Bài tập 3.41 trang 180 SBT Toán 12

Bài tập 3.41 trang 180 SBT Toán 12

Ta có: \(\sin x = \frac{{2x}}{\pi } \Rightarrow \left[ {\begin{array}{*{20}{l}}
{x = 0}\\
{x = \frac{\pi }{2}}\\
{x =  - \frac{\pi }{2}}
\end{array}} \right.\)

Khi đó \(V = \pi \int \limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} \left| {{{\sin }^2}x - {{\left( {\frac{{2x}}{\pi }} \right)}^2}} \right|dx\)

Dễ thấy \(f\left( x \right) = \left| {{{\sin }^2}x - {{\left( {\frac{{2x}}{\pi }} \right)}^2}} \right|\) 

là hàm số chẵn nên:

\(\begin{array}{l}
V = 2\pi \int \limits_0^{\frac{\pi }{2}} \left| {{{\sin }^2}x - {{\left( {\frac{{2x}}{\pi }} \right)}^2}} \right|dx\\
 = 2\pi \int \limits_0^{\frac{\pi }{2}} \left( {{{\sin }^2}x - {{\left( {\frac{{2x}}{\pi }} \right)}^2}} \right)dx\\
 = 2\pi \int \limits_0^{\frac{\pi }{2}} {\sin ^2}xdx - \frac{8}{\pi }\int \limits_0^{\frac{\pi }{2}} {x^2}dx\\
 = \pi \int \limits_0^{\frac{\pi }{2}} \left( {1 - \cos 2x} \right)dx - \frac{8}{\pi }\int \limits_0^{\frac{\pi }{2}} {x^2}dx\\
 = \pi \left. {\left( {x - \frac{{\sin 2x}}{2}} \right)} \right|_0^{\frac{\pi }{2}} - \left. {\frac{8}{\pi }.\frac{{{x^3}}}{3}} \right|_0^{\frac{\pi }{2}}\\
 = \pi \left( {\frac{\pi }{2} - 0} \right) - \frac{8}{\pi }.\frac{1}{3}.{\left( {\frac{\pi }{2}} \right)^3}\\
 = \frac{{{\pi ^2}}}{2} - \frac{{{\pi ^2}}}{3} = \frac{{{\pi ^2}}}{6}
\end{array}\)

Chọn D.

-- Mod Toán 12