Bài tập 3.16 trang 170 SBT Toán 12

a)

\(\begin{array}{l}
\int\limits_0^1 {\left( {{y^3} + 3{y^2} - 2} \right)dy}  = \int\limits_0^1 {{y^3}dy}  + \smallint 013{y^2}dy - \int\limits_0^1 {2dy} \\
 = \frac{1}{4}\left. {{y^4}} \right|_0^1 + \left. {{y^3}} \right|_0^1 - 2\left. y \right|_0^1 = \frac{1}{4} + 1 - 2 = \frac{3}{4}
\end{array}\)

b) ${\displaystyle \frac{\mathrm{}}{}}$

\(\begin{array}{l}
\int \limits_1^4 (t + \frac{1}{{\sqrt t }} - \frac{1}{{{t^2}}})dt = \left. {\left( {\frac{{{t^2}}}{2} + 2\sqrt t  + \frac{1}{4}} \right)} \right|_1^4\\
 = \frac{{{4^2}}}{2} + 2\sqrt 4  + \frac{1}{4} - \frac{1}{2} - 2\sqrt 1  - \frac{1}{1} = \frac{{35}}{4}
\end{array}\)

c) \(\int \limits_0^{\frac{\pi }{2}} (2\cos x - \sin 2x)dx = \left. {\left( {2\sin x + \frac{{\cos 2x}}{2}} \right)} \right|_0^{\frac{\pi }{2}} = 2.1 - \frac{1}{2} - 2.0 - \frac{1}{2} = 1\)

d)

\(\begin{array}{l}
\int \limits_0^1 {({3^s} - {2^s})^2}ds = \int \limits_0^1 \left( {{3^{2s}} - {{2.6}^s} + {2^{2s}}} \right)ds\\
 = \int \limits_0^1 \left( {{9^s} - {{2.6}^s} + {4^s}} \right)ds = \left. {\left( {\frac{{{9^s}}}{{\ln 9}} - 2.\frac{{{6^s}}}{{\ln 6}} + \frac{{{4^s}}}{{\ln 4}}} \right)} \right|_0^1\\
 = \frac{9}{{\ln 9}} - 2.\frac{6}{{\ln 6}} + \frac{4}{{\ln 4}} - \frac{1}{{\ln 9}} + 2.\frac{1}{{\ln 6}} - \frac{1}{{\ln 4}}\\
 = \frac{8}{{\ln 9}} - \frac{{10}}{{\ln 6}} + \frac{3}{{\ln 4}} = \frac{4}{{\ln 3}} - \frac{{10}}{{\ln 6}} + \frac{3}{{2\ln 2}}
\end{array}\)

e)  

\(\begin{array}{l}
\int \limits_0^{\frac{\pi }{3}} \cos 3xdx + \int \limits_{\frac{\pi }{3}}^{\frac{{3\pi }}{2}} \cos 3xdx + \int \limits_{\frac{{3\pi }}{2}}^{\frac{{5\pi }}{2}} \cos 3xdx\\
 = \int\limits_0^{\frac{{5\pi }}{2}} {\cos 3xdx}  = \left. {\frac{{\sin 3x}}{3}} \right|_0^{\frac{{5\pi }}{2}} = \frac{{\sin \frac{{15\pi }}{2}}}{3} - \frac{{\sin 0}}{3} =  - \frac{1}{3}
\end{array}\)

-- Mod Toán 12 HỌC247