85 câu trắc nghiệm Toán 12 Cánh diều Bài 3: Tích phân có đáp án - Đề 2
30 câu hỏi
A. \[3\].
B. \[2\].
C. \[6\].
D. \[4\].
Chọn A
\[\int_0^1 {\left[ {f\left( x \right) + 2x} \right]{\rm{d}}x} = 4 \Leftrightarrow \int_0^1 {f\left( x \right){\rm{d}}x} + \int_0^1 {2x{\rm{d}}x} = 4 \Leftrightarrow \int_0^1 {f\left( x \right){\rm{d}}x} = 4 - 1 = 3\]
A. \(1\).
B. \(5\).
C. \( - 1\).
D. \(6\).
Chọn D
Ta có: \[\int\limits_1^2 {[f(x) + g(x)]dx} = \int\limits_1^2 {f(x)dx} + \int\limits_1^2 {g(x)dx} = 2 + 3 = 5\].
\(7\).
\(3\).
\(5\).
\(4\).
Chọn D
\(\int\limits_0^1 {\left[ {f\left( x \right) + 2x} \right]} {\rm{d}}x = 5\)\( \Leftrightarrow \int\limits_0^1 {f\left( x \right){\rm{d}}x} + \int\limits_0^1 {{\rm{2xd}}x} = 5\)
\[\int\limits_0^1 {f\left( x \right){\rm{d}}x} + \left. {{x^2}} \right|_0^1 = 5 \Leftrightarrow \int\limits_0^1 {f\left( x \right){\rm{d}}x} + 1 = 5 \Leftrightarrow \int\limits_0^1 {f\left( x \right){\rm{d}}x} = 4\].122
\(8\).
\( - 4\).
\(4\).
\( - 8\).
Chọn B
Ta có: \(\int\limits_1^2 {\left[ {f\left( x \right) - g\left( x \right)} \right]\,} {\rm{d}}x = \int\limits_1^2 {f\left( x \right)\,} {\rm{d}}x - \int\limits_1^2 {g\left( x \right)\,} {\rm{d}}x = 2 - 6 = - 4\).
\( - 7\).
\(7\).
\( - 1\).
\(1\).
Chọn C
Ta có \[\int\limits_0^1 {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} = \int\limits_0^1 {f\left( x \right)dx + \int\limits_0^1 {g\left( x \right)dx = 3 + \left( { - 4} \right) = - 1} } \].
\(6\).
\( - 6\).
\( - 2\).
\(2\).
Chọn C
\[\int_0^1 {\left[ {f(x) + g(x)} \right]{\rm{d}}x} = \int_0^1 {f(x){\rm{d}}x} + \int_0^1 {g(x){\rm{d}}x} = 2 + ( - 4) = - 2\].
\[ - 1\].
\[1\].
\[ - 5\].
\[5\].
Chọn C
\[\int\limits_0^1 {\left[ {f\left( x \right) - g\left( x \right)} \right]} {\rm{d}}x = \int\limits_0^1 {f\left( x \right)} {\rm{d}}x - \int\limits_0^1 {g\left( x \right)} {\rm{d}}x = - 2 - 3 = - 5\].
\( - 8\)
\(1\)
\( - 3\)
\(12\)
Chọn A
Có \(\int\limits_0^1 {\left[ {f\left( x \right) - 2g\left( x \right)} \right]{\rm{d}}x} = \int\limits_0^1 {f\left( x \right){\rm{d}}x} - 2\int\limits_0^1 {g\left( x \right){\rm{d}}x} \)\( = 2 - 2.5 = - 8\).
\[\int\limits_a^b {\left[ {f(x) + 2g(x)} \right]} {\rm{d}}x = \int\limits_a^b {f(x)} {\rm{d}}x\,{\rm{ + 2}}\,\int\limits_a^b {g(x)} {\rm{d}}x\].
\[\int\limits_a^b {\frac{{f(x)}}{{g(x)}}} {\rm{d}}x = \frac{{\int\limits_a^b {f(x)} {\rm{d}}x}}{{\,\int\limits_a^b {g(x)} {\rm{d}}x}}\,\].
