Bài tập 3.5 trang 164 SBT Toán 12

Bài tập 3.5 trang 164 SBT Toán 12

a) \(I = \mathop \smallint \nolimits \left( {1 - 2x} \right){e^x}dx\)

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

Ta có:

\({I = \left( {1 - 2x} \right){e^x} + \mathop \smallint \nolimits 2{e^x}dx + C = {e^x} - 2x{e^x} + 2{e^x} + C = \left( {3 - 2x} \right){e^x} + C}\)

b) \(J = \mathop \smallint \nolimits x{e^{ - x}}dx\)

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

Ta có:

\({I =  - x{e^{ - x}} + \mathop \smallint \nolimits {e^{ - x}}dx + C =  - x{e^{ - x}} - {e^{ - x}} + C =  - \left( {1 + x} \right){e^{ - x}} + C}\)

c) \(G = \mathop \smallint \nolimits x\ln \left( {1 - x} \right)dx\)

Đặt \(\left\{ \begin{array}{l}
u = \ln \left( {1 - x} \right)\\
dv = xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \frac{1}{{x - 1}}dx\\
v = \frac{{{x^2}}}{2}
\end{array} \right.\)

Ta có: 

\(\begin{array}{l}
G = \frac{{{x^2}}}{2}\ln \left( {1 - x} \right) - \frac{1}{2}\int {\frac{{{x^2}}}{{x - 1}}} dx\\
 = \frac{{{x^2}}}{2}\ln \left( {1 - x} \right) - \frac{1}{2}\int {\left( {x + 1 + \frac{1}{{x - 1}}} \right)} dx\\
 = \frac{{{x^2}}}{2}\ln \left( {1 - x} \right) - \frac{1}{2}\left[ {\frac{{{x^2}}}{2} + x + \ln \left( {1 - x} \right)} \right] + C\\
 = \frac{{{x^2}}}{2}\ln \left( {1 - x} \right) - \frac{1}{2}\ln \left( {1 - x} \right) - \frac{1}{4}{x^2} - \frac{1}{2}x + C
\end{array}\)

d) Ta có:

\(H = \mathop \smallint \nolimits x{\sin ^2}xdx = \mathop \smallint \nolimits x.\frac{{1 - cos2x}}{2}dx = \frac{{{x^2}}}{4} - \frac{1}{2}\mathop \smallint \nolimits^ x\cos 2xdx = \frac{{{x^2}}}{4} - \frac{1}{2}I\)

Đặt \(\left\{ \begin{array}{l}
u = x\\
dv = \cos 2xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v = \frac{1}{2}\sin 2x
\end{array} \right.\)

Suy ra:

\({I = \frac{1}{2}x\sin 2x - \frac{1}{2}\mathop \smallint \nolimits \sin 2xdx = \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x + C}\)

Vậy \(H = \frac{{{x^2}}}{4} - \frac{1}{2}\left( {\frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x} \right) + C\)

-- Mod Toán 12