Tích phân (phương pháp từng phần)

Đặt $\left\{ \begin{array}{l}u = f\left( x \right)\\{\rm{d}}v = g'\left( x \right){\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = f'\left( x \right){\rm{d}}x\\v = g\left( x \right)\end{array} \right.,$ khi đó $I = \left. {f\left( x \right).g\left( x \right)} \right|_a^b - \int\limits_a^b {f'\left( x \right).g\left( x \right){\rm{d}}x} .$

Đặt $\left\{ \begin{array}{l}u = {x^2}\\{\rm{d}}v = \cos x\,{\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = 2x\,{\rm{d}}x\\v = \sin x\end{array} \right.,$ khi đó $I = \left. {{x^2}\sin x} \right|_0^{\dfrac{\pi }{2}} - 2\int\limits_0^{\dfrac{\pi }{2}} {x\sin x\,{\rm{d}}x} .$

Đặt $\left\{ \begin{array}{l}u = g\left( x \right)\\{\rm{d}}v = f'\left( x \right){\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = g'\left( x \right){\rm{d}}x\\v = f\left( x \right)\end{array} \right..$

Khi đó $\int\limits_0^1 {g\left( x \right).f'\left( x \right){\rm{d}}x}  = \left. {\left[ {g\left( x \right).f\left( x \right)} \right]} \right|_0^1 - \int\limits_0^1 {g'\left( x \right).f\left( x \right){\rm{d}}x}  \Leftrightarrow \left. {\left[ {g\left( x \right).f\left( x \right)} \right]} \right|_0^1 = 3.$

Mặt khác $I = \int\limits_0^1 {{{\left[ {f\left( x \right).g\left( x \right)} \right]}^\prime }\,{\rm{d}}x}  = \left. {\left[ {f\left( x \right).g\left( x \right)} \right]} \right|_0^1\,\, \Rightarrow \,\,I = 3.$

Đặt  $\left\{ \begin{array}{l}\ln \left( {x - 1} \right) = u\\\dfrac{1}{{{{\left( {x - 1} \right)}^2}}}dx = dv\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{1}{{x - 1}}dx = du\\ - \dfrac{1}{{x - 1}} = v\end{array} \right.$

 Ta có $\int\limits_2^{e + 1} {\dfrac{{\ln \left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}dx = \ln \left( {x - 1} \right).\left. {\left( { - \dfrac{1}{{x - 1}}} \right)} \right|} _2^{e + 1} + \int\limits_2^{e + 1} {\dfrac{1}{{{{\left( {x - 1} \right)}^2}}}dx}$

                               $=  - \dfrac{1}{e} - \left. {\dfrac{1}{{x - 1}}} \right|_2^{e + 1} =  - \dfrac{1}{e} - \dfrac{1}{e} + 1 = 1 - 2.{e^{ - 1}}$

Suy ra $a = 1;\,\,\,b =  - 2 \Rightarrow a + b =  - 1.$

Xét $\int\limits_0^1 {\left( {1 + 3x} \right)f'\left( x \right)dx}  = 2019$

Đặt $\left\{ \begin{array}{l}1 + 3x = u\\f'\left( x \right)dx = dv\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3dx = du\\f\left( x \right) = v\end{array} \right.$

Suy ra $\int\limits_0^1 {\left( {1 + 3x} \right)f'\left( x \right)dx}  = \left. {\left( {1 + 3x} \right)f\left( x \right)} \right|_0^1 - 3\int\limits_0^1 {f\left( x \right)dx}$

$= 4f\left( 1 \right) - f\left( 0 \right) - 3\int\limits_0^1 {f\left( x \right)dx}  = 2020 - 3\int\limits_0^1 {f\left( x \right)dx}  = 2019$

$\Leftrightarrow \int\limits_0^1 {f\left( x \right)dx}  = \dfrac{1}{3}$

Xét $\int\limits_0^{\dfrac{1}{3}} {f\left( {3x} \right)dx}$, đặt $3x = t \Leftrightarrow 3dx = dt \Leftrightarrow dx = \dfrac{{dt}}{3}$.

