Bài tập 3.44 trang 180 SBT Toán 12

a) Đặt \(t = \sqrt y  \Rightarrow {t^2} = y \Rightarrow 2tdt = dy\)

\(\begin{array}{l}
 \Rightarrow \int \limits_0^1 {(y - 1)^2}\sqrt y dy = \int \limits_0^1 {\left( {{t^2} - 1} \right)^2}.t.2tdt\\
 = 2\int \limits_0^1 {t^2}\left( {{t^4} - 2{t^2} + 1} \right)dt\\
 = 2\int \limits_0^1 \left( {{t^6} - 2{t^4} + {t^2}} \right)dt\\
 = 2\left. {\left( {\frac{{{t^7}}}{7} - 2.\frac{{{t^5}}}{5} + \frac{{{t^3}}}{3}} \right)} \right|_0^1\\
 = 2\left( {\frac{1}{7} - \frac{2}{5} + \frac{1}{3}} \right) = \frac{{16}}{{105}}
\end{array}\)

b) Đặt \(u = \sqrt[3]{{{{(z - 1)}^2}}} \Rightarrow {u^3} = {\left( {z - 1} \right)^2}\)

\( \Rightarrow z = 1 + {u^{\frac{3}{2}}} \Rightarrow dz = \frac{3}{2}{u^{\frac{1}{2}}}du\)

\(\begin{array}{l}
 \Rightarrow \int \limits_1^2 ({z^2} + 1)\sqrt[3]{{{{(z - 1)}^2}}}dz\\
 = \int \limits_0^1 \left[ {{{\left( {1 + {u^{\frac{3}{2}}}} \right)}^2} + 1} \right].u.\frac{3}{2}{u^{\frac{1}{2}}}du\\
 = \frac{3}{2}\int \limits_0^1 {u^{\frac{3}{2}}}\left( {2 + 2{u^{\frac{3}{2}}} + {u^3}} \right)du\\
 = \frac{3}{2}\int \limits_0^1 \left( {2{u^{\frac{3}{2}}} + 2{u^3} + {u^{\frac{9}{2}}}} \right)du\\
 = \frac{3}{2}\left. {\left( {2.\frac{2}{5}{u^{\frac{5}{2}}} + 2.\frac{{{u^4}}}{4} + \frac{2}{{11}}{u^{\frac{{11}}{2}}}} \right)} \right|_0^1\\
 = \frac{3}{2}\left( {\frac{4}{5} + \frac{1}{2} + \frac{2}{{11}}} \right) = \frac{{489}}{{220}}
\end{array}\)

c) Đặt \(t = \sqrt {4 + 5\ln x}  \Rightarrow {t^2} = 4 + 5\ln x\)

\( \Rightarrow 2tdt = \frac{5}{x}dx \Rightarrow \frac{{dx}}{x} = \frac{2}{5}tdt\)

${\displaystyle }$\(\begin{array}{l}
 \Rightarrow \int \limits_1^e \frac{{\sqrt {4 + 5\ln x} }}{x}dx = \int \limits_2^3 t.\frac{2}{5}tdt\\
 = \frac{2}{5}\int\limits_2^3 {{t^2}dt}  = \frac{2}{5}.\left. {\frac{{{t^3}}}{3}} \right|_2^3\\
 = \frac{2}{5}\left( {\frac{{27}}{3} - \frac{8}{3}} \right) = \frac{{38}}{{15}}
\end{array}\)

d) Xét hàm số \(f\left( t \right) = {t^5}\) xác định và liên tục trên R.

Khi đó \(\int \limits_0^{\frac{\pi }{2}} f\left( {\sin \varphi } \right)d\varphi  = \int \limits_0^{\frac{\pi }{2}} f\left( {\cos \varphi } \right)d\varphi \)

Hay \(\int \limits_0^{\frac{\pi }{2}} {\sin ^5}\varphi d\varphi  = \int \limits_0^{\frac{\pi }{2}} {\cos ^5}\varphi d\varphi \)

\(\begin{array}{l}
 \Rightarrow \int \limits_0^{\frac{\pi }{2}} {\cos ^5}\varphi d\varphi  - \int \limits_0^{\frac{\pi }{2}} {\sin ^5}\varphi d\varphi  = 0\\
 \Rightarrow \int \limits_0^{\frac{\pi }{2}} \left( {{{\cos }^5}\varphi  - {{\sin }^5}\varphi } \right)d\varphi  = 0
\end{array}\)

-- Mod Toán 12 HỌC247