Câu 1:
Biết \(F\left( x \right) = \left( {ax + b} \right).{e^x}\) là nguyên hàm của hàm số \(y = \left( {2x + 3} \right).{e^x}.\) Tính tổng a + b.
A. a+b=2
B. a+b=3
C. a+b=4
D. a+b=5
Câu 2:
Tính tích phân \(I = \int\limits_1^{{2^{1000}}} {\frac{{\ln x}}{{{{\left( {x + 1} \right)}^2}}}dx} .\)
A. \(I = - \frac{{\ln {2^{1000}}}}{{1 + {2^{1000}}}} + 1000\ln \frac{2}{{1 + {2^{1000}}}}.\)
B. \(I = - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \ln \frac{{{2^{1000}}}}{{1 + {2^{1000}}}}.\)
C. \(I = \frac{{\ln {2^{1000}}}}{{1 + {2^{1000}}}} - 1000\ln \frac{2}{{1 + {2^{1000}}}}.\)
D. \(I = \frac{{1000\ln 2}}{{1 + {2^{1000}}}} - \ln \frac{{{2^{1000}}}}{{1 + {2^{1000}}}}.\)
Câu 3:
Biết \(I = \int\limits_0^4 {x\ln (2x + 1)dx} = \frac{a}{b}\ln 3 - c,\) trong đó a, b, c là các số nguyên dương và \(\frac{b}{c}\) là phân số tối giản. Tính \(S=a+b+c\)
A. S=60
B. S=70
C. S=72
D. S=68
Câu 4:
Tìm hàm số f(x) biết \(f\left( x \right) = \int {\frac{{5 + 4x}}{{{x^2}}}.lnxdx} .\)
A. \(f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\left( {\ln x + 1} \right) + C\)
B. \(f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\left( {\ln x - 1} \right) + C\)
C. \(f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\ln x - \frac{5}{x}\)
D. \(f\left( x \right) = 2\ln x - \frac{5}{x}\left( {\ln x + 1} \right) + C\)
Câu 5:
Tìm nguyên hàm của hàm số \(f\left( x \right) = x\sin x\cos x.\)
A. \(\int {f(x)dx = } \frac{1}{2}\left( {\frac{1}{4}\sin 2x + \frac{x}{2}\cos 2x} \right) + C.\)
B. \(\int {f(x)dx = } - \frac{1}{2}\left( {\frac{1}{4}\sin 2x - \frac{x}{2}\cos 2x} \right) + C.\)
C. \(\int {f(x)dx = } \frac{1}{2}\left( {\frac{1}{4}\sin 2x - \frac{x}{2}\cos 2x} \right) + C.\)
D. \(\int {f(x)dx = } - \frac{1}{2}\left( {\frac{1}{4}\sin 2x + \frac{x}{2}\cos 2x} \right) + C.\)
Câu 6:
Biết rằng \(\int\limits_0^1 {x\cos 2xdx = \frac{1}{4}\left( {a\sin 2 + b\cos 2 + c} \right)}\), với \(a,b,c \in \mathbb{Z}\) Mệnh đề nào sau đây là đúng?
A. \(a+b+c =1\)
B. \(a-b+c =0\)
C. \(a+2b+c =1\)
D. \(2a+b+c =-1\)
Câu 7:
Biết rằng \(\int\limits_0^1 {3{e^{\sqrt {1 + 3x} }}} dx = \frac{a}{5}{e^2} + \frac{b}{2}e + c\left( {a,b,c \in\mathbb{R} } \right).\) Tính \(T = a + \frac{b}{2} + \frac{c}{3}.\)
A. \(T = 9\)
B. \(T =10\)
C. \(T =5\)
D. \(T =6\)
Câu 8:
Xét tích phân \(I = \int\limits_0^1 {\left( {2{x^2} - 4} \right){e^{2x}}dx} .\) Nếu đặt \(u = 2{x^2} - 4,\,\,dv = {e^{2x}}dx,\) ta được tích phân \(I = \left. {\phi \left( x \right)} \right|_0^1 - \int\limits_0^1 {2x{e^{2x}}dx} ,\)trong đó:
A. \(\phi \left( x \right) = \left( {2{x^2} - 4} \right){e^{2x}}.\)
B. \(\phi \left( x \right) = \left( {{x^2} - 2} \right){e^x}.\)
C. \(\phi \left( x \right) = \frac{1}{2}\left( {2{x^2} - 4} \right){e^x}.\)
D. \(\phi \left( x \right) = \left( {{x^2} - 2} \right){e^{2x}}.\)
Câu 9:
Giả sử \(\int\limits_1^2 {(2x - 1)\ln xdx = a\ln 2 + b,(a,b \in \mathbb{Q})\). Tính tổng S=a+b.
A. \(S=\frac{5}{2}\)
B. S=2
C. S=1
D. \(S=\frac{3}{2}\)
Câu 10:
Biết \(I = \int\limits_0^1 {\ln (3x + 1)dx = a\ln 2 + b,} \)(với \(a,b \in \mathbb{Q}).\) Tính S=3a-b.
A. \(S = 7.\)
B. \(S = 11.\)
C. \(S = 8.\)
D. \(S = 9.\)
Câu 11:
Cho biết \(\int\limits_1^2 {\ln \left( {9 - {{\rm{x}}^2}} \right)d{\rm{x}}} = a\ln 5 + b\ln 2 + c,\) với a, b, c là các số nguyên. Tính \(S = \left| a \right| + \left| b \right| + \left| c \right|.\)
A. S = 34
B. S = 18
C. S = 26
D. S = 13
Câu 12:
Cho tích phân \(I = \int\limits_0^\pi {{x^2}\cos xdx} \) và \(u = {x^2},dv = \cos xdx\). Khẳng định nào sau đây đúng?
