(Đúng sai) 24 bài tập Tích phân (có lời giải)- Đề 2

Ta có: \(\int\limits_0^2 {\left[ {2f\left( x \right) - 1)} \right]dx}  = \int\limits_0^2 {2f\left( x \right)dx}  - \int\limits_0^2 {dx}  = 2.4 - 2 = 6\).

B-Đúng 

Ta có\[\int_0^2 {\left[ {f\left( x \right) + 3g\left( x \right)} \right]} dx = \int_0^2 f \left( x \right)dx + 3\int_0^2 g \left( x \right)dx = 3 + 3.7 = 24\].

Ta có\[\int_0^2 {\left[ {f\left( x \right) + 3g\left( x \right)} \right]} dx = \int_0^2 f \left( x \right)dx + 3\int_0^2 g \left( x \right)dx = 3 + 3.7 = 24\].

Ta có \(\int\limits_0^1 {\left[ {f\left( x \right) + 2x} \right]} dx = 3 \Leftrightarrow \int\limits_0^1 {f\left( x \right)} dx + 2\int\limits_0^1 x dx = 3 \Leftrightarrow \int\limits_0^1 {f\left( x \right)} dx + 2.\frac{{{x^2}}}{2}\left| {\begin{array}{*{20}{c}}1\\0\end{array}} \right. = 3\).

Suy ra \(\int\limits_0^1 {f\left( x \right){\rm{d}}x}  = 3 - {x^2}\left| {\begin{array}{*{20}{c}}1\\0\end{array}} \right. = 3 - \left( {1 - 0} \right) = 2\).

Ta có : \(\int\limits_3^2 {2f\left( x \right){\rm{d}}x}  =  - \int\limits_2^3 {2f\left( x \right){\rm{d}}x}  =  - 2\int\limits_2^3 {f\left( x \right){\rm{d}}x}  =  - 2.\left( { - 6} \right) = 12.\)

Ta có: \(\int\limits_2^1 {\left[ {f\left( x \right) - g\left( x \right)} \right]{\mkern 1mu} } {\rm{d}}x =  - \int\limits_1^2 {\left[ {f\left( x \right) - g\left( x \right)} \right]\,} {\rm{d}}x =  - \int\limits_1^2 {f\left( x \right)\,} {\rm{d}}x + \int\limits_1^2 {g\left( x \right)\,} {\rm{d}}x =  - 2 + 6 = 4\).

Ta có \[\int\limits_1^0 {\left[ {f\left( x \right) + g\left( x \right)} \right]dx}  =  - \int\limits_0^1 {\left[ {f\left( x \right) + g\left( x \right)} \right]dx}  =  - \int\limits_0^1 {f\left( x \right)dx - \int\limits_0^1 {g\left( x \right)dx =  - 3 + 4 = 1} } \].

Ta có

\[\int\limits_0^3 {f(x)} \]dx =\[\int\limits_0^1 {f(x)} \]dx +\[\int\limits_1^3 {f(x)} \]dx\[ \Rightarrow \int\limits_1^3 {f(x)} \]dx =\[\int\limits_0^3 {f(x)} \]dx \[ - \int\limits_0^1 {f(x)} \]dx = 5+ 1= 6

\[\int\limits_1^3 {f\left( x \right){\rm{d}}x} \]\[ = \int\limits_1^2 {f\left( x \right){\rm{d}}x}  + \int\limits_2^3 {f\left( x \right){\rm{d}}x} \]\[ =  - 3 + 4\]\[ = 1\].

