Giải bài 3.44 trang 180 - SBT Đại số và Giải tích lớp 12
Tính các tích phân sau:
a) \(\int\limits_{0}^{1}{{{\left( y-1 \right)}^{2}}\sqrt{y}dy}\) , đặt \(t = \sqrt{y}\)
b) \(\int\limits_{1}^{2}{\left( {{z}^{2}}+1 \right)\sqrt[3]{{{\left( z-1 \right)}^{2}}}dz}\) , đặt \(u=\sqrt[3]{z-1}\)
c) \(\int\limits_{1}^{e}{\dfrac{\sqrt{4+5\ln x}}{x}dx}\)
d) \(\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\cos }^{5}}\varphi -\sin {{\,}^{5}}\varphi \right)d\varphi }\)
(Đề thi tốt nghiệp THPT năm 2011)
a)
Ta có:
\(\begin{aligned} & \int\limits_{0}^{1}{{{\left( y-1 \right)}^{2}}\sqrt{y}dy}=\int\limits_{0}^{1}{{{\left( {{t}^{2}}-1 \right)}^{2}}t.2tdt} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int\limits_{0}^{1}{\left( 2{{t}^{6}}-4{{t}^{4}}+2{{t}^{2}} \right)dt} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \dfrac{2}{7}{{t}^{7}}-\dfrac{4}{5}{{t}^{5}}+\dfrac{2}{3}{{t}^{3}} \right)\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{16}{105} \\ \end{aligned}\)
b)
Đặt \(u=\sqrt[3]{z-1}\Rightarrow {{u}^{3}}=z-1\Rightarrow \left\{ \begin{aligned} & 3{{u}^{2}}du=dz \\ & z={{u}^{3}}+1 \\ \end{aligned} \right.\)
Ta có:
\(\begin{aligned} & \int\limits_{1}^{2}{\left( {{z}^{2}}+1 \right)\sqrt[3]{{{\left( z-1 \right)}^{2}}}dz}=\int\limits_{0}^{1}{\left[ {{\left( {{u}^{3}}+1 \right)}^{2}}+1 \right].{{u}^{2}}.3{{u}^{2}}du} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int\limits_{0}^{1}{\left( 3{{u}^{10}}+6{{u}^{7}}+6{{u}^{4}} \right)du} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \dfrac{3}{11}{{u}^{11}}+\dfrac{3}{4}{{u}^{8}}+\dfrac{6}{5}{{u}^{5}} \right)\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{489}{220} \\ \end{aligned}\)
c) Đặt \(u=\sqrt{4+5\ln x}\Rightarrow {{u}^{2}}=4+5\ln x\Rightarrow 2udu=\dfrac{5dx}{x}\)
\(\int\limits_{1}^{e}{\dfrac{\sqrt{4+5\ln x}}{x}dx}=\int\limits_{2}^{3}{2{{u}^{2}}du}=\dfrac{2}{3}{{u}^{3}}\left| \begin{aligned} & 3 \\ & 2 \\ \end{aligned} \right.=\dfrac{38}{3}\)
d) \(\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\cos }^{5}}\varphi -\sin {{\,}^{5}}\varphi \right)d\varphi }=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{5}}\varphi d\varphi }-\int\limits_{0}^{\dfrac{\pi }{2}}{\sin {{\,}^{5}}\varphi d\varphi }=0\)(dựa theo kết quả bài 3.32 trang 172)