A. \(I = {x^2}\sin x\left| {\begin{array}{*{20}{c}}\pi \\0\end{array}} \right. - \int\limits_0^\pi {x\sin xdx} \)
B. \(I = {x^2}\sin x\left| {\begin{array}{*{20}{c}}\pi \\0\end{array}} \right. + \int\limits_0^\pi {x\sin xdx} \)
C. \(I = {x^2}\sin x\left| {\begin{array}{*{20}{c}}\pi \\0\end{array}} \right. + 2\int\limits_0^\pi {x\sin xdx} \)
D. \(I = {x^2}\sin x\left| {\begin{array}{*{20}{c}}\pi \\0\end{array}} \right. - 2\int\limits_0^\pi {x\sin xdx} \)
Câu 13:
Cho hàm số \(y = f\left( x \right)\) thỏa mãn \(f'\left( x \right) = \left( {x + 1} \right){e^x}\) và \(\int {f\left( x \right)} dx = \left( {ax + b} \right){e^x} + c\), với a, b, c là các hằng số. Khi đó:
A. \(a + b = 2\)
B. \(a + b = 3\)
C. \(a + b = 0\)
D. \(a + b = 1\)
Câu 14:
Tìm nguyên hàm của hàm số \(f\left( x \right) = x\ln \left( {x + 2} \right)\).
A. \(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^2}}}{2}\ln \left( {x + 2} \right) - \frac{{{x^2} + 4x}}{4} + C\).
B. \(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^2} - 4}}{2}\ln \left( {x + 2} \right) - \frac{{{x^2} - 4x}}{4} + C\).
C. \(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^2}}}{2}\ln \left( {x + 2} \right) - \frac{{{x^2} + 4x}}{2} + C\).
D. \(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^2} - 4}}{2}\ln \left( {x + 2} \right) - \frac{{{x^2} + 4x}}{2} + C\).
Câu 15:
Cho \(I = \int\limits_0^{\frac{\pi }{4}} {\left( {x - 1} \right)\sin 2xdx} \). Tìm đẳng thức đúng?
A. \(I = - \left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. + \int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
B. \(I = - \left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. - \int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
C. \(I = - \frac{1}{2}\left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. + \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
D. \(I = - \frac{1}{2}\left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. - \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
Câu 16:
Giả sử hàm số f có đạo hàm liên tục trên đoạn \(\left[ {0;1} \right],\) thỏa mãn điều kiện \(f\left( 1 \right) = 6\) và \(\int\limits_0^1 {xf'\left( x \right)d{\rm{x}}} = 5.\) Khi đó \(\int\limits_0^1 {f\left( x \right)d{\rm{x}}} \) bằng:
A. 1
B. -1
C. 11
D. 3
Câu 17:
Cho hàm số \(y = f\left( x \right)\) thỏa mãn hệ thức \(\int {f\left( x \right)} \sin xdx = - f\left( x \right)\cos x + \int {{\pi ^x}\cos xdx} \). Hỏi \(y = f\left( x \right)\) là hàm số nào trong các hàm số sau:
A. \(f\left( x \right) = - \frac{{{\pi ^x}}}{{\ln \pi }}\)
B. \(f\left( x \right) = \frac{{{\pi ^x}}}{{\ln \pi }}\)
C. \(f\left( x \right) = {\pi ^x}.\ln x\)
D. \(f\left( x \right) = - {\pi ^x}.\ln x\)
Câu 18:
Tính tích phân \(I = \int\limits_0^1 {3x.{e^{2x}}} dx.\)
A. \(I = \frac{{3{e^2} + 3}}{{16}}\)
B. \(I = \frac{{2{e^2} + 2}}{9}\)
C. \(I = \frac{{3{e^2} + 3}}{4}\)
D. \(I = \frac{{2{e^2} + 2}}{3}\)
Biết \(F\left( x \right) = \left( {ax + b} \right).{e^x}\) là nguyên hàm của hàm số \(y = \left( {2x + 3} \right).{e^x}.\) Tính tổng a + b.
A. a+b=2
B. a+b=3
C. a+b=4
D. a+b=5
Ta có:
\(I = \int\limits_0^2 {{e^x}\left( {2x + {e^x}} \right)dx = } \int\limits_0^2 {{e^{2x}}dx + \int\limits_0^2 {2x.{e^x}dx = } } \left. {\frac{{{e^{2x}}}}{2}} \right|_0^2 + 2\int\limits_0^2 {x{e^x}dx = \frac{{{e^x}}}{2} - \frac{1}{2} + 2\int\limits_0^2 {x{e^x}dx} }\)
Đặt
\(\left\{ \begin{array}{l} u = x\\ dv = {e^x}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = dx\\ v = {e^x} \end{array} \right.\)
\(\Rightarrow I = \frac{{{e^4}}}{2} - \frac{1}{2} + \left. {\left( {2x.{e^x}} \right)} \right|_0^2 - 2\int\limits_0^2 {{e^x}dx} = \frac{{{e^4}}}{2} - \frac{1}{2} + \left. {\left( {2x.{e^2}} \right)} \right|_0^2 - \left. {\left( {2{e^x}} \right)} \right|_0^2\)
\(= \frac{{{e^4}}}{2} + 2{e^2} + \frac{3}{2}\)
\(\Rightarrow \left\{ \begin{array}{l} a = \frac{1}{2}\\ b = 2\\ c = \frac{3}{2} \end{array} \right. \Rightarrow S = a + b + c = 4.\)
\(I = \int\limits_0^2 {{e^x}\left( {2x + {e^x}} \right)dx = } \int\limits_0^2 {{e^{2x}}dx + \int\limits_0^2 {2x.{e^x}dx = } } \left. {\frac{{{e^{2x}}}}{2}} \right|_0^2 + 2\int\limits_0^2 {x{e^x}dx = \frac{{{e^x}}}{2} - \frac{1}{2} + 2\int\limits_0^2 {x{e^x}dx} }\)
Đặt
\(\left\{ \begin{array}{l} u = x\\ dv = {e^x}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = dx\\ v = {e^x} \end{array} \right.\)
