int }_0^2\left(f\left(x\right)-g\left(x\right)\right)dx}$

${\displaystyle {\int }_2^4\left(f\left(x\right)-g\left(x\right)\right)dx}$

${\displaystyle {\int }_0^2\left(3f\left(x\right)-4g\left(x\right)\right)dx}$

${\displaystyle {\int }_2^4\left(4f\left(x\right)-3g\left(x\right)\right)dx}$

${\displaystyle {\int }_{\text{−}\pi }^{\pi }\frac{\text{sin}t}{1+t^2}dt}$ $\text{(}Hint\text{:}\text{sin}\left(\text{−}t\right)=\text{−}\text{sin}\left(t\right)\text{)}$

${\displaystyle {\int }_{\text{−}\sqrt{\pi }}^{\sqrt{\pi }}\frac{t}{1+\text{cos}t}dt}$

${\displaystyle {\int }_1^3\left(2-x\right)dx}$ ( Hint: Look at the graph of f .)

${\displaystyle {\int }_2^4{\left(x-3\right)}^{3}dx}$ ( Hint: Look at the graph of f .)

${\displaystyle {\int }_0^1\left(1+x+x^2+x^3\right)dx}$

${\displaystyle {\int }_0^1\left(1-x+x^2-x^3\right)dx}$

${\displaystyle {\int }_0^1{\left(1-x\right)}^{2}dx}$

${\displaystyle {\int }_0^1{\left(1-2x\right)}^{3}dx}$

${\displaystyle {\int }_0^1\left(6x-\frac{4}{3}{x}^2\right)dx}$

${\displaystyle {\int }_0^1\left(7-5x^3\right)dx}$

Show that ${\displaystyle {\int }_0^3\left(x^2-6x+9\right)dx}\ge 0.$

Show that ${\displaystyle {\int }_{\mathrm{-2}}^3\left(x-3\right)\left(x+2\right)dx}\le 0.$

Show that ${\displaystyle {\int }_0^1\sqrt{1+x^3}dx}\le {\displaystyle {\int }_0^1\sqrt{1+x^2}dx}.$

Show that ${\displaystyle {\int }_1^2\sqrt{1+x}dx}\le {\displaystyle {\int }_1^2\sqrt{1+x^2}dx}.$

Show that ${\displaystyle {\int }_0^{\pi \text{/}2}\text{sin}tdt}\ge \frac{\pi }{4}.$ $\text{(}Hint\text{:}\text{sin}t\ge \frac{2t}{\pi }$ over $\left[0,\frac{\pi }{2}\right]\text{)}$

Show that ${\displaystyle {\int }_{\text{−}\pi \text{/}4}^{\pi \text{/}4}\text{cos}tdt}\ge \pi \sqrt{2}\text{/}4.$

$f\left(x\right)=x^2,a=\mathrm{-1},b=1$

$f\left(x\right)=x^5,a=\mathrm{-1},b=1$

$f\left(x\right)=\sqrt{4-x^2},a=0,b=2$

$f\left(x\right)=\left(3-\left|x\right|\right),a=\mathrm{-3},b=3$