20 câu trắc nghiệm Toán 12 Cánh diều Bài 3. Tích phân (Đúng sai - Trả lời ngắn) có đáp án

Chọn B

$\int\limits_1^2 {f\left( x \right)dx} = \left. {F\left( x \right)} \right|_1^2 = F\left( 2 \right) - F\left( 1 \right) = 4 - \left( { - 2} \right) = 6$.

Chọn B

$\int\limits_0^4 {f\left( x \right)dx} = \int\limits_0^2 {f\left( x \right)dx} + \int\limits_2^4 {f\left( x \right)dx}$$\Rightarrow \int\limits_0^2 {f\left( x \right)dx} = \int\limits_0^4 {f\left( x \right)dx} - \int\limits_2^4 {f\left( x \right)dx} = 5 - \left( { - 1} \right) = 6$.

Chọn C

$\int\limits_1^4 {\left[ {\frac{1}{3}f\left( x \right) - 5} \right]dx}$$= \frac{1}{3}\int\limits_1^4 {f\left( x \right)dx} - \left. {5x} \right|_1^4$$= \frac{1}{3}.3 - 15 = - 14$.

Chọn D

$I = \int\limits_0^{2024} {{3^x}dx} = \left. {\frac{{{3^x}}}{{\ln 3}}} \right|_0^{2024} = \frac{{{3^{2024}} - 1}}{{\ln 3}}$.

Chọn B

$\int\limits_{ - 1}^{16} {f\left( x \right)dx} = \int\limits_{ - 1}^1 {f\left( x \right)dx} + \int\limits_1^{16} {f\left( x \right)dx}$$= \int\limits_{ - 1}^1 {\left( {{x^2} + x + 1} \right)dx} + \int\limits_1^{16} {\left( {\frac{1}{{\sqrt[4]{{{x^3}}}}} + 2} \right)dx}$

$= \left. {\left( {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + x} \right)} \right|_{ - 1}^1 + \left. {\left( {4\sqrt[4]{x} + 2x} \right)} \right|_1^{16}$$= \frac{8}{3} + 34 = \frac{{110}}{3}$.

Suy ra $a = 110;b = 3$. Do đó $a + b = 113$.

Chọn C

$I = \int\limits_0^2 {f\left( {3x} \right)dx}$$= \frac{1}{3}\int\limits_0^2 {f\left( {3x} \right)d\left( {3x} \right)}$$= \frac{1}{3}\int\limits_0^6 {f\left( u \right)du} =$$= \frac{1}{3}.12 = 4$.

Chọn C

$\int\limits_a^a {f\left( x \right)dx} = \left. {F\left( x \right)} \right|_a^a = F\left( a \right) - F\left( a \right) = 0$.

Chọn A

$\int\limits_0^{\frac{\pi }{2}} {\left( {4 - \sin x} \right)dx} = \left. {4x + \cos x} \right|_0^{\frac{\pi }{2}} = 2\pi - 1$.

Suy ra $a = 2;b = - 1$. Do đó $a + b = 1$.

Chọn A

$\int\limits_0^8 {\left| {{x^2} - 6x} \right|dx}$$= \int\limits_0^6 {\left| {{x^2} - 6x} \right|dx} + \int\limits_6^8 {\left| {{x^2} - 6x} \right|dx}$$= \int\limits_0^6 {\left( { - {x^2} + 6x} \right)dx} + \int\limits_6^8 {\left( {{x^2} - 6x} \right)dx}$

$= \left. {\left( {\frac{{ - {x^3}}}{3} + \frac{{6{x^2}}}{2}} \right)} \right|_0^6 + \left. {\left( {\frac{{{x^3}}}{3} - \frac{{6{x^2}}}{2}} \right)} \right|_6^8 = \frac{{152}}{3}$.

Chọn C

Thời điểm vật dừng lại là $v\left( t \right) = 0 \Leftrightarrow 27 - 9\sqrt t = 0 \Leftrightarrow t = 9$ giây.

Quãng đường mà ô tô di chuyển được từ thời điểm $t = 0$ đến thời điểm mà vật dừng lại là:

$s = \int\limits_0^9 {\left( {27 - 9\sqrt t } \right)dt} = \left. {\left( {27t - 6t\sqrt t } \right)} \right|_0^9 = 81$ m.

a) $\int\limits_{ - 1}^1 {f\left( x \right)dx} = \int\limits_{ - 1}^0 {f\left( x \right)dx} + \int\limits_0^1 {f\left( x \right)dx}$$= \int\limits_0^1 {f\left( x \right)dx} + \int\limits_0^1 {f\left( x \right)dx} = 2\int\limits_0^1 {f\left( x \right)dx} = 2.5 = 10$.

b) $\int\limits_{ - 1}^1 {g\left( x \right)dx} = \int\limits_{ - 1}^0 {g\left( x \right)dx} + \int\limits_0^1 {g\left( x \right)dx}$$= - \int\limits_0^1 {g\left( x \right)dx} + \int\limits_0^1 {g\left( x \right)dx} = 0$.

c) $\int\limits_{ - 1}^1 {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} = \int\limits_{ - 1}^1 {f\left( x \right)dx} + \int\limits_{ - 1}^1 {g\left( x \right)dx} = 10$.

d) $\int\limits_{ - 1}^1 {\left[ {f\left( x \right) - g\left( x \right)} \right]dx} = \int\limits_{ - 1}^1 {f\left( x \right)dx} - \int\limits_{ - 1}^1 {g\left( x \right)dx} = 10$.