\(\Rightarrow I = \frac{{{e^4}}}{2} - \frac{1}{2} + \left. {\left( {2x.{e^x}} \right)} \right|_0^2 - 2\int\limits_0^2 {{e^x}dx} = \frac{{{e^4}}}{2} - \frac{1}{2} + \left. {\left( {2x.{e^2}} \right)} \right|_0^2 - \left. {\left( {2{e^x}} \right)} \right|_0^2\)
\(= \frac{{{e^4}}}{2} + 2{e^2} + \frac{3}{2}\)
\(\Rightarrow \left\{ \begin{array}{l} a = \frac{1}{2}\\ b = 2\\ c = \frac{3}{2} \end{array} \right. \Rightarrow S = a + b + c = 4.\)
Tính tích phân \(I = \int\limits_1^{{2^{1000}}} {\frac{{\ln x}}{{{{\left( {x + 1} \right)}^2}}}dx} .\)
A. \(I = - \frac{{\ln {2^{1000}}}}{{1 + {2^{1000}}}} + 1000\ln \frac{2}{{1 + {2^{1000}}}}.\)
B. \(I = - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \ln \frac{{{2^{1000}}}}{{1 + {2^{1000}}}}.\)
C. \(I = \frac{{\ln {2^{1000}}}}{{1 + {2^{1000}}}} - 1000\ln \frac{2}{{1 + {2^{1000}}}}.\)
D. \(I = \frac{{1000\ln 2}}{{1 + {2^{1000}}}} - \ln \frac{{{2^{1000}}}}{{1 + {2^{1000}}}}.\)
Xét tích phân: \(I = \int\limits_1^{{2^{1000}}} {\frac{{\ln x}}{{{{\left( {x + 1} \right)}^2}}}dx}\)
Đặt: \(\left\{ \begin{array}{l} u = \ln x\\ dv = \frac{1}{{{{(x + 1)}^2}}}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{1}{x}dx\\ v = - \frac{1}{{x + 1}} \end{array} \right.\)
Khi đó:
\(I = - \frac{{\ln x}}{{x + 1}}\left| \begin{array}{l} ^{{2^{1000}}}\\ _1 \end{array} \right. + \int\limits_1^{{2^{1000}}} {\frac{1}{{x + 1}}.\frac{1}{x}dx}\)
\(= - \frac{{\ln {2^{1000}}}}{{1 + {2^{1000}}}} + \int\limits_1^{{2^{1000}}} {\frac{1}{{x + 1}}.\frac{1}{x}dx} = - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \int\limits_1^{{2^{1000}}} {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)dx}\)
\(= - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \left( {\ln \left| x \right| - \ln \left| {x + 1} \right|} \right)\left| \begin{array}{l} ^{{2^{1000}}}\\ _1 \end{array} \right. = - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \ln \left| {\frac{x}{{x + 1}}} \right|\left| \begin{array}{l} ^{{2^{1000}}}\\ _1 \end{array} \right.\)
\(= - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \ln \frac{{{2^{1000}}}}{{1 + {2^{1000}}}}.\)
Đặt: \(\left\{ \begin{array}{l} u = \ln x\\ dv = \frac{1}{{{{(x + 1)}^2}}}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{1}{x}dx\\ v = - \frac{1}{{x + 1}} \end{array} \right.\)
Khi đó:
\(I = - \frac{{\ln x}}{{x + 1}}\left| \begin{array}{l} ^{{2^{1000}}}\\ _1 \end{array} \right. + \int\limits_1^{{2^{1000}}} {\frac{1}{{x + 1}}.\frac{1}{x}dx}\)
\(= - \frac{{\ln {2^{1000}}}}{{1 + {2^{1000}}}} + \int\limits_1^{{2^{1000}}} {\frac{1}{{x + 1}}.\frac{1}{x}dx} = - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \int\limits_1^{{2^{1000}}} {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)dx}\)
\(= - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \left( {\ln \left| x \right| - \ln \left| {x + 1} \right|} \right)\left| \begin{array}{l} ^{{2^{1000}}}\\ _1 \end{array} \right. = - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \ln \left| {\frac{x}{{x + 1}}} \right|\left| \begin{array}{l} ^{{2^{1000}}}\\ _1 \end{array} \right.\)
\(= - \frac{{1000\ln 2}}{{1 + {2^{1000}}}} + \ln \frac{{{2^{1000}}}}{{1 + {2^{1000}}}}.\)
Biết \(I = \int\limits_0^4 {x\ln (2x + 1)dx} = \frac{a}{b}\ln 3 - c,\) trong đó a, b, c là các số nguyên dương và \(\frac{b}{c}\) là phân số tối giản. Tính \(S=a+b+c\)
A. S=60
B. S=70
C. S=72
D. S=68
Đặt \(\left\{ \begin{array}{l} u = \ln (2x + 1)\\ dv = xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{2}{{2x + 1}}dx\\ v = \frac{{{x^2}}}{2} \end{array} \right. \Rightarrow I = \left. {\left[ {\frac{{{x^2}}}{2}\ln (2x + 1)} \right]} \right|_0^4 - \int\limits_0^4 {\frac{{{x^2}}}{{2x + 1}}dx}\) \(\Rightarrow I = \left. {\left[ {\frac{{{x^2}}}{2}\ln (2x + 1)} \right]} \right|_0^4 - \int\limits_0^4 {\left( {\frac{x}{2} - \frac{1}{4} + \frac{1}{{4(2x + 1)}}} \right)dx}\)
\(= \left. {\left[ {\frac{{{x^2}}}{2}\ln (2x + 1)} \right]} \right|_0^4 - \left. {\left( {\frac{{{x^2}}}{4} - \frac{1}{4}x + \frac{1}{8}\ln (2x + 1)} \right)} \right|_0^4\)
\(\Rightarrow I = \frac{{63}}{4}\ln 3 - 3 \Rightarrow \left\{ \begin{array}{l} a = 63\\ b = 4\\ c = 3 \end{array} \right. \Rightarrow S = a + b + c = 70.\)
Cách khác: \(\left\{ \begin{array}{l} u = \ln (2x + 1)\\ dv = xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{2}{{2x + 1}}dx\\ v = \frac{{{x^2} - \frac{1}{4}}}{2} = \frac{{(2x + 1)(2x - 1)}}{8} \end{array} \right.\)
