Giải bài 4 trang 113 SGK Giải tích 12

LG a

\(\int_{0}^{\dfrac{\pi}{2}}(x+1)\sin xdx\)

Phương pháp giải:

Phương pháp tích phân từng phần: \(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b – \int\limits_a^b {vdu} \).

Đặt \(\left\{ \begin{array}{l}u = x + 1\\dv = \sin xdx\end{array} \right.\)

Lời giải chi tiết:

Đặt \(\left\{ \begin{array}{l}u = x + 1\\dv = \sin xdx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = dx\\v = – \cos x\end{array} \right.\)

\(\begin{array}{l}\Rightarrow \int\limits_0^{\frac{\pi }{2}} {\left( {x + 1} \right)\sin xdx} \\= \left. { – \left( {x + 1} \right)\cos x} \right|_0^{\frac{\pi }{2}} + \int\limits_0^{\frac{\pi }{2}} {\cos xdx} \\= \left. { – \left( {x + 1} \right)\cos x} \right|_0^{\frac{\pi }{2}} + \left. {\sin x} \right|_0^{\frac{\pi }{2}}\end{array}\)

\( =  – \left( {\frac{\pi }{2} + 1} \right)\cos \frac{\pi }{2} + \left( {0 + 1} \right)\cos 0 \)\(+ \sin \frac{\pi }{2} – \sin 0\)

\(=0+1+1-0=2\)

LG b

\(\int_{1}^{e}x^{2}\ln xdx\)

Phương pháp giải:

Phương pháp tích phân từng phần: \(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b – \int\limits_a^b {vdu} \).

Đặt \(\left\{ \begin{array}{l}u = \ln x\\dv = {x^2}dx\end{array} \right.\)

Lời giải chi tiết:

Đặt \(\left\{ \begin{array}{l}u = \ln x\\dv = {x^2}dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \frac{{dx}}{x}\\v = \frac{{{x^3}}}{3}\end{array} \right.\)

\(\begin{array}{l}\Rightarrow \int\limits_1^e {{x^2}\ln x} dx \\= \left. {\left( {\ln x.\frac{{{x^3}}}{3}} \right)} \right|_1^e – \frac{1}{3}\int\limits_1^e {{x^2}dx} \\= \left. {\left( {\ln x.\frac{{{x^3}}}{3}} \right)} \right|_1^e – \left. {\frac{{{x^3}}}{9}} \right|_1^e\end{array}\)

\(\begin{array}{l}
= \ln e.\frac{{{e^3}}}{3} – \ln 1.\frac{{{1^3}}}{3} – \left( {\frac{{{e^3}}}{9} – \frac{{{1^3}}}{9}} \right)\\
= \frac{{{e^3}}}{3} – 0 – \frac{{{e^3}}}{9} + \frac{1}{9}\\
= \frac{{2{e^3}}}{9} + \frac{1}{9}\\
= \frac{1}{9}\left( {2{e^3} + 1} \right)
\end{array}\)

LG c

\(\int_{0}^{1}\ln(1+x)dx\);      

Phương pháp giải:

Phương pháp tích phân từng phần: \(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b – \int\limits_a^b {vdu} \).

Đặt \(\left\{ \begin{array}{l}u = \ln \left( {1 + x} \right)\\dv = dx\end{array} \right.\)

Lời giải chi tiết:

Đặt \(\left\{ \begin{array}{l}u = \ln \left( {1 + x} \right)\\dv = dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \frac{{dx}}{{1 + x}}\\v = x\end{array} \right.\)

\(\begin{array}{l}\Rightarrow \int\limits_0^1 {\ln \left( {x + 1} \right)dx} \\= \left. {\left( {x.\ln \left( {1 + x} \right)} \right)} \right|_0^1 – \int\limits_0^1 {\frac{x}{{x + 1}}dx} \\= \left. {\left( {x.\ln \left( {1 + x} \right)} \right)} \right|_0^1 – \int\limits_0^1 {\frac{{x + 1 – 1}}{{x + 1}}dx} \\= \left. {\left( {x.\ln \left( {1 + x} \right)} \right)} \right|_0^1 – \int\limits_0^1 {\left( {1 – \frac{1}{{x + 1}}} \right)dx} \\= \left. {\left( {x.\ln \left( {1 + x} \right)} \right)} \right|_0^1 – \left. {\left( {x – \ln \left| {x + 1} \right|} \right)} \right|_0^1\end{array}\)

\(\begin{array}{l}
= 1.\ln \left( {1 + 1} \right) – 0.\ln \left( {0 + 1} \right)\\
– \left( {1 – \ln |1+1| – 0 + \ln |0+1|} \right)\\
= \ln 2 – 1 + \ln 2\\
= 2\ln 2 – 1
\end{array}\)

LG d

\(\int_{0}^{1}(x^{2}-2x-1)e^{-x}dx\)

Phương pháp giải:

Phương pháp tích phân từng phần: \(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b – \int\limits_a^b {vdu} \).

Đặt \(\left\{ \begin{array}{l}u = {x^2} – 2x – 1\\dv = {e^{ – x}}dx\end{array} \right.\)

Lời giải chi tiết:

Đặt \(\left\{ \begin{array}{l}u = {x^2} – 2x + 1\\dv = {e^{ – x}}dx\end{array} \right. \)\(\Rightarrow \left\{ \begin{array}{l}du = \left( {2x – 2} \right)dx\\v = – {e^{ – x}}\end{array} \right.\)

\(\begin{array}{l}\Rightarrow \int\limits_0^1 {\left( {{x^2} – 2x – 1} \right){e^{ – x}}dx} \\= \left. { – {e^{ – x}}\left( {{x^2} – 2x – 1} \right)} \right|_0^1 \\+ 2\int\limits_0^1 {\left( {x – 1} \right){e^{ – x}}dx} \\= \left. { – {e^{ – x}}\left( {{x^2} – 2x – 1} \right)} \right|_0^1 + 2{I_1}\\= 2{e^{ – 1}} – 1 + 2{I_1}\end{array}\)

Đặt \(\left\{ \begin{array}{l}u = x – 1\\dv = {e^{ – x}}\end{array} \right. \)\(\Rightarrow \left\{ \begin{array}{l}du = dx\\v = – {e^{ – x}}\end{array} \right.\).

\(\begin{array}{l}\Rightarrow {I_1} = \left. { – {e^{ – x}}\left( {x – 1} \right)} \right|_0^1 + \int\limits_0^1 {{e^{ – x}}dx} \\= \left. { – {e^{ – x}}\left( {x – 1} \right)} \right|_0^1\left. { – {e^{ – x}}} \right|_0^1\\= – 1 – \left( {{e^{ – 1}} – 1} \right) =- {e^{ – 1}}\end{array}\).

Vậy \(I = 2{e^{ – 1}} – 1 – 2{e^{ – 1}} =  – 1\).

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