Tích phân \(A = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x}}{{\sin x + \cos x}}dx} \) bằng

Hướng dẫn giải

Khi vật dừng lại thì \(v\left( t \right) = 160 - 10t = 0 \Leftrightarrow t = 16\)

Do đó \(S = \int\limits_0^{16} {v\left( t \right)dt} = \int\limits_0^{16} {\left( {160 - 10t} \right)dt} \)

\( = \left( {160t - 5{t^2}} \right)\left| {_{\scriptstyle\atop\scriptstyle0}^{\scriptstyle16\atop\scriptstyle}} \right. = 1280\left( m \right)\).

Hướng dẫn giải

Đặt \(t = \frac{x}{2} \Rightarrow x = 2t \Rightarrow dx = 2dt.\)

Đổi cận \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 4 \Rightarrow t = 2\end{array} \right..\) Do đó \(\int\limits_0^4 {xf'\left( {\frac{x}{2}} \right)dx} = \int\limits_0^2 {4tf'\left( t \right)dt} = \int\limits_0^2 {4xf'\left( x \right)dx} .\)

Đặt \(\left\{ \begin{array}{l}u = 4x\\dv = f'\left( x \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = 4dx\\v = f\left( x \right)\end{array} \right..\)

Suy ra

\(\int\limits_0^2 {4xf'\left( x \right)dx} = \left[ {4xf\left( x \right)} \right]\left| {_{\scriptstyle\atop\scriptstyle0}^{\scriptstyle2\atop\scriptstyle}} \right. - \int\limits_0^2 {4f\left( x \right)dx} = 8f\left( 2 \right) - 4\int\limits_0^2 {f\left( x \right)dx} = 8.16 - 4.4 = 112.\)

Chọn A.