Đặt: \(\left\{ \begin{array}{l} u = \ln (2x + 1)\\ dv = xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{2}{{2x + 1}}dx\\ v = \frac{{{x^2} - \frac{1}{4}}}{2} = \frac{{(2x + 1)(2x - 1)}}{8} \end{array} \right.\)
\(\Rightarrow I = \left. {\left[ {\frac{{4{x^2} - 1}}{8}\ln (2x + 1)} \right]} \right|_0^4 - \int\limits_0^4 {\frac{{2x - 1}}{4}dx}\)
\(\Rightarrow I = \frac{{63}}{8}\ln 9 - \left. {\frac{{({x^2} - x)}}{4}} \right|_0^4 = \frac{{63}}{4}\ln 3 - 3 \Rightarrow \left\{ \begin{array}{l} a = 63\\ b = 4\\ c = 3 \end{array} \right.\)
\(\Rightarrow S = a + b + c = 70.\)
\(= \left. {\left[ {\frac{{{x^2}}}{2}\ln (2x + 1)} \right]} \right|_0^4 - \left. {\left( {\frac{{{x^2}}}{4} - \frac{1}{4}x + \frac{1}{8}\ln (2x + 1)} \right)} \right|_0^4\)
\(\Rightarrow I = \frac{{63}}{4}\ln 3 - 3 \Rightarrow \left\{ \begin{array}{l} a = 63\\ b = 4\\ c = 3 \end{array} \right. \Rightarrow S = a + b + c = 70.\)
Cách khác: \(\left\{ \begin{array}{l} u = \ln (2x + 1)\\ dv = xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{2}{{2x + 1}}dx\\ v = \frac{{{x^2} - \frac{1}{4}}}{2} = \frac{{(2x + 1)(2x - 1)}}{8} \end{array} \right.\)
Đặt: \(\left\{ \begin{array}{l} u = \ln (2x + 1)\\ dv = xdx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{2}{{2x + 1}}dx\\ v = \frac{{{x^2} - \frac{1}{4}}}{2} = \frac{{(2x + 1)(2x - 1)}}{8} \end{array} \right.\)
\(\Rightarrow I = \left. {\left[ {\frac{{4{x^2} - 1}}{8}\ln (2x + 1)} \right]} \right|_0^4 - \int\limits_0^4 {\frac{{2x - 1}}{4}dx}\)
\(\Rightarrow I = \frac{{63}}{8}\ln 9 - \left. {\frac{{({x^2} - x)}}{4}} \right|_0^4 = \frac{{63}}{4}\ln 3 - 3 \Rightarrow \left\{ \begin{array}{l} a = 63\\ b = 4\\ c = 3 \end{array} \right.\)
\(\Rightarrow S = a + b + c = 70.\)
Tìm hàm số f(x) biết \(f\left( x \right) = \int {\frac{{5 + 4x}}{{{x^2}}}.lnxdx} .\)
A. \(f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\left( {\ln x + 1} \right) + C\)
B. \(f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\left( {\ln x - 1} \right) + C\)
C. \(f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\ln x - \frac{5}{x}\)
D. \(f\left( x \right) = 2\ln x - \frac{5}{x}\left( {\ln x + 1} \right) + C\)
Ta có \(f\left( x \right) = \int {\frac{{5 + 4x}}{{{x^2}}}} \ln xdx = \int {\frac{{5\ln x}}{{{x^2}}}dx + \int {\frac{{4.\ln x}}{x}} dx = 2{{\ln }^2}x + \int {\frac{{5\ln x}}{{{x^2}}}} dx + C}\)
Đặt \(\left\{ {\begin{array}{*{20}{c}} {u = \ln x}\\ {dv = \frac{{dx}}{{{x^2}}}} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {du = \frac{{dx}}{x}}\\ {v = - \frac{1}{x}} \end{array}} \right.\)
\(\Rightarrow \int {\frac{{5\ln x}}{{{x^2}}}dx} = - \frac{{5\ln x}}{x} + 5.\int {\frac{{dx}}{{{x^2}}}} = - \frac{{5\ln x}}{x} - \frac{5}{x} + C\)
\(\Rightarrow f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\left( {\ln x + 1} \right) + C.\)
Đặt \(\left\{ {\begin{array}{*{20}{c}} {u = \ln x}\\ {dv = \frac{{dx}}{{{x^2}}}} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {du = \frac{{dx}}{x}}\\ {v = - \frac{1}{x}} \end{array}} \right.\)
\(\Rightarrow \int {\frac{{5\ln x}}{{{x^2}}}dx} = - \frac{{5\ln x}}{x} + 5.\int {\frac{{dx}}{{{x^2}}}} = - \frac{{5\ln x}}{x} - \frac{5}{x} + C\)
\(\Rightarrow f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\left( {\ln x + 1} \right) + C.\)
Tìm nguyên hàm của hàm số \(f\left( x \right) = x\sin x\cos x.\)
A. \(\int {f(x)dx = } \frac{1}{2}\left( {\frac{1}{4}\sin 2x + \frac{x}{2}\cos 2x} \right) + C.\)
B. \(\int {f(x)dx = } - \frac{1}{2}\left( {\frac{1}{4}\sin 2x - \frac{x}{2}\cos 2x} \right) + C.\)
C. \(\int {f(x)dx = } \frac{1}{2}\left( {\frac{1}{4}\sin 2x - \frac{x}{2}\cos 2x} \right) + C.\)
D. \(\int {f(x)dx = } - \frac{1}{2}\left( {\frac{1}{4}\sin 2x + \frac{x}{2}\cos 2x} \right) + C.\)
Ta có: \(f\left( x \right) = \frac{1}{2}x.\sin 2x\)
Ta tính \(I = \frac{1}{2}\int {x.\sin 2x{\rm{d}}x}\).
Đặt: \(\left\{ \begin{array}{l} u = x\\ {\rm{d}}v = \sin 2x \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {\rm{d}}u = {\rm{d}}x\\ v = - \frac{1}{2}\cos 2x \end{array} \right.\)
\(\begin{array}{l} I = \frac{1}{2}\left( { - \frac{1}{2} \cdot x.\cos 2x - \int {\left( { - \frac{1}{2}} \right)\cos 2x{\rm{d}}x} } \right)\\ = \frac{1}{2}\left( { - \frac{1}{2}x.\cos 2x + \frac{1}{2} \cdot \frac{1}{2}\sin 2x} \right) + C \end{array}\)
\(= \frac{1}{2}\left( {\frac{1}{4}\sin 2x - \frac{1}{2}x.\cos 2x} \right) + C.\)
Ta tính \(I = \frac{1}{2}\int {x.\sin 2x{\rm{d}}x}\).
