ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Sử dụng phương pháp đổi biến số để tính tích phân

A.\[\mathop \smallint \limits_a^b f'\left( x \right){e^{f\left( x \right)}}dx = 0\]

B.\[\mathop \smallint \limits_a^b f'\left( x \right){e^{f\left( x \right)}}dx = 1\]

C.\[\mathop \smallint \limits_a^b f'\left( x \right){e^{f\left( x \right)}}dx = - 1\]

D. \[\mathop \smallint \limits_a^b f'\left( x \right){e^{f\left( x \right)}}dx = 2\]

A.\[\mathop \smallint \limits_{ - 1}^2 f\left( {2x} \right)d{\rm{x}} = 2\]

B. \[\mathop \smallint \limits_{ - 3}^3 f\left( {x + 1} \right)d{\rm{x}} = 2\]

C. \[\mathop \smallint \limits_{ - 1}^2 f\left( {2x} \right)d{\rm{x}} = 1\]

D. \[\mathop \smallint \limits_0^6 \frac{1}{2}f\left( {x - 2} \right)d{\rm{x}} = 1\]

A.\[\mathop \smallint \limits_{ - a}^a f\left( x \right)dx = 0\]

B. \[\mathop \smallint \limits_{ - a}^a f\left( x \right)dx = 1\]

C. \[\mathop \smallint \limits_{ - a}^a f\left( x \right)dx = - 1\]

D. \[\mathop \smallint \limits_{ - a}^a f\left( x \right)dx = a\]

A.\[I = \frac{{ - 1}}{2}\]

B. \[I = - \frac{1}{4}\]

C. \[I = \frac{1}{4}\]

D. \[I = - 2\]

A.\[I = - \frac{1}{4}{\pi ^4}\]

B. \[I = - {\pi ^4}\]

C. \[I = 0\]

D. \[I = - \frac{1}{4}\]Trả lời:

A.\[I = 2\mathop \smallint \limits_8^9 \sqrt u du\]

B. \[I = \frac{1}{2}\mathop \smallint \limits_8^9 \sqrt u du\]

C. \[I = \mathop \smallint \limits_9^8 \sqrt u du\]

D. \[I = \mathop \smallint \limits_8^9 \sqrt u du\]

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

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

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

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

A.\[a = 2\]

B. \[a = \frac{1}{2}\]

C. \[a = \sqrt 2 \]

D. \[a = 4\]

A.\[ - \frac{2}{3}\]

B. \[4\sqrt 3 - 1\]

C. \[\frac{{2\sqrt 3 }}{3}\]

D. Kết quả khác

A.\[I = \mathop \smallint \limits_1^0 \left( {1 - u} \right)du\]

B. \[I = \mathop \smallint \limits_0^1 \left( {1 - u} \right){e^{ - u}}du\]

C. \[I = \mathop \smallint \limits_1^0 \left( {1 - u} \right){e^{ - u}}du\]

D. \[I = \mathop \smallint \limits_1^0 \left( {1 - u} \right){e^{2u}}du\]

A.\[I = \frac{2}{3}\mathop \smallint \limits_1^2 tdt\]

B. \[I = \frac{2}{3}\mathop \smallint \limits_1^2 {t^2}dt\]

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

D. \[I = \frac{{14}}{9}\]

A.\[2a + b = 1\]

B. \[{a^2} + {b^2} = 4\]

C. \[a - b = 1\]

D. \[ab = \frac{1}{2}\]

A.\[I = - \mathop \smallint \limits_{\sqrt 2 }^{\frac{2}{{\sqrt 3 }}} \frac{{{t^2}}}{{{t^2} - 1}}dt\]

B. \[I = \mathop \smallint \limits_2^3 \frac{{{t^2}}}{{{t^2} + 1}}dt\]

C. \[I = \mathop \smallint \limits_{\sqrt 2 }^{\frac{2}{{\sqrt 3 }}} \frac{{{t^2}}}{{{t^2} - 1}}dt\]

D. \[I = \mathop \smallint \limits_2^3 \frac{t}{{{t^2} + 1}}dt\]

A.\[I = - 16\mathop \smallint \limits_0^{\frac{\pi }{4}} {\cos ^2}tdt\]

B. \[I = 8\mathop \smallint \limits_0^{\frac{\pi }{4}} \left( {1 + \cos 2t} \right)dt\]

C. \[I = 16\mathop \smallint \limits_0^{\frac{\pi }{4}} {\sin ^2}tdt\]

D. \[I = 8\mathop \smallint \limits_0^{\frac{\pi }{4}} \left( {1 - \cos 2t} \right)dt\]

A.\[I = \mathop \smallint \limits_0^{\frac{\pi }{6}} dt\]

B. \[I = \mathop \smallint \limits_0^{\frac{\pi }{6}} tdt\]

C. \[I = \mathop \smallint \limits_0^{\frac{\pi }{6}} \frac{{dt}}{t}\]

D. \[I = \mathop \smallint \limits_0^{\frac{\pi }{3}} dt\]

A.\[a = 1\]

B. \[a = - \frac{1}{3}\]

C. \[a = 2\]

D. \[a = --2\]

A.\[I = \frac{1}{2}\mathop \smallint \limits_0^1 {e^t}\left( {1 - t} \right)dt\]

B. \[I = 2\left[ {\mathop \smallint \limits_0^1 {e^t}dt + \mathop \smallint \limits_0^1 t{e^t}dt} \right]\]

C. \[I = 2\mathop \smallint \limits_0^1 {e^t}\left( {1 - t} \right)dt\]

D. \[I = \frac{1}{2}\mathop \smallint \limits_0^1 {e^t}\left( {1 - {t^2}} \right)dt\]Trả lời:

A.\[S = 6\]

B. \[S = 5\]

C. \[S = 7\]

D. \[S = 8\]

A.\[\frac{{13}}{{9\pi }}\]

B. \[\frac{{14}}{9}\]

C. \[\frac{{14}}{{9\pi }}\]

D. \[\frac{{14\pi }}{9}\]

A.\(\frac{1}{2}\)

B. \( - \frac{1}{2}\)

C. 2

D. -2