85 câu trắc nghiệm Toán 12 Chân trời sáng tạo Bài 2. Tích phân có đáp án - Đề 1

Chọn A

Theo định nghĩa tích phân ta có: $\int\limits_1^2 {f\left( x \right){\rm{d}}x} = F\left( 2 \right) - F\left( 1 \right) = 6$.

Chọn D

Ta có $\int\limits_1^2 {f\left( x \right){\rm{d}}x} = F\left( 2 \right) - F\left( 1 \right) = 3 - \left( { - 2} \right) = 5$.

A. Chọn C

$$\int\limits_2^5 {{\rm{[}}f(x) + g(x){\rm{]d}}x} = \int\limits_2^5 {f(x){\rm{d}}x} + \int\limits_2^5 {g(x){\rm{d}}x = 3 - 2 = 1.}$$

Chọn A

Ta có: $\int\limits_2^5 {3f\left( x \right)dx} = 3\int\limits_2^5 {f\left( x \right)dx} = 3.2 = 6$.

Chọn B

Ta có $\int\limits_1^3 {\left[ {f\left( x \right) + 2x} \right]dx} = \int\limits_1^3 {f\left( x \right)dx} + \int\limits_1^3 {2xdx} = 2 + \left. {{x^2}} \right|_1^3 = 2 + \left( {9 - 1} \right) = 10$.

Chọn A

Ta có: $\int\limits_0^2 {\left[ {\frac{1}{2}f\left( x \right) + 2} \right]} {\rm{d}}x = \frac{1}{2}\int\limits_0^2 {f\left( x \right){\rm{d}}x} + \int\limits_0^2 {2{\rm{d}}x} = 2 + 4 = 6$.

Chọn A

Ta có: $\int\limits_5^{ - 1} {f\left( x \right)} {\rm{d}}x = - \int\limits_{ - 1}^5 {f\left( x \right)} {\rm{d}}x = - \left( { - 3} \right) = 3$.

Chọn B

$\int\limits_0^2 {\left[ {\frac{1}{2}f\left( x \right) + 2} \right]} {\rm{d}}x = \frac{1}{2}\int\limits_0^2 {f\left( x \right)} {\rm{d}}x + \int\limits_0^2 2 {\rm{d}}x = 2 + 4 = 6$.

Chọn A

$\int\limits_5^{ - 1} {f\left( x \right){\rm{d}}x} = - \int\limits_{ - 1}^5 {f\left( x \right){\rm{d}}x} = 3$.

Chọn A

Ta có $\int\limits_0^3 {\left[ {\frac{1}{3}f\left( x \right) + 2} \right]{\rm{d}}x} = \frac{1}{3}\int\limits_0^3 {f\left( x \right)dx + \int\limits_0^3 {2dx = \frac{1}{3} \cdot 6 + 6 = 8} }$.

Chọn B

Ta có $\int\limits_{ - 1}^5 {f\left( x \right){\rm{d}}x = } \int\limits_{ - 1}^2 {f\left( x \right){\rm{d}}x + } \int\limits_2^5 {f\left( x \right){\rm{d}}x = 2 - 5 = - 3}$.

Chọn B

Ta có: $\int\limits_{ - 1}^5 {f\left( x \right){\rm{d}}x} = \int\limits_{ - 1}^2 {f\left( x \right){\rm{d}}x} + \int\limits_2^5 {f\left( x \right){\rm{d}}x} = 2 + \left( { - 5} \right) = - 3$.

Chọn D

Ta có $\int\limits_0^3 {\left[ {\frac{1}{3}f\left( x \right) + 2} \right]{\rm{d}}x} = \frac{1}{3}\int\limits_0^3 {f\left( x \right){\rm{d}}x} + \int\limits_0^3 {{\rm{2d}}x} = 2 + 6 = 8$.

Chọn B

Ta có $\int\limits_0^2 {\left[ {4x - f\left( x \right)} \right]{\rm{d}}x}$$=$$\int\limits_0^2 {4x{\rm{d}}x} - \int\limits_0^2 {f\left( x \right){\rm{d}}x = } \left. {2{x^2}} \right|_0^2 - 3 = 8 - 3 = 5$

Chọn C

Ta có $\int\limits_0^2 {\left[ {2x - f\left( x \right)} \right]{\rm{d}}x} = \int\limits_0^2 {2x{\rm{d}}x} - \int\limits_0^2 {f\left( x \right){\rm{d}}x}$$= \left. {{x^2}} \right|_0^2 - 3 = 4 - 3 = 1$.

Chọn B

Ta có $\int\limits_1^3 {f\left( x \right){\rm{d}}x} = \int\limits_1^2 {f\left( x \right){\rm{d}}x} + \int\limits_2^3 {f\left( x \right){\rm{d}}x} = - 2 + 1 = - 1$.