Đặt: \(\left\{ \begin{array}{l} u = x\\ {\rm{d}}v = \sin 2x \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {\rm{d}}u = {\rm{d}}x\\ v = - \frac{1}{2}\cos 2x \end{array} \right.\)
\(\begin{array}{l} I = \frac{1}{2}\left( { - \frac{1}{2} \cdot x.\cos 2x - \int {\left( { - \frac{1}{2}} \right)\cos 2x{\rm{d}}x} } \right)\\ = \frac{1}{2}\left( { - \frac{1}{2}x.\cos 2x + \frac{1}{2} \cdot \frac{1}{2}\sin 2x} \right) + C \end{array}\)
\(= \frac{1}{2}\left( {\frac{1}{4}\sin 2x - \frac{1}{2}x.\cos 2x} \right) + C.\)
Biết rằng \(\int\limits_0^1 {x\cos 2xdx = \frac{1}{4}\left( {a\sin 2 + b\cos 2 + c} \right)}\), với \(a,b,c \in \mathbb{Z}\) Mệnh đề nào sau đây là đúng?
A. \(a+b+c =1\)
B. \(a-b+c =0\)
C. \(a+2b+c =1\)
D. \(2a+b+c =-1\)
Đặt \(\left\{ {\begin{array}{*{20}{c}} {u = x}\\ {dv = \cos 2xdx} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {du = dx}\\ {v = \frac{{\sin 2x}}{2}} \end{array}} \right.\) .
Khi đó \(I = \frac{{x.\sin 2x}}{2}\left| {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right. - \frac{1}{2}\int\limits_0^1 {\sin 2xdx = \frac{{\sin 2}}{2} + \frac{1}{4}\cos 2x\left| {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right.}\)
\(= \frac{{\sin 2}}{2} + \frac{{\cos 2}}{4} - \frac{1}{4} = \frac{1}{4}\left( {2.\sin 2 + \cos 2 - 1} \right) \Rightarrow \left\{ {\begin{array}{*{20}{c}} {a = 2}\\ {b = 1}\\ {c = - 1} \end{array}} \right. \Rightarrow a - b + c = 0.\)
Khi đó \(I = \frac{{x.\sin 2x}}{2}\left| {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right. - \frac{1}{2}\int\limits_0^1 {\sin 2xdx = \frac{{\sin 2}}{2} + \frac{1}{4}\cos 2x\left| {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right.}\)
\(= \frac{{\sin 2}}{2} + \frac{{\cos 2}}{4} - \frac{1}{4} = \frac{1}{4}\left( {2.\sin 2 + \cos 2 - 1} \right) \Rightarrow \left\{ {\begin{array}{*{20}{c}} {a = 2}\\ {b = 1}\\ {c = - 1} \end{array}} \right. \Rightarrow a - b + c = 0.\)
Biết rằng \(\int\limits_0^1 {3{e^{\sqrt {1 + 3x} }}} dx = \frac{a}{5}{e^2} + \frac{b}{2}e + c\left( {a,b,c \in\mathbb{R} } \right).\) Tính \(T = a + \frac{b}{2} + \frac{c}{3}.\)
A. \(T = 9\)
B. \(T =10\)
C. \(T =5\)
D. \(T =6\)
Đặt \(t = \sqrt {1 + 3x} \Rightarrow {t^2} = 1 + 3x \Rightarrow 2tdt = 3dx \Rightarrow \left\{ {\begin{array}{*{20}{c}} {x = 0,t = 1}\\ {x = 1,t = 2} \end{array}} \right.\)
\(\Rightarrow \int\limits_0^1 {3{e^{\sqrt {1 + 3x} }}dx} = I = 2\int\limits_1^2 {t.{e^t}} dt\)
Đặt \(\left\{ {\begin{array}{*{20}{c}} {u = t}\\ {dv = {e^t}dt} \end{array} \Rightarrow \left\{ {\begin{array}{*{20}{c}} {du = dt}\\ {v = {e^t}} \end{array}} \right.} \right.\)
\(\begin{array}{l} \Rightarrow I = 2t.{e^t}\left| {\begin{array}{*{20}{c}} 2\\ 1 \end{array}} \right. - 2\int\limits_1^2 {{e^t}dt} = 2t.{e^t}\left| {\begin{array}{*{20}{c}} 2\\ 1 \end{array} - 2{e^t}\left| {\begin{array}{*{20}{c}} 2\\ 1 \end{array}} \right.} \right. = 2{e^2}\\ \Rightarrow \left\{ {\begin{array}{*{20}{c}} {a = 10}\\ {b = 0}\\ {c = 0} \end{array}} \right. \Rightarrow T = 10 \end{array}\)
\(\Rightarrow \int\limits_0^1 {3{e^{\sqrt {1 + 3x} }}dx} = I = 2\int\limits_1^2 {t.{e^t}} dt\)
Đặt \(\left\{ {\begin{array}{*{20}{c}} {u = t}\\ {dv = {e^t}dt} \end{array} \Rightarrow \left\{ {\begin{array}{*{20}{c}} {du = dt}\\ {v = {e^t}} \end{array}} \right.} \right.\)
\(\begin{array}{l} \Rightarrow I = 2t.{e^t}\left| {\begin{array}{*{20}{c}} 2\\ 1 \end{array}} \right. - 2\int\limits_1^2 {{e^t}dt} = 2t.{e^t}\left| {\begin{array}{*{20}{c}} 2\\ 1 \end{array} - 2{e^t}\left| {\begin{array}{*{20}{c}} 2\\ 1 \end{array}} \right.} \right. = 2{e^2}\\ \Rightarrow \left\{ {\begin{array}{*{20}{c}} {a = 10}\\ {b = 0}\\ {c = 0} \end{array}} \right. \Rightarrow T = 10 \end{array}\)
Xét tích phân \(I = \int\limits_0^1 {\left( {2{x^2} - 4} \right){e^{2x}}dx} .\) Nếu đặt \(u = 2{x^2} - 4,\,\,dv = {e^{2x}}dx,\) ta được tích phân \(I = \left. {\phi \left( x \right)} \right|_0^1 - \int\limits_0^1 {2x{e^{2x}}dx} ,\)trong đó:
A. \(\phi \left( x \right) = \left( {2{x^2} - 4} \right){e^{2x}}.\)
B. \(\phi \left( x \right) = \left( {{x^2} - 2} \right){e^x}.\)
C. \(\phi \left( x \right) = \frac{1}{2}\left( {2{x^2} - 4} \right){e^x}.\)
D. \(\phi \left( x \right) = \left( {{x^2} - 2} \right){e^{2x}}.\)
Đặt: \(\left\{ \begin{array}{l} u = 2{x^2} - 4\,\\ \,dv = {e^{2x}}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = 4xdx\\ v = \frac{1}{2}{e^{2x}} \end{array} \right.\)
Vậy: \(I = \left. {\left( {{x^2} - 2} \right){e^{2x}}} \right|_0^1 - \int\limits_0^1 {2x{e^{2x}}dx}\)
\(\Rightarrow \phi \left( x \right) = \left( {{x^2} - 2} \right){e^{2x}}.\)
Vậy: \(I = \left. {\left( {{x^2} - 2} \right){e^{2x}}} \right|_0^1 - \int\limits_0^1 {2x{e^{2x}}dx}\)
\(\Rightarrow \phi \left( x \right) = \left( {{x^2} - 2} \right){e^{2x}}.\)
Giả sử \(\int\limits_1^2 {(2x - 1)\ln xdx = a\ln 2 + b,(a,b \in \mathbb{Q})\). Tính tổng S=a+b.