\(\mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx = 3;\;\mathop \smallint \limits_0^3 f\left( x \right)dx = 1;\;{\rm{\;}}\mathop \smallint \limits_{ - 1}^3 f\left( x \right)dx = \mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx + \mathop \smallint \limits_0^3 f\left( x \right)dx = 3 + 1 = 4\)

D-Sai

\(\int\limits_{ - 2}^5 {\left[ {f\left( x \right) - 4g\left( x \right) - 1} \right]{\rm{d}}} x\)\( = \int\limits_{ - 2}^5 {f\left( x \right){\rm{d}}x}  - \int\limits_{ - 2}^5 {4g\left( x \right)} {\rm{d}}x - \int\limits_{ - 2}^5 {{\rm{d}}x} \)\[ = \int\limits_{ - 2}^5 {f\left( x \right){\rm{d}}x}  - 4\int\limits_{ - 2}^5 {g\left( x \right)} {\rm{d}}x - \int\limits_{ - 2}^5 {{\rm{d}}x} \]

\[ = \int\limits_{ - 2}^5 {f\left( x \right){\rm{d}}x}  + 4\int\limits_5^{ - 2} {g\left( x \right)} {\rm{d}}x - \int\limits_{ - 2}^5 {{\rm{d}}x} \]\[ = 8 + 4.3 - x\left| \begin{array}{l}5\\ - 2\end{array} \right.\]\[ = 8 + 4.3 - 7\]\[ = 13\].

 Biết \[\int\limits_1^2 {f\left( x \right)dx}  = 2\]. Giá trị của \[\int\limits_2^1 {3f\left( x \right)dx}  =  - 6\].

           Ta có : \(\int\limits_2^1 {3f\left( x \right)dx}  =  - \int\limits_1^2 {3f\left( x \right)} dx =  - 3\int\limits_1^2 {f\left( x \right)} dx =  - 3.2 =  - 6\).

Biết \(\int\limits_1^2 {f\left( x \right){\mkern 1mu} } {\rm{d}}x =  - 1\) và \(\int\limits_1^2 {g\left( x \right){\mkern 1mu} } {\rm{d}}x = 3\), khi đó \(\int\limits_2^1 {\left[ {f\left( x \right) - g\left( x \right)} \right]{\mkern 1mu} } {\rm{d}}x = 5\)         

           Ta có: \(\int\limits_1^2 {f\left( x \right){\mkern 1mu} } {\rm{d}}x =  - 1 \Leftrightarrow \int\limits_2^1 {f\left( x \right){\mkern 1mu} } {\rm{d}}x = 1;\)     \(\int\limits_1^2 {g\left( x \right){\mkern 1mu} } {\rm{d}}x = 3 \Leftrightarrow \int\limits_2^1 {g\left( x \right){\mkern 1mu} } {\rm{d}}x =  - 3\)

           \(\int\limits_2^1 {\left[ {f\left( x \right) - g\left( x \right)} \right]\,} {\rm{d}}x = \int\limits_2^1 {f\left( x \right)\,} {\rm{d}}x - \int\limits_2^1 {g\left( x \right)\,} {\rm{d}}x = 1 - \left( { - 3} \right) = 4\).

C-Đúng

Nếu \(\int\limits_1^2 {f\left( x \right){\rm{d}}x}  =  - 2\) và \(\int\limits_2^3 {f\left( x \right){\rm{d}}x}  = 1\) thì \(\int\limits_1^3 {f\left( x \right){\rm{d}}x}  =  - 1\).

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

D-Đúng 

Nếu \(\int\limits_0^2 {\left( {f\left( x \right) + 3{x^2}} \right){\rm{d}}x}  = 10\) thì \(\int\limits_0^2 {f\left( x \right){\rm{d}}x}  = 2\).

Ta có:

\(\,\,\,\,\,\,\int\limits_0^2 {\left( {f\left( x \right) + 3{x^2}} \right){\rm{d}}x}  = 10\) \( \Leftrightarrow \int\limits_0^2 {f\left( x \right)} {\rm{d}}x + \int\limits_0^2 {3{x^2}} {\rm{d}}x = 10\) \( \Leftrightarrow \int\limits_0^2 {f\left( x \right)} {\rm{d}}x = 10 - \int\limits_0^2 {3{x^2}} {\rm{d}}x\)

\( \Leftrightarrow \int\limits_0^2 {f\left( x \right)} {\rm{d}}x = 10 - {x^3}\left| \begin{array}{l}2\\0\end{array} \right.\,\) \( \Leftrightarrow \int\limits_0^2 {f\left( x \right)} {\rm{d}}x = 10 - 8 = 2\).