\[\int\limits_a^b {\left[ {f(x).g(x)} \right]} {\rm{d}}x = \int\limits_a^b {f(x)} {\rm{d}}x{\rm{ }}{\rm{.}}\,\int\limits_a^b {g(x)} {\rm{d}}x\].
\[\,\int\limits_a^b {{f^2}(x)} {\rm{d}}x{\rm{ = }}{\left[ {\int\limits_a^b {f(x){\rm{d}}x} } \right]^2}\].
Theo tính chất tích phân ta có
\[\int\limits_a^b {\left[ {f(x) + g(x)} \right]} {\rm{d}}x = \int\limits_a^b {f(x)} {\rm{d}}x\,{\rm{ + }}\,\int\limits_a^b {g(x)} {\rm{d}}x;\,\int\limits_a^b {kf(x)} {\rm{d}}x = k\int\limits_a^b {f(x)} {\rm{d}}x\], với \(k \in \mathbb{R}\).
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\(I = 5\).
\(I = - 3\).
\(I = 3\).
\(I = - 5\).
chọn D
Ta có: \(\int\limits_{ - 2}^4 {f\left( t \right){\rm{d}}t} = \int\limits_{ - 2}^4 {f\left( x \right){\rm{d}}x} \), \(\int\limits_2^4 {f\left( y \right){\rm{d}}y} = \int\limits_2^4 {f\left( x \right){\rm{d}}x} \).
Khi đó: \[\int\limits_{ - 2}^2 {f\left( x \right){\rm{d}}x} + \int\limits_2^4 {f\left( x \right){\rm{d}}x} = \int\limits_{ - 2}^4 {f\left( x \right){\rm{d}}x} \].
\( \Rightarrow \int\limits_2^4 {f\left( x \right){\rm{d}}x} = \int\limits_{ - 2}^4 {f\left( x \right){\rm{d}}x} - \int\limits_{ - 2}^2 {f\left( x \right){\rm{d}}x} = - 4 - 1 = - 5\).
Vậy \(\int\limits_2^4 {f\left( y \right){\rm{d}}y} = - 5\).
\(16\).
\( - 18\).
\(24\).
\(10\).
chọn C
Ta có
\[\int_0^2 {\left[ {f\left( x \right) + 3g\left( x \right)} \right]} dx = \int_0^2 f \left( x \right)dx + 3\int_0^2 g \left( x \right)dx = 3 + 3.7 = 24\].
1.
4.
6.
5.
C. 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
Vậy \[\int\limits_1^3 {f(x)} \]dx = 6
12.
7.
1.
\[ - 12\].
Chọn C
\[\int\limits_1^3 {f\left( x \right){\rm{d}}x} \]\[ = \int\limits_1^2 {f\left( x \right){\rm{d}}x} + \int\limits_2^3 {f\left( x \right){\rm{d}}x} \]\[ = - 3 + 4\]\[ = 1\].
\(1.\)
\(7.\)
\( - 9.\)
\(9.\)
Chọn C
Ta có \[\int\limits_{ - 1}^2 {f'\left( x \right)} dx = \left. {f\left( x \right)} \right|_{ - 1}^2 = f\left( 2 \right) - f\left( { - 1} \right) = - 1 - 8 = - 9.\]
\(I = 5\).
\(I = 36\).
\(I = \frac{9}{4}\).
\(I = 13\).
chọn D.
Ta có: \(I = \int\limits_0^4 {f(x){\rm{d}}x} = \int\limits_0^2 {f(x){\rm{d}}x} + \int\limits_2^4 {f(x){\rm{d}}x} = 9 + 4 = 13.\)
\(6\)
\(4\)
\(2\)
\(0\)
Chọn D
Có \(\mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx = 3;\;\mathop \smallint \limits_0^3 f\left( x \right)dx = 1;\;{\rm{\;}}\mathop \smallint \limits_{ - 1}^3 f\left( x \right)dx = \mathop \smallint \limits_{ - 1}^0 f\left( x \right)dx + \mathop \smallint \limits_0^3 f\left( x \right)dx = 3 + 1 = 4\)
\(4\).