A. \(S=\frac{5}{2}\)
B. S=2
C. S=1
D. \(S=\frac{3}{2}\)
Đặt \(\left\{ {\begin{array}{*{20}{c}} {u = \ln x}\\ {dv = (2x - 1)dx} \end{array} \Rightarrow \left\{ {\begin{array}{*{20}{c}} {du = \frac{{dx}}{x}}\\ {v = {x^2} - x} \end{array}} \right.} \right.\)
\(\Rightarrow I = \int\limits_1^2 {(2x - 1)\ln xdx = } \left. {({x^2} - x)\ln x} \right|_1^2 - \int\limits_1^2 {(x - 1)dx}\)
\(\Leftrightarrow I = \left. {({x^2} - x)\ln x} \right|_1^2 - \left. {\left( {\frac{{{x^2}}}{2} - x} \right)} \right|_1^2 = 2\ln 2 - \frac{1}{2} \Rightarrow \left\{ {\begin{array}{*{20}{c}} {a = 2}\\ {b = - \frac{1}{2}} \end{array}} \right.\)
\(\Rightarrow a + b = \frac{3}{2}\)
\(\Rightarrow I = \int\limits_1^2 {(2x - 1)\ln xdx = } \left. {({x^2} - x)\ln x} \right|_1^2 - \int\limits_1^2 {(x - 1)dx}\)
\(\Leftrightarrow I = \left. {({x^2} - x)\ln x} \right|_1^2 - \left. {\left( {\frac{{{x^2}}}{2} - x} \right)} \right|_1^2 = 2\ln 2 - \frac{1}{2} \Rightarrow \left\{ {\begin{array}{*{20}{c}} {a = 2}\\ {b = - \frac{1}{2}} \end{array}} \right.\)
\(\Rightarrow a + b = \frac{3}{2}\)
Biết \(I = \int\limits_0^1 {\ln (3x + 1)dx = a\ln 2 + b,} \)(với \(a,b \in \mathbb{Q}).\) Tính S=3a-b.
A. \(S = 7.\)
B. \(S = 11.\)
C. \(S = 8.\)
D. \(S = 9.\)
Ta có \(\int\limits_0^1 {\ln (3x + 1)dx} = x\ln (3x + 1)\left| \begin{array}{l}^1\\{}_0\end{array} \right. - \int\limits_0^1 {x.\frac{3}{{3x + 1}}} dx = \ln 4 - \left( {x - \frac{1}{3}\ln (3x + 1)} \right)\left| \begin{array}{l}^1\\{}_0\end{array} \right.\)
\( = \ln 4 - 1 + \frac{1}{3}\ln 4 = \frac{4}{3}\ln 4 - 1 = \frac{8}{3}\ln 2 - 1 \Rightarrow a = \frac{8}{3};b = - 1 \Rightarrow S = 3a - b = 9.\)
\( = \ln 4 - 1 + \frac{1}{3}\ln 4 = \frac{4}{3}\ln 4 - 1 = \frac{8}{3}\ln 2 - 1 \Rightarrow a = \frac{8}{3};b = - 1 \Rightarrow S = 3a - b = 9.\)
Cho biết \(\int\limits_1^2 {\ln \left( {9 - {{\rm{x}}^2}} \right)d{\rm{x}}} = a\ln 5 + b\ln 2 + c,\) với a, b, c là các số nguyên. Tính \(S = \left| a \right| + \left| b \right| + \left| c \right|.\)
A. S = 34
B. S = 18
C. S = 26
D. S = 13
\(\int\limits_1^2 {\ln \left( {9 - {x^2}} \right)d{\rm{x}}} = \left. {x\ln \left( {9 - {x^2}} \right)} \right|_1^2 + 2\int\limits_1^2 {\frac{{{x^2}dx}}{{9 - {x^2}}}} = 2\ln 5 - 3\ln 2 + 2\int\limits_1^2 {\frac{{{x^2}dx}}{{9 - {x^2}}}} .\)
\(\begin{array}{l}\int\limits_1^2 {\frac{{{x^2}dx}}{{9 - {x^2}}}} = \int\limits_1^2 {\frac{3}{2}\left( {\frac{1}{{3 - x}} + \frac{1}{{3 + x}}} \right)d{\rm{x}}} = \left. {\left( { - \frac{{3\ln \left| {3 - x} \right|}}{2} + \frac{{3\ln \left| {3 + x} \right|}}{2} - x} \right)} \right|_1^2\\ = \frac{3}{2}\ln 5 + \frac{3}{2}\ln 2 - \frac{3}{2}\ln 4 - 1\end{array}\)
\( \Rightarrow \int\limits_1^2 {\ln \left( {9 - {x^2}} \right)d{\rm{x}}} = 5\ln 5 - 6\ln 2 - 2 \Rightarrow S = 13.\)
\(\begin{array}{l}\int\limits_1^2 {\frac{{{x^2}dx}}{{9 - {x^2}}}} = \int\limits_1^2 {\frac{3}{2}\left( {\frac{1}{{3 - x}} + \frac{1}{{3 + x}}} \right)d{\rm{x}}} = \left. {\left( { - \frac{{3\ln \left| {3 - x} \right|}}{2} + \frac{{3\ln \left| {3 + x} \right|}}{2} - x} \right)} \right|_1^2\\ = \frac{3}{2}\ln 5 + \frac{3}{2}\ln 2 - \frac{3}{2}\ln 4 - 1\end{array}\)
\( \Rightarrow \int\limits_1^2 {\ln \left( {9 - {x^2}} \right)d{\rm{x}}} = 5\ln 5 - 6\ln 2 - 2 \Rightarrow S = 13.\)
Cho tích phân \(I = \int\limits_0^\pi {{x^2}\cos xdx} \) và \(u = {x^2},dv = \cos xdx\). Khẳng định nào sau đây đúng?