Đáp án: a) Đúng; b) Sai; c) Đúng; d) Đúng.

a) $f\left( x \right) = {\left( {{x^2} + x + C} \right)^\prime } = 2x + 1$.

b) $g\left( x \right) = {\left( {{x^4} + {x^3} + C} \right)^\prime } = 4{x^3} + 3{x^2}$. Suy ra $g\left( 0 \right) = 0$.

c) $\int\limits_0^1 {g\left( x \right)dx} = \int\limits_0^1 {\left( {4{x^3} + 3{x^2}} \right)dx} = \left. {\left( {{x^4} + {x^3}} \right)} \right|_0^1 = 2$.

d) $\int\limits_0^1 {f\left( x \right)g\left( x \right)dx} = \int\limits_0^1 {\left( {2x + 1} \right)\left( {4{x^3} + 3{x^2}} \right)dx}$$= \int\limits_0^1 {\left( {8{x^4} + 10{x^3} + 3{x^2}} \right)dx}$$= \left. {\left( {\frac{8}{5}{x^5} + \frac{{10}}{4}{x^4} + {x^3}} \right)} \right|_0^1 = \frac{{51}}{{10}}$.

Đáp án: a) Đúng; b) Sai; c) Sai; d) Đúng.

a) Quãng đường xe di chuyển được phải là nguyên hàm của $v\left( t \right)$, $v'\left( t \right) = 2{,}01 - 0{,}05t\left( {0 \le t \le 10} \right)$ là công thức tính gia tốc của vật.

b) Quãng đường xe di chuyển được trong 3 giây là $\int\limits_0^3 {\left( {2{,}01t - 0{,}025{t^2}} \right)dt} = 8{,}82$ m.

c) Quãng đường xe di chuyển được trong giây thứ 9 là:

$s\left( 9 \right) - s\left( 8 \right) = \int\limits_8^9 {\left( {2{,}01t - 0{,}025{t^2}} \right)dt} \approx 15{,}277$ m.

d) Ta có $v\left( t \right) = 2{,}01t - 0{,}025{t^2}\left( {0 \le t \le 10} \right)$ $\Rightarrow \mathop {\max }\limits_{\left[ {0;10} \right]} v\left( t \right) = 17{,}6$ m/s khi t = 10 giây.

Gia tốc vật khi đó là $a\left( {10} \right) = v'\left( {10} \right) = 2{,}01 - 0{,}05.10 = 1{,}51\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$.

Đáp án: a) Sai; b) Đúng; c) Đúng; d) Đúng.

a) $\int\limits_1^2 {f'\left( x \right)dx} = \left. {f\left( x \right)} \right|_1^2 = f\left( 2 \right) - f\left( 1 \right) = - 1 - 0 = - 1$.

b) $\int\limits_1^4 {\left[ {3 + f'\left( x \right)} \right]dx} = \left. {3x} \right|_1^4 + \left. {f\left( x \right)} \right|_1^4 = 12 - 3 + f\left( 4 \right) - f\left( 1 \right) = 9 + f\left( 4 \right)$.

c) $\int\limits_1^2 {\left| {f'\left( x \right)} \right|dx} = - \int\limits_1^2 {f'\left( x \right)dx} = \left. { - f\left( x \right)} \right|_1^2 = - f\left( 2 \right) + f\left( 1 \right)$.

d) $\int\limits_1^4 {\left| {f'\left( x \right)} \right|dx} = 5$$\Leftrightarrow \int\limits_1^2 {\left| {f'\left( x \right)} \right|dx} + \int\limits_2^4 {\left| {f'\left( x \right)} \right|dx} = 5$$\Leftrightarrow - f\left( 2 \right) + f\left( 1 \right) + \left. {f\left( x \right)} \right|_2^4 = 5$$\Leftrightarrow - f\left( 2 \right) + f\left( 1 \right) + f\left( 4 \right) - f\left( 2 \right) = 5$$\Leftrightarrow 1 + 0 + f\left( 4 \right) + 1 = 5 \Leftrightarrow f\left( 4 \right) = 3$.

Đáp án: a) Sai; b) Sai; c) Đúng; d) Đúng.

a) $\int\limits_0^\pi {f\left( x \right)dx} = \int\limits_0^\pi {\sin 2xdx}$$= \left. { - \frac{1}{2}\cos 2x} \right|_0^\pi = - \frac{1}{2} + \frac{1}{2} = 0$.

b) Có $F\left( x \right) = \int {f\left( x \right)dx} = \int {\sin 2x} dx = - \frac{1}{2}\cos 2x + C$.