\(7\).
\(3\).
\(6\).
Chọn D
Theo tính chất của tích phân, ta có: \(\int\limits_0^3 {f\left( x \right)} \,{\rm{d}}x + \int\limits_3^4 {f\left( x \right)} \,{\rm{d}}x = \int\limits_0^4 {f\left( x \right)} \,{\rm{d}}x\).
Suy ra: \(\int\limits_0^3 {f\left( x \right)} \,{\rm{d}}x\)\( = \int\limits_0^4 {f\left( x \right)} \,{\rm{d}}x - \int\limits_3^4 {f\left( x \right)} \,{\rm{d}}x\)\( = 10 - 4\)\( = 6\).
Vậy \(\int\limits_0^3 {f\left( x \right)} \,{\rm{d}}x = 6\).
\(\ln 7.\)
\(1 + \frac{1}{2}\ln 7.\)
\(\ln 3.\)
\(1 + \ln 7.\)
Chọn B
Ta có: \(\int\limits_1^4 {F'\,\left( x \right)} {\rm{d}}x = \int\limits_1^4 {\frac{1}{{2x - 1}}} {\rm{d}}x = \left. {\frac{1}{2}\ln |2x - 1|} \right|_1^4 = \frac{1}{2}\ln 7\).
Lại có: \(\int\limits_1^4 {F'\,\left( x \right)} {\rm{d}}x = \left. {F\left( x \right)} \right|_1^4 = F\left( 4 \right) - F\left( 1 \right)\).
Suy ra \(F\left( 4 \right) - F\left( 1 \right) = \frac{1}{2}\ln 7\). Do đó \(F\left( 4 \right) = F\left( 1 \right) + \frac{1}{2}\ln 7 = 1 + \frac{1}{2}\ln 7\).
\(I = 17\).
\(I = 1\).
\(I = 11\).
\(I = 7\).
Chọn D
Ta có: \(I = \int\limits_1^{12} {f\left( x \right)} \,{\rm{d}}x = \int\limits_1^8 {f\left( x \right)} \,{\rm{d}}x + \int\limits_8^{12} {f\left( x \right)} \,{\rm{d}}x\). \( = \int\limits_1^8 {f\left( x \right)} \,{\rm{d}}x + \int\limits_4^{12} {f\left( x \right)} \,{\rm{d}}x - \int\limits_4^8 {f\left( x \right)} \,{\rm{d}}x = 9 + 3 - 5 = 7\).
\(P = 10\).
\(P = 4\).
\(P = 7\).
\(P = - 6\).
Chọn B
Ta có \(\int\limits_0^{10} {f\left( x \right)dx} = \int\limits_0^2 {f\left( x \right)dx} + \int\limits_2^6 {f\left( x \right)dx} + \int\limits_6^{10} {f\left( x \right)dx} \)
Suy ra \(\int\limits_0^2 {f\left( x \right)dx} + \int\limits_6^{10} {f\left( x \right)dx} = \int\limits_0^{10} {f\left( x \right)dx} - \int\limits_2^6 {f\left( x \right)dx} = 7 - 3 = 4\).
7.
6.
8.
9.
Chọn B
\[\int\limits_1^3 {\left[ {f\left( x \right) + 3g\left( x \right)} \right]} {\rm{d}}x = 10\] \( \Leftrightarrow \) \(\int\limits_1^3 {f\left( x \right)} {\rm{d}}x + 3\int\limits_1^3 {g\left( x \right){\rm{d}}x = 10} \) \(\left( 1 \right)\).
\[\int\limits_1^3 {\left[ {2f\left( x \right) - g\left( x \right)} \right]} {\rm{d}}x = 6\] \( \Leftrightarrow \)\(2\int\limits_1^3 {f\left( x \right)} {\rm{d}}x - \int\limits_1^3 {g\left( x \right)} {\rm{d}}x = 6\) \(\left( 2 \right)\).