A. \(I = {x^2}\sin x\left| {\begin{array}{*{20}{c}}\pi \\0\end{array}} \right. - \int\limits_0^\pi {x\sin xdx} \)
B. \(I = {x^2}\sin x\left| {\begin{array}{*{20}{c}}\pi \\0\end{array}} \right. + \int\limits_0^\pi {x\sin xdx} \)
C. \(I = {x^2}\sin x\left| {\begin{array}{*{20}{c}}\pi \\0\end{array}} \right. + 2\int\limits_0^\pi {x\sin xdx} \)
D. \(I = {x^2}\sin x\left| {\begin{array}{*{20}{c}}\pi \\0\end{array}} \right. - 2\int\limits_0^\pi {x\sin xdx} \)
\(\begin{array}{l}\left\{ \begin{array}{l}u = {x^2}\\dv = \cos xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = 2xdx\\v = \sin xdx\end{array} \right.\\ \Rightarrow I = \int\limits_0^\pi {{x^2}\cos xdx} = \left. {{x^2}\sin x} \right|_0^\pi - 2\int\limits_0^\pi {x\sin xdx} .\end{array}\)
Cho hàm số \(y = f\left( x \right)\) thỏa mãn \(f'\left( x \right) = \left( {x + 1} \right){e^x}\) và \(\int {f\left( x \right)} dx = \left( {ax + b} \right){e^x} + c\), với a, b, c là các hằng số. Khi đó:
A. \(a + b = 2\)
B. \(a + b = 3\)
C. \(a + b = 0\)
D. \(a + b = 1\)
\(f'\left( x \right) = \left( {x + 1} \right){e^x} \Rightarrow f\left( x \right) = x{e^x} + C'.\)
Khi đó đặt \(I = \int {\left( {x{e^x} + C'} \right)} dx = \int {x{e^x}dx} + C'x\)
Đặt \(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = {e^x}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = {e^x}}\end{array} \Rightarrow I = Cx + x{e^x}} \right. - \int {{e^x}dx = C'x + x{e^x} - {e^x}} = \left( {x - 1} \right){e^x} + C'x + c\)
Suy ra C’=0.
Do đó \(a = 1,b = - 1 \Rightarrow a + b = 0.\)
Khi đó đặt \(I = \int {\left( {x{e^x} + C'} \right)} dx = \int {x{e^x}dx} + C'x\)
Đặt \(\left\{ {\begin{array}{*{20}{c}}{u = x}\\{dv = {e^x}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = {e^x}}\end{array} \Rightarrow I = Cx + x{e^x}} \right. - \int {{e^x}dx = C'x + x{e^x} - {e^x}} = \left( {x - 1} \right){e^x} + C'x + c\)
Suy ra C’=0.
Do đó \(a = 1,b = - 1 \Rightarrow a + b = 0.\)
Tìm nguyên hàm của hàm số \(f\left( x \right) = x\ln \left( {x + 2} \right)\).
A. \(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^2}}}{2}\ln \left( {x + 2} \right) - \frac{{{x^2} + 4x}}{4} + C\).
B. \(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^2} - 4}}{2}\ln \left( {x + 2} \right) - \frac{{{x^2} - 4x}}{4} + C\).
C. \(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^2}}}{2}\ln \left( {x + 2} \right) - \frac{{{x^2} + 4x}}{2} + C\).
D. \(\int {f\left( x \right){\rm{d}}x} = \frac{{{x^2} - 4}}{2}\ln \left( {x + 2} \right) - \frac{{{x^2} + 4x}}{2} + C\).
Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x + 2} \right)\\{\rm{d}}v = x{\rm{d}}x\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\rm{d}}u = \frac{{{\rm{d}}x}}{{x + 2}}\\v = \frac{{{x^2}}}{2}\end{array} \right.\).
