Dùng phương pháp tích phân từng phần để tính các tích phân sau:
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 \(\int\limits_0^{{\pi \over 2}} {x\cos xdx }\)
\( = \left. {x\sin x} \right|_0^{\frac{\pi }{2}} - \int\limits_0^{\frac{\pi }{2}} {\sin xdx} \) \( = \frac{\pi }{2}\sin \frac{\pi }{2} - 0 + \left. {\cos x} \right|_0^{\frac{\pi }{2}}\) \( = \frac{\pi }{2} + \cos \frac{\pi }{2} - \cos 0 = \frac{\pi }{2} - 1\)