Đặt \(X = \int\limits_1^3 {f\left( x \right)} {\rm{d}}x\), \(Y = \int\limits_1^3 {g\left( x \right)} {\rm{d}}x\).
Từ \(\left( 1 \right)\) và \(\left( 2 \right)\) ta có hệ phương trình:\[\left\{ \begin{array}{l}X + 3Y = 10\\2X - Y = 6\end{array} \right.\] \( \Leftrightarrow \) \[\left\{ \begin{array}{l}X = 4\\Y = 2\end{array} \right.\]. \(\)
Do đó ta được:\(\int\limits_1^3 {f\left( x \right)} {\rm{d}}x = 4\) và\(\int\limits_1^3 {g\left( x \right){\rm{d}}x} = 2\).
Vậy \[\int\limits_1^3 {\left[ {f\left( x \right) + g\left( x \right)} \right]} {\rm{d}}x = 4 + 2 = 6\].
\(P = 4\)
\(P = 10\)
\(P = 7\)
\(P = - 4\)
Chọn A
Ta có: \(\int\limits_0^{10} {f\left( x \right)dx} = \int\limits_0^2 {f\left( x \right)dx} + \int\limits_2^6 {f\left( x \right)dx} + \int\limits_6^{10} {f\left( x \right)dx} \).
\( \Rightarrow 7 = P + 3 \Rightarrow P = 4\).
\(9\).
\(6\).
\(7\).
\(8\).
Chọn B
Ta có: \(\int\limits_1^3 {\left[ {f\left( x \right) + 3g\left( x \right)} \right]} {\rm{dx = 10}}\)\( \Leftrightarrow \int\limits_1^3 {f\left( x \right)} {\rm{dx + 3}}\int\limits_1^3 {g\left( x \right)} {\rm{dx = 10}}\).
\(\int\limits_1^3 {\left[ {2f\left( x \right) - g\left( x \right)} \right]} {\rm{dx = 6}}\)\( \Leftrightarrow 2\int\limits_1^3 {f\left( x \right)} {\rm{dx - }}\int\limits_1^3 {g\left( x \right)} {\rm{dx = 6}}\).
Đặt \(u = \int\limits_1^3 {f\left( x \right)} {\rm{dx; v = }}\int\limits_1^3 {g\left( x \right)} {\rm{dx}}\).
Ta được hệ phương trình: \(\left\{ \begin{array}{l}u + 3v = 10\\2u - v = 6\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}u = 4\\v = 2\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}\int\limits_1^3 {f\left( x \right)} {\rm{dx = 4}}\\\int\limits_1^3 {g\left( x \right)} {\rm{dx = 2}}\end{array} \right.\)
Vậy \(\int\limits_1^3 {\left[ {f\left( x \right) + g\left( x \right)} \right]} {\rm{dx = 6}}\).
8.
7.
9.
6.
Chọn D
Đặt \(a = \int\limits_1^3 {f\left( x \right){\rm{d}}x} \) và \(b = \int\limits_1^3 {g\left( x \right){\rm{d}}x} \).
Khi đó, \(\int\limits_1^3 {\left[ {f\left( x \right) + 3g\left( x \right)} \right]{\rm{d}}x} = a + 3b\), \(\int\limits_1^3 {\left[ {2f\left( x \right) - g\left( x \right)} \right]{\rm{d}}x} = 2a - b\).
Theo giả thiết, ta có \(\left\{ \begin{array}{l}a + 3b = 10\\2a - b = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 4\\b = 2\end{array} \right.\).
Vậy \(I = a + b = 6\).