Có $F\left( 0 \right) = \frac{1}{2} \Rightarrow C = 1$. Suy ra $F\left( x \right) = - \frac{1}{2}\cos 2x + 1$.

Do đó $F\left( {\frac{\pi }{2}} \right) = - \frac{1}{2}\cos \pi + 1 = \frac{3}{2}$.

c) $\int\limits_0^{\frac{\pi }{2}} {\left( {\cos x - f\left( x \right)} \right)dx} = \int\limits_0^{\frac{\pi }{2}} {\left( {\cos x - \sin 2x} \right)dx}$$= \int\limits_0^{\frac{\pi }{2}} {\cos xdx} - \int\limits_0^{\frac{\pi }{2}} {\sin 2xdx}$$= \left. {\sin x} \right|_0^{\frac{\pi }{2}} + \left. {\frac{1}{2}\cos 2x} \right|_0^{\frac{\pi }{2}} = 1 - \frac{1}{2} - \frac{1}{2} = 0$.

d) $\int\limits_{ - \pi }^\pi {\left| {f\left( x \right)} \right|dx} = \int\limits_{ - \pi }^\pi {\left| {\sin 2x} \right|dx}$$= 2\int\limits_0^\pi {\sin 2xdx} = 2.0 = 0$.

Đáp án: a) Đúng; b) Sai; c) Sai; d) Sai.

$\int\limits_0^1 {\left( {2f\left( x \right) + {e^{2x}}} \right)dx} = 2\int\limits_0^1 {f\left( x \right)dx} + \left. {\frac{{{e^{2x}}}}{2}} \right|_0^1 = 4 + \frac{{{e^2} - 1}}{2} = \frac{{{e^2} + 7}}{2}$.

Suy ra $a = 2;b = 7;c = 2 \Rightarrow a + b + c = 11$.

Trả lời: 11.

$\int\limits_0^1 {\frac{{xdx}}{{{{\left( {x + 2} \right)}^2}}}} = \int\limits_0^1 {\frac{{\left( {x + 2} \right) - 2}}{{{{\left( {x + 2} \right)}^2}}}} dx$$= \int\limits_0^1 {\frac{1}{{x + 2}}dx - \int\limits_0^1 {\frac{2}{{{{\left( {x + 2} \right)}^2}}}dx} }$

$= \left. {\ln \left| {x + 2} \right|} \right|_0^1 + \left. {\frac{2}{{x + 2}}} \right|_0^1 = \ln 3 - \ln 2 + \frac{2}{3} - 1 = \frac{{ - 1}}{3} + \ln 3 - \ln 2$.

Suy ra $a = - \frac{1}{3};b = - 1;c = 1$. Do đó $3a + b + c = - 1 - 1 + 1 = - 1$.

Trả lời: −1.

$\left\{ \begin{array}{l}\int\limits_2^7 {\left[ {2f\left( x \right) + 3g\left( x \right)} \right]dx} = 2\\\int\limits_2^7 {\left[ {f\left( x \right) - 2g\left( x \right)} \right]dx} = 4\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}2\int\limits_2^7 {f\left( x \right)dx} + 3\int\limits_2^7 {g\left( x \right)dx} = 2\\\int\limits_2^7 {f\left( x \right)dx} - 2\int\limits_2^7 {g\left( x \right)dx} = 4\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}\int\limits_2^7 {f\left( x \right)dx} = \frac{{16}}{7}\\\int\limits_2^7 {g\left( x \right)dx} = - \frac{6}{7}\end{array} \right.$.

Do đó $\int\limits_2^7 {\left[ {f\left( x \right) - g\left( x \right)} \right]dx}$$= \int\limits_2^7 {f\left( x \right)dx} - \int\limits_2^7 {g\left( x \right)dx} = \frac{{16}}{7} + \frac{6}{7} = \frac{{22}}{7} \approx 3{,}14$.

Trả lời: 3,14.

Ta có $\int\limits_0^2 {f\left( x \right)dx} = F\left( 2 \right) - F\left( 0 \right)$$\Rightarrow F\left( 2 \right) = \int\limits_0^2 {f\left( x \right)dx} + F\left( 0 \right) = 3 + 2 = 5$.

Trả lời: 5.

Khi máy bay đạt vận tốc 200 m/s, suy ra ${t^2} + 10t = 200 \Leftrightarrow \left[ \begin{array}{l}t = 10\\t = - 20\end{array} \right.$.

Vì $t > 0$ nên $t = 10$.

Quãng đường mãy bay di chuyển trên đường băng là $\int\limits_0^{10} {\left( {{t^2} + 10t} \right)dt} = \frac{{2500}}{3} \approx 833$ m.

Trả lời: 833.