Bài 18 Trang 161 SGK Đại số và Giải tích 12 Nâng cao

LG a

\(\int\limits_1^2 {{x^5}} \ln xdx;\)

Lời giải chi tiết:

Đặt 

\(\left\{ \matrix{
u = \ln x \hfill \cr 
dv = {x^5}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = {{dx} \over x} \hfill \cr 
v = {{{x^6}} \over 6} \hfill \cr} \right.\)

\(\int\limits_1^2 {{x^5}} \ln xdx = \left. {{{{x^6}} \over 6}\ln x} \right|_1^2 – {1 \over 6}\int\limits_1^2 {{x^5}} dx \) \(= \left. {\left( {{{{x^6}} \over 6}\ln x – {{{x^6}} \over {36}}} \right)} \right|_1^2 \) 

\( = \left( {\dfrac{{64}}{6}\ln 2 – \dfrac{1}{6}\ln 1} \right) – \dfrac{1}{6}.\left. {\dfrac{{{x^6}}}{6}} \right|_1^2\) \( = \dfrac{{32}}{3}\ln 2 – \dfrac{1}{6}\left( {\dfrac{{64}}{6} – \dfrac{1}{6}} \right)\) \( = \dfrac{{32}}{3}\ln 2 – \dfrac{7}{4}\)

LG b

\(\int\limits_0^1 {\left( {x + 1} \right)} {e^x}dx;\)  

Lời giải chi tiết:

Đặt 

\(\left\{ \matrix{
u = x + 1 \hfill \cr 
dv = {e^x}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = dx \hfill \cr 
v = {e^x} \hfill \cr} \right.\)

\(\int\limits_0^1 {\left( {x + 1} \right)} {e^x}dx \) \(= \left. {\left( {x + 1} \right){e^x}} \right|_0^1 – \int\limits_0^1 {{e^x}dx } \) \( = 2e – 1 – \left. {{e^x}} \right|_0^1\) \( = 2e – 1 – \left( {e – 1} \right) = e\)

LG c

\(\int\limits_0^\pi  {{e^x}} \cos xdx;\)

Lời giải chi tiết:

Đặt \(I = \int\limits_0^\pi  {{e^x}\cos xdx} \)

Đặt

\(\left\{ \matrix{
u = {e^x} \hfill \cr 
dv = \cos xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = {e^x}dx \hfill \cr 
v = {\mathop{\rm s}\nolimits} {\rm{inx}} \hfill \cr} \right.\)

Suy ra \(I = \left. {{e^x}{\mathop{\rm s}\nolimits} {\rm{inx}}} \right|_0^\pi  – \int\limits_0^\pi  {{e^x}\sin {\rm{x}}dx}  \) \( = {e^\pi }\sin \pi  – {e^0}\sin 0 – \int\limits_0^\pi  {{e^x}\sin xdx} \) \( = 0 – \int\limits_0^\pi  {{e^x}\sin xdx} \) \(=  – \int\limits_0^\pi  {{e^x}\sin {\rm{x}}dx} \) 

Đặt 

\(\left\{ \matrix{
u = {e^x} \hfill \cr 
dv = \sin {\rm{x}}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = {e^x}dx \hfill \cr 
v = – \cos x \hfill \cr} \right.\)

Do đó \(I =  – \left[ {\left. {\left( { – {e^x}\cos x} \right)} \right|_0^\pi  + \int\limits_0^\pi  {{e^x}\cos xdx} } \right] \) \(= {e^\pi }\cos \pi  – {e^0}.\cos 0 – I\)

\( \Rightarrow 2I =  – {e^\pi } – 1 \Rightarrow I =  – {1 \over 2}\left( {{e^\pi } + 1} \right)\)

LG d

\(\int\limits_0^{{\pi  \over 2}} {x\cos xdx.} \)

Lời giải chi tiết:

Đặt 

\(\left\{ \matrix{
u = x \hfill \cr 
dv = \cos xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = dx \hfill \cr 
v = {\mathop{\rm s}\nolimits} {\rm{inx}} \hfill \cr} \right.\)

Do đó \(\int\limits_0^{{\pi  \over 2}} {x\cos xdx }\) 

\( = \left. {x\sin x} \right|_0^{\frac{\pi }{2}} – \int\limits_0^{\frac{\pi }{2}} {\sin xdx} \) \( = \frac{\pi }{2}\sin \frac{\pi }{2} – 0 + \left. {\cos x} \right|_0^{\frac{\pi }{2}}\) \( = \frac{\pi }{2} + \cos \frac{\pi }{2} – \cos 0 = \frac{\pi }{2} – 1\)

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