Chọn D

Ta có: $\int\limits_0^1 {2f\left( x \right){\rm{d}}x} = 2\int\limits_0^1 {f\left( x \right){\rm{d}}x} = 2.4 = 8$.

Chọn C

Ta có: $\int\limits_1^3 {2f\left( x \right){\rm{d}}x} = 2\int\limits_1^3 {f\left( x \right){\rm{d}}x} = 2.3 = 6$.

Chọn A

Ta có: $\int\limits_1^2 {\left[ {2 + f\left( x \right)} \right]{\rm{d}}x} = \left( {2x + {x^2}} \right)\left| \begin{array}{l}2\\1\end{array} \right. = 8 - 3 = 5$

Chọn D

Ta có ${\int }_{1}^{5}3f(x)dx=3{\int }_{1}^{5}f(x)dx=3.4=12$

Chọn C

Ta có ${\int }_{1}^{2}(2+f(x))dx={\int }_{1}^{2}f(x)dx=2x\{\begin{matrix}2 \\ 1\end{matrix}+F(x)\{\begin{matrix}2 \\ 1\end{matrix}=2x\{\begin{matrix}2 \\ 1\end{matrix}+{x}^{3}\{\begin{matrix}2 \\ 1\end{matrix}=9$

Chọn B

Ta có : $\int\limits_1^2 {3f\left( x \right)} dx = 3\int\limits_1^2 {f\left( x \right)} dx$$= 3.2 = 6$.

Chọn D

Ta có $\int\limits_1^3 {\left[ {1 + f(x)} \right]} {\rm{d}}x = \left[ {x + F(x)} \right]\left| {\mathop {}\limits_1^3 } \right. = \left[ {x + {x^3})} \right]\left| {\mathop {}\limits_1^3 } \right. = 30 - 2 = 28$.

Chọn C

Ta có : $\int\limits_2^3 {2f\left( x \right){\rm{d}}x} = 2\int\limits_2^3 {f\left( x \right){\rm{d}}x} = 12.$.

Chọn A

Ta có $\int\limits_1^3 {\left[ {1 + f(x)} \right]dx} = \left. {\left( {x + F\left( x \right)} \right)} \right|_1^3 = \left. {\left( {x + {x^2}} \right)} \right|_1^3 = 12 - 2 = 10.$

Chọn B

Ta có $\int\limits_2^3 {\left[ {f\left( x \right) - g\left( x \right)} \right]dx} = \int\limits_2^3 {f\left( x \right)dx} - \int\limits_2^3 {g\left( x \right)dx} = 4 - 1 = 3$

Chọn A

Ta có

$\int\limits_0^1 {\left[ {f\left( x \right) + 2{\rm{x}}} \right]} d{\rm{x = 2}} \Leftrightarrow \int\limits_0^1 {f\left( x \right)} d{\rm{x + }}\int\limits_0^1 {2{\rm{x}}} d{\rm{x = 2}}$$\, \Leftrightarrow \int\limits_0^1 {f\left( x \right)} d{\rm{x}} = 2 - \left. {{x^2}} \right|_0^1$$\, \Leftrightarrow \int\limits_0^1 {f\left( x \right)} d{\rm{x}} = 2 - 1$ $\, \Leftrightarrow \int\limits_0^1 {f\left( x \right)} d{\rm{x}} = 1$

Chọn A

Ta có: $\int\limits_2^3 {\left[ {f\left( x \right) + g\left( x \right)} \right]} dx = \int\limits_2^3 {f\left( x \right)dx} + \int\limits_2^3 {g\left( x \right)dx} = 4$.

Chọn D

Ta có $\int\limits_0^1 {\left[ {f\left( x \right) + 2x} \right]} dx = 3 \Leftrightarrow \int\limits_0^1 {f\left( x \right)} dx + 2\int\limits_0^1 x dx = 3 \Leftrightarrow \int\limits_0^1 {f\left( x \right)} dx + 2.\frac{{{x^2}}}{2}\left| {\begin{array}{cccccccccccccccccccc}1\\0\end{array}} \right. = 3$.

Suy ra $\int\limits_0^1 {f\left( x \right){\rm{d}}x} = 3 - {x^2}\left| {\begin{array}{cccccccccccccccccccc}1\\0\end{array}} \right. = 3 - \left( {1 - 0} \right) = 2$.

Chọn B

Ta có: $\int\limits_1^2 {\left[ {f\left( x \right) - g\left( x \right)} \right]} \,{\rm{d}}x = \int\limits_1^2 {f\left( x \right)} \,{\rm{d}}x - \int\limits_1^2 {g\left( x \right){\rm{d}}x} = 3 - 2 = 1$.