Bài tập 18 trang 219 SBT Toán 12

a) \(\begin{array}{l}
\int\limits_{ - 1}^2 {\left( {5{x^2} - x + {e^{0,5x}}} \right)dx} \\
= \left. {\left( {\dfrac{{5{x^3}}}{3} - \dfrac{{{x^2}}}{2} + \dfrac{1}{{0,5}}{e^{0,5x}}} \right)} \right|_{ - 1}^2\\
= \dfrac{{34}}{3} + 2e - \left( { - \dfrac{{13}}{6} + 2{e^{ - \dfrac{1}{2}}}} \right)\\
= \dfrac{{27}}{2} + 2e - \dfrac{2}{{\sqrt e }}
\end{array}\)

b) \(\begin{array}{l}
\int\limits_{0,5}^2 {\left( {2\sqrt x - \dfrac{3}{{{x^3}}} + \cos x} \right)dx} \\
= \int\limits_{0,5}^2 {\left( {2{x^{\dfrac{1}{2}}} - 3{x^{ - 3}} + \cos x} \right)dx} \\
= \left. {\left( {2.\dfrac{{{x^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}} - 3.\dfrac{{{x^{ - 2}}}}{{ - 2}} + \sin x} \right)} \right|_{0,5}^2\\
= \left. {\left( {\dfrac{1}{3}{x^{\dfrac{3}{2}}} + \dfrac{3}{{2{x^2}}} + \sin x} \right)} \right|_{0,5}^2\\
= \dfrac{{7\sqrt 2 }}{3} - \dfrac{{45}}{8} + \sin 2 - \sin \dfrac{1}{2}
\end{array}\)

c) Đặt \(t = \sqrt {2x + 3}  \Rightarrow {t^2} = 2x + 3\) \( \Rightarrow 2tdt = 2dx \Rightarrow dx = tdt\)

Đổi cận \(x = 1 \Rightarrow t = \sqrt 5 ,\) \(x = 2 \Rightarrow t = \sqrt 7 \)

Khi đó \(I = \int\limits_{\sqrt 5 }^{\sqrt 7 } {\dfrac{{tdt}}{t}}  = \int\limits_{\sqrt 5 }^{\sqrt 7 } {dt} \) \( = \left. t \right|_{\sqrt 5 }^{\sqrt 7 } = \sqrt 7  - \sqrt 5 \)

d) Đổi biến  \(t = \root 3 \of {3{x^3} + 4} \)

\(\Rightarrow {t^3} = 3{x^3} + 4 \Rightarrow 3{t^2}dt = 9{x^2}dx \) \(\Rightarrow {x^2}dx = {1 \over 3}{t^2}dt\)

Ta có 

\(\eqalign{
& \int\limits_1^2 {\root 3 \of {3{x^3} + 4} } {x^2}dx = {1 \over 3}\int\limits_{\root 3 \of 7 }^{\root 3 \of {28} } {{t^3}dt} \cr & = {1 \over {12}}{t^4}\left| {\matrix{{\root 3 \of {28} } \cr {\root 3 \of 7 } \cr} } \right. = {{7\root 3 \of 7 (4\root 3 \of 4 - 1)} \over {12}} \cr} \) 

e) \(\eqalign{
& \int\limits_{ - 2}^2 {(x - 2)|x|dx} \cr 
& = \int\limits_{ - 2}^0 {(2x - {x^2})dx + \int\limits_0^2 {({x^2} - 2x)dx} } \cr 
& = - {{20} \over 3} - {4 \over 3} = - 8 \cr} \)

g) \(\eqalign{& \int\limits_1^0 {x\cos xdx = x\sin x\left| {\matrix{0 \cr 1 \cr} } \right.} - \int\limits_1^0 {\sin xdx} \cr & = - \sin 1 + \cos x\left| {\matrix{0 \cr 1 \cr} } \right. = 1 - (\sin 1 + \cos 1) \cr} \)

h) 

Ta có:  

\(\eqalign{
& 1 + \sin 2x + \cos 2x \cr 
& = 1 + 2\sin x\cos x + 2{\cos ^2}x - 1 \cr 
& = 2\cos x(\sin x + \cos x) \cr} \)

\(\begin{array}{l}
\Rightarrow I = \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} {\dfrac{{2\cos x\left( {\sin x + \cos x} \right)}}{{\sin x + \cos x}}dx} \\
= \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} {2\cos xdx} = 2\left. {\sin x} \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}\\
= 2\left( {1 - \dfrac{1}{2}} \right) = 1
\end{array}\)

i) Áp dụng phương pháp tính tích phân từng phần hai lần, cả hai lần đều đặt \({e^x}dx = dv \Rightarrow v = {e^x}\) . Ta có:

\(\eqalign{& I = \int\limits_0^{{\pi \over 2}} {{e^x}\sin xdx} = {e^x}\sin x\left| {\matrix{{{\pi \over 2}} \cr 0 \cr} } \right. - \int\limits_0^{{\pi \over 2}} {{e^x}\cos xdx} \cr & = {e^{{\pi \over 2}}} - \left[ {{e^x}\cos x\left| {\matrix{{{\pi \over 2}} \cr 0 \cr} + \int\limits_0^{{\pi \over 2}} {{e^x}\sin xdx} } \right.} \right] \cr & = {e^{{\pi \over 2}}} + 1 - I \cr & \Rightarrow I = {{{e^{{\pi \over 2}}} + 1} \over 2} \cr} \)

-- Mod Toán 12 HỌC247