Ta có:
\(\begin{array}{l}\int {f\left( x \right){\rm{d}}x} = \frac{{{x^2}}}{2}\ln \left( {x + 2} \right) - \frac{1}{2}\int {\frac{{{x^2}}}{{x + 2}}{\rm{d}}x} = \frac{{{x^2}}}{2}\ln \left( {x + 2} \right) - \frac{1}{2}\int {\left( {x - 2 + \frac{4}{{x + 2}}} \right){\rm{d}}x} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^2}}}{2}\ln \left( {x + 2} \right) - \frac{1}{2}\left( {\frac{{{x^2}}}{2} - 2x + 4\ln \left( {x + 2} \right)} \right) + C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^2} - 4}}{2}\ln \left( {x + 2} \right) - \frac{{{x^2} - 4x}}{4} + C.\end{array}\)
Ta có:
\(\begin{array}{l}\int {f\left( x \right){\rm{d}}x} = \frac{{{x^2}}}{2}\ln \left( {x + 2} \right) - \frac{1}{2}\int {\frac{{{x^2}}}{{x + 2}}{\rm{d}}x} = \frac{{{x^2}}}{2}\ln \left( {x + 2} \right) - \frac{1}{2}\int {\left( {x - 2 + \frac{4}{{x + 2}}} \right){\rm{d}}x} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^2}}}{2}\ln \left( {x + 2} \right) - \frac{1}{2}\left( {\frac{{{x^2}}}{2} - 2x + 4\ln \left( {x + 2} \right)} \right) + C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x^2} - 4}}{2}\ln \left( {x + 2} \right) - \frac{{{x^2} - 4x}}{4} + C.\end{array}\)
Cho \(I = \int\limits_0^{\frac{\pi }{4}} {\left( {x - 1} \right)\sin 2xdx} \). Tìm đẳng thức đúng?
A. \(I = - \left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. + \int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
B. \(I = - \left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. - \int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
C. \(I = - \frac{1}{2}\left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. + \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
D. \(I = - \frac{1}{2}\left( {x - 1} \right)\cos 2x\left| {\begin{array}{*{20}{c}}{\frac{\pi }{4}}\\0\end{array}} \right. - \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2x} dx\)
Đặt \(\left\{ {\begin{array}{*{20}{c}}{u = x - 1}\\{dv = \sin 2xdx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = dx}\\{v = - \frac{1}{2}\cos 2x}\end{array} \Rightarrow I = } \right.\left. { - \frac{1}{2}\left( {x - 1} \right)\cos 2x} \right|_0^{\frac{\pi }{4}} + \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2xdx} .\)
Giả sử hàm số f có đạo hàm liên tục trên đoạn \(\left[ {0;1} \right],\) thỏa mãn điều kiện \(f\left( 1 \right) = 6\) và \(\int\limits_0^1 {xf'\left( x \right)d{\rm{x}}} = 5.\) Khi đó \(\int\limits_0^1 {f\left( x \right)d{\rm{x}}} \) bằng:
A. 1
B. -1
C. 11
D. 3
Đặt \(\left\{ \begin{array}{l}u = x\\dv = {f'}\left( x \right)d{\rm{x}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = d{\rm{x}}\\v = f\left( x \right)\end{array} \right.\)
\(\Rightarrow \int\limits_0^1 {x{f'}\left( x \right)d{\rm{x}}} = \left. {xf\left( x \right)} \right|_0^1 - \int\limits_0^1 {f\left( x \right)d{\rm{x}}} = f\left( 1 \right) - \int\limits_0^1 {f\left( x \right)d{\rm{x}}} = 5\)
\( \Rightarrow \int\limits_0^1 {f\left( x \right)d{\rm{x}}} = f\left( 1 \right) - 5 = 6 - 5 = 1.\)
\(\Rightarrow \int\limits_0^1 {x{f'}\left( x \right)d{\rm{x}}} = \left. {xf\left( x \right)} \right|_0^1 - \int\limits_0^1 {f\left( x \right)d{\rm{x}}} = f\left( 1 \right) - \int\limits_0^1 {f\left( x \right)d{\rm{x}}} = 5\)
\( \Rightarrow \int\limits_0^1 {f\left( x \right)d{\rm{x}}} = f\left( 1 \right) - 5 = 6 - 5 = 1.\)
Cho hàm số \(y = f\left( x \right)\) thỏa mãn hệ thức \(\int {f\left( x \right)} \sin xdx = - f\left( x \right)\cos x + \int {{\pi ^x}\cos xdx} \). Hỏi \(y = f\left( x \right)\) là hàm số nào trong các hàm số sau:
A. \(f\left( x \right) = - \frac{{{\pi ^x}}}{{\ln \pi }}\)
B. \(f\left( x \right) = \frac{{{\pi ^x}}}{{\ln \pi }}\)
C. \(f\left( x \right) = {\pi ^x}.\ln x\)
D. \(f\left( x \right) = - {\pi ^x}.\ln x\)
Đặt: \(\left\{ {\begin{array}{*{20}{c}}{u = f\left( x \right)}\\{dv = \sin xdx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = f'\left( x \right)dx}\\{v = - \cos x}\end{array}} \right. \Rightarrow \int {f\left( x \right)} \sin xdx\)
\( = - f\left( x \right).\cos x + \int {f'\left( x \right)} \cos xdx.\)
Nên suy ra \(f'\left( x \right) = {\pi ^x} \Rightarrow f\left( x \right) = \int {{\pi ^x}} dx = \frac{{{\pi ^x}}}{{\ln \pi }}\)
\( = - f\left( x \right).\cos x + \int {f'\left( x \right)} \cos xdx.\)
Nên suy ra \(f'\left( x \right) = {\pi ^x} \Rightarrow f\left( x \right) = \int {{\pi ^x}} dx = \frac{{{\pi ^x}}}{{\ln \pi }}\)
Tính tích phân \(I = \int\limits_0^1 {3x.{e^{2x}}} dx.\)
A. \(I = \frac{{3{e^2} + 3}}{{16}}\)
B. \(I = \frac{{2{e^2} + 2}}{9}\)
C. \(I = \frac{{3{e^2} + 3}}{4}\)
D. \(I = \frac{{2{e^2} + 2}}{3}\)
Đặt \(\left\{ {\begin{array}{*{20}{c}}{u = 3x}\\{dv = {e^{2x}}dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = 3dx}\\{v = \frac{{{e^{2x}}}}{2}}\end{array}} \right. \Rightarrow I = \frac{{3x.{e^{2x}}}}{2}\left| {\begin{array}{*{20}{c}}1\\0\end{array}} \right. - \frac{3}{2}\int\limits_0^1 {{e^{2x}}dx} = \frac{{3x.{e^{2x}}}}{2}\left| {\begin{array}{*{20}{c}}1\\0\end{array}} \right. - \frac{3}{4}{e^{2x}}\left| {\begin{array}{*{20}{c}}1\\0\end{array} = \frac{{3{e^2} + 3}}{4}} \right..\)