ĐGTD ĐH Bách khoa - Tư duy Toán học - Ứng dụng tích phân vào tính diện tích

$S={\int }_a^{b}f\left(x\right)dx$

$S={\int }_0^{b}f\left(x\right)dx$

$S={\int }_b^a\left|f\left(x\right)\right|dx$

$S={\int }_a^b\left|f\left(x\right)\right|dx$

$S={\int }_a^b\left(f\left(x\right)-g\left(x\right)\right)dx$

$S={\int }_a^b\left(g\left(x\right)-f\left(x\right)\right)dx$

$S={\int }_a^b\left|f\left(x\right)-g\left(x\right)\right|dx$

$S={\int }_a^b\left|f\left(x\right)\right|dx-{\int }_a^b\left|g\left(x\right)\right|dx$

$S=\left|{\int }_{-1}^1\left(3x-x^3\right)dx\right|$

$S={\int }_{-1}^0\left(3x-x^3\right)dx+{\int }_0^1\left(x^3-3x\right)dx$

$S={\int }_{-1}^1\left(3x-x^3\right)dx$

$S={\int }_{-1}^0\left(x^3-3x\right)dx+{\int }_0^1\left(3x-x^3\right)dx$

$S={\int }_0^1{x}^{3}dx+{\int }_1^2\left(x-2\right)d\text{x}$

$S=\left|{\int }_0^2\left(x^3+x-2\right)dx\right|$

$S=\frac{1}{2}+{\int }_0^1{x}^{3}d\text{x}$

$S={\int }_0^1\left|x^3+x-2\right|d\text{x}$

$S={\int }_{-3}^{-1}\left|x^2-1\right|dx$

$S={\int }_{-1}^{-3}\left|x^2-1\right|dx$

$S={\int }_{-3}^0\left|x^2-1\right|dx$

$S={\int }_{-3}^{-1}\left(1-x^2\right)dx$

$S=\underset{0}{\overset{e}{\int }}\left|e^x+x\right|dx$

$S=\underset{0}{\overset{e}{\int }}\left|e^x-x\right|dx$

$S=\underset{e}{\overset{0}{\int }}\left|e^x-x\right|dx$

$S=\underset{e}{\overset{0}{\int }}\left|e^x+x\right|dx$

S=2+e

S=2-e

S=e-2

S=e-1

$S={\int }_{-2}^{2}f\left(x\right)dx$

$S={\int }_1^{-2}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx$

$S={\int }_{-2}^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx$

$S={\int }_{-2}^{1}f\left(x\right)dx-{\int }_1^{2}f\left(x\right)dx$

$\frac{5}{2}$

$\frac{35}{6}$

$\frac{-5}{2}$

$\frac{-35}{6}$

$S=\underset{-1}{\overset{0}{\int }}f\left(x\right)dx+\underset{0}{\overset{1}{\int }}\left|f\right(x\left)\right|dx$

$\underset{-1}{\overset{1}{\int }}\left|f\left(x\right)\right|dx$

${\int }_{-1}^{1}f\left(x\right)dx$

$S=\left|\underset{-1}{\overset{1}{\int }}f\left(x\right)dx\right|$

S=b-a

S=b+a

S=-b+a

S=-b-a

$S=\frac{16}{3}$

$S=\frac{161}{6}$

$S=\frac{1}{6}$

$S=\frac{5}{6}$

$\pi +\frac{4}{3}$

$\frac{\pi }{2}-1$

$\frac{\pi }{2}$

$\frac{\pi }{2}+\frac{1}{3}$

0

16

4

8

$\frac{253}{12}$

$\frac{253}{24}$

$-\frac{125}{24}$

$-\frac{125}{12}$

$\frac{1}{2}$

$\frac{3}{5}$

$\frac{2}{5}$

$\frac{1}{3}$

$S=\frac{8}{3}$

$S=\frac{4}{3}$

S = 4

$S=\frac{16}{9}$

3

-4

6

-2

$\frac{9}{2}$

$\frac{18}{5}$

4

5

2ln3

ln3

ln18

2ln2

$S=\frac{8}{3}$

S = 1

$S=\frac{4}{3}$

S = 2

$OM=2\sqrt{5}$

$OM=3\sqrt{10}$

$\mathrm{OM}=15$

$\mathrm{OM}=10$

$\frac{128}{3}{m}^2$

$\frac{131}{3}{m}^2$

$\frac{28}{3}{m}^2$

$\frac{26}{3}m$

$506\left(c{m}^2\right)$

$747\left(c{m}^2\right)$

$507\left(c{m}^2\right)$

$746\left(c{m}^2\right)$

1.954.000 đồng

2.123.000 đồng

1.946.000 đồng

2.145.000 đồng

30kg

40kg

50kg

45kg