\(I = 7\)
\(I = 5 + \frac{\pi }{2}\)
\(I = 3\)
\(I = 5 + \pi \)
Chọn A
Ta có
\(I = \int\limits_0^{\frac{\pi }{2}} {\left[ {f\left( x \right)\, + 2\sin x} \right]\,{\rm{d}}x = \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\,{\rm{d}}x\,{\rm{ + 2}}\int\limits_0^{{\textstyle{\pi \over 2}}} {\sin x\,{\rm{d}}x} } } \)\( = \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)} \,{\rm{d}}x\,\, - \left. {2\cos x} \right|_0^{\frac{\pi }{2}} = 5 - 2\left( {0 - 1} \right) = 7\).
\(I = \frac{{17}}{2}\)
\(I = \frac{5}{2}\)
\(I = \frac{7}{2}\)
\(I = \frac{{11}}{2}\)
Lời giải
Chọn A
Ta có: \(I = \int\limits_{ - 1}^2 {\left[ {x + 2f\left( x \right) - 3g\left( x \right)} \right]{\rm{d}}x} \)\( = \)\(\left. {\frac{{{x^2}}}{2}} \right|_{ - 1}^2 + 2\int\limits_{ - 1}^2 {f\left( x \right){\rm{d}}x} - 3\int\limits_{ - 1}^2 {g\left( x \right){\rm{d}}x} \)\( = \)\(\frac{3}{2} + 2.2 - 3\left( { - 1} \right)\)\( = \)\(\frac{{17}}{2}\).
\(13\).
\(27\).
\( - 11\).
\(3\).
Chọn A
\(I = \int\limits_{ - 2}^5 {\left[ {f\left( x \right) - 4g\left( x \right) - 1} \right]{\rm{d}}} x\)\( = \int\limits_{ - 2}^5 {f\left( x \right){\rm{d}}x} - \int\limits_{ - 2}^5 {4g\left( x \right)} {\rm{d}}x - \int\limits_{ - 2}^5 {{\rm{d}}x} \)\[ = \int\limits_{ - 2}^5 {f\left( x \right){\rm{d}}x} - 4\int\limits_{ - 2}^5 {g\left( x \right)} {\rm{d}}x - \int\limits_{ - 2}^5 {{\rm{d}}x} \]
\[ = \int\limits_{ - 2}^5 {f\left( x \right){\rm{d}}x} + 4\int\limits_5^{ - 2} {g\left( x \right)} {\rm{d}}x - \int\limits_{ - 2}^5 {{\rm{d}}x} \]\[ = 8 + 4.3 - x\left| \begin{array}{l}5\\ - 2\end{array} \right.\]\[ = 8 + 4.3 - 7\]\[ = 13\].
\(\frac{5}{2}\)
\(\frac{7}{2}\)
\(\frac{{17}}{2}\)
\(\frac{{11}}{2}\)
Chọn A
Ta có \[\int\limits_{ - 1}^2 {\left[ {x + 2f(x) + 3g(x)} \right]dx} = \int\limits_{ - 1}^2 {xdx} + 2\int\limits_{ - 1}^2 {f(x)dx + 3\int\limits_{ - 1}^2 {g(x)dx = \frac{3}{2}} } + 4 - 3 = \frac{5}{2}\]
\[12\].
\[0\].
\[8\].
\[10\]
Chọn D
\[\int\limits_0^2 {\left[ {f\left( x \right) - 5g\left( x \right) + x} \right]{\rm{d}}x} = \int\limits_0^2 {f\left( x \right)dx - 5\int\limits_0^2 {{\rm{g}}\left( x \right){\rm{d}}x + \int\limits_0^2 {x{\rm{d}}x} } } \]\[ = 3 + 5 + 2 = 10\]
\[ - 140\].
\[ - 130\].
\[ - 120\].
\[ - 133\].
Chọn D
\(\int\limits_0^5 {\left[ {4f\left( x \right) - 3{x^2}} \right]{\rm{d}}x} = 4\int\limits_0^5 {f\left( x \right){\rm{d}}x} - \int\limits_0^5 {3{x^2}{\rm{d}}x} = - 8 - \left. {{x^3}} \right|_0^5 = - 8 - 125 = - 133\).
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