ĐGTD ĐH Bách khoa - Tư duy Toán học - Tích phân (đổi biến số)

$\underset{-1}{\overset{2}{\int }}f\left(2x\right)d\text{x}=2$

$\underset{-3}{\overset{3}{\int }}f\left(x+1\right)d\text{x}=2$

$\underset{-1}{\overset{2}{\int }}f\left(2x\right)d\text{x}=1$

$\underset{0}{\overset{6}{\int }}\frac{1}{2}f\left(x-2\right)d\text{x}=1$

$I=\frac{-1}{2}$

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

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

I = -2

$I=2\underset{8}{\overset{9}{\int }}\sqrt{u}du$

$I=\frac{1}{2}\underset{8}{\overset{9}{\int }}\sqrt{u}du$

$I=\underset{9}{\overset{8}{\int }}\sqrt{u}du$

$I=\underset{8}{\overset{9}{\int }}\sqrt{u}du$

$I=-16\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\cos }^{2}tdt$

$I=8\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(1+\cos 2t\right)dt$

$I=16\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\sin }^{2}tdt$

$I=8\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(1-\cos 2t\right)dt$

$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=0$

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

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

$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\mathrm{a}$

$I=\left(\frac{u^3}{3}+u\right)\left|{}_1^2\right.$

$I=\frac{4}{3}\left(u^3+u\right)\left|{}_1^2\right.$

$I=2\left(\frac{u^3}{3}+u\right)\left|{}_1^2\right.$

$I=\frac{1}{3}\left(\frac{u^3}{3}+u\right)\left|{}_1^2\right.$

$I=\underset{1}{\overset{0}{\int }}\left(1-u\right)du$

$I=\underset{0}{\overset{1}{\int }}\left(1-u\right)e^{-u}du$

$I=\underset{1}{\overset{0}{\int }}\left(1-u\right)e^{-u}du$

$I=\underset{1}{\overset{0}{\int }}\left(1-u\right)e^{2u}du$

$\sqrt{11}-\sqrt{5}$

6

$\sqrt{5}-\sqrt{11}$

3

a = 2

$a=\frac{1}{2}$

$a=\sqrt{2}$

a = 4

$-\frac{2}{3}$

$4\sqrt{3}-1$

$\frac{2\sqrt{3}}{3}$

Một kết quả khác

$I=\frac{2}{3}\underset{1}{\overset{2}{\int }}tdt$

$I=\frac{2}{3}\underset{1}{\overset{2}{\int }}{t}^{\mathit{2}}dt$

$I=\left(\frac{2}{9}{t}^3+2\right)\left|{}_1^2\right.$

$I=\frac{14}{9}$

2a + b = 1

$a^2+b^2=4$

a - b = 1

$ab=\frac{1}{2}$

$f\left(t\right)=\frac{2}{t^2}-\frac{1}{t}$

$f\left(t\right)=-\frac{1}{t^2}+\frac{2}{t}$

$f\left(t\right)=\frac{2}{t^2}+\frac{1}{t}$

$f\left(t\right)=-\frac{2}{t^2}+\frac{1}{t}$

n = 3

n = 4

n = 6

n = 5

$I=\underset{0}{\overset{\frac{\pi }{6}}{\int }}dt$

$I=\underset{0}{\overset{\frac{\pi }{6}}{\int \mathrm{t}}}dt$

$I=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{\mathrm{dt}}{\mathrm{t}}$

$I=\underset{0}{\overset{\frac{\pi }{3}}{\int }}dt$

a = 1

$a=-\frac{1}{3}$

a = 2

a = -2

$I=\frac{1}{2}\underset{0}{\overset{1}{\int }}{e}^t\left(1-t\right)dt$

$I=2\left[\underset{0}{\overset{1}{\int }}{e}^{t}dt+\underset{0}{\overset{1}{\int }}t{e}^{t}dt\right]$

$I=2\underset{0}{\overset{1}{\int }}{e}^t\left(1-t\right)dt$

$I=\frac{1}{2}\underset{0}{\overset{1}{\int }}{e}^t\left(1-t^2\right)dt$

$a=\frac{1}{3}$

$a=-\frac{1}{3}$

$a=-\frac{2}{3}$

$a=\frac{2}{3}$

$I=\frac{4}{3}\underset{1}{\overset{2}{\int }}\left(2u^2+1\right)du$

$I=\frac{2}{3}\underset{1}{\overset{2}{\int }}\left(u^2-1\right)du$

$I=\frac{4}{3}\underset{1}{\overset{2}{\int }}\left(u^2-1\right)du$

$I=\frac{4}{3}\underset{1}{\overset{2}{\int }}\left(2u^2-1\right)du$

$I=\frac{1}{n+1}$

$I=\frac{1}{n-1}$

$I=\frac{1}{2n}$

$I=-\frac{n}{n+1}$

$\frac{1}{2}$

$-\frac{1}{2}$

2

-2

$I=\frac{14}{3}$

$I=\frac{28}{3}$

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

$I=2$

I = 6

I = 4

I = 10

I = 2

$\frac{2}{3}.$

4.

$\frac{3}{2}.$

6.

16160

4040

2020

8080

S = 6

S = 5

S = 7

S = 8

3(1-e)

3e

0

3(e-1)

I = 5

$I=\frac{5}{2}\pi$

$I=5\pi$

$I=10\pi$