Đổi cận : $\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = \dfrac{1}{3} \Rightarrow t = 1\end{array} \right.$.

Suy ra $\int\limits_0^{\dfrac{1}{3}} {f\left( {3x} \right)dx}  = \dfrac{1}{3}\int\limits_0^1 {f\left( t \right)dt}  = \dfrac{1}{3}.\int\limits_0^1 {f\left( x \right)dx}  = \dfrac{1}{3}.\dfrac{1}{3} = \dfrac{1}{9}$

Ta có: $I = \int\limits_0^1 {xf'\left( x \right)dx}$

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

$\Rightarrow I = \left. {\left[ {xf\left( x \right)} \right]} \right|_0^1 - \int\limits_0^1 {f\left( x \right)dx}$ $= f\left( 1 \right) + \int\limits_1^0 {f\left( x \right)dx} .$

Đặt $\left\{ \begin{array}{l}u = \ln x\\dv = dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{{dx}}{x}\\v = x\end{array} \right.$

$\begin{array}{l} \Rightarrow \int\limits_1^a {\ln xdx}  = \left. {x\ln x} \right|_1^a - \int\limits_1^a {dx}  = \left. {\left( {x\ln x - x} \right)} \right|_1^a\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = a\ln a - a - 1\ln 1 + 1 = 1 - a + a\ln a\\ \Rightarrow 1 - a + a\ln a = 1 + 2a\\ \Leftrightarrow 3a = a\ln a \Leftrightarrow \ln a = 3 \Leftrightarrow a = {e^3} \approx 20,1 \in \left( {18;21} \right)\end{array}$

Đặt $t = 2x \Rightarrow dt = 2dx$.

Đổi cận: $\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 1 \Rightarrow t = 2\end{array} \right.$, khi đó ta có: $\int\limits_0^1 {x.f'\left( {2x} \right)dx}  = \dfrac{1}{4}\int\limits_0^2 {tf'\left( t \right)dt}$.

Đặt $\left\{ \begin{array}{l}u = t\\dv = f'\left( t \right)dt\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = dt\\v = f\left( t \right)\end{array} \right.$.

$\begin{array}{l} \Rightarrow \int\limits_0^2 {tf'\left( t \right)dt}  = \left. {tf\left( t \right)} \right|_0^2 - \int\limits_0^2 {f\left( t \right)dt} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2f\left( 2 \right) - \int\limits_0^2 {f\left( x \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2.16 - 4 = 28\end{array}$

Vậy $\int\limits_0^1 {x.f'\left( {2x} \right)dx}  = \dfrac{1}{4}.28 = 7$.

Ta có: $I = \int\limits_0^{\frac{\pi }{2}} {\sin 2x.f'\left( {\sin x} \right){\rm{d}}x}  = 2\int\limits_0^{\frac{\pi }{2}} {\sin x\cos x.f'\left( {\sin x} \right){\rm{d}}x}$.

Đặt $t = \sin x \Rightarrow du = \cos xdx$.

Đổi cận: $\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = \dfrac{\pi }{2} \Rightarrow t = 1\end{array} \right.$, khi đó ta có $I = 2\int\limits_0^1 {tf'\left( t \right)dt}$.

Đặt $\left\{ \begin{array}{l}u = t\\dv = f'\left( t \right)dt\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = dt\\v = f\left( t \right)\end{array} \right.$.

$\begin{array}{l} \Rightarrow I = 2\left( {\left. {tf\left( t \right)} \right|_0^1 - \int\limits_0^1 {f\left( t \right)dt} } \right)\\ \Leftrightarrow I = 2\left( {f\left( 1 \right) - \int\limits_0^1 {f\left( t \right)dt} } \right)\\ \Leftrightarrow I = 2\left( {1 - \dfrac{1}{3}} \right) = \dfrac{4}{3}\end{array}$

Đặt $\left\{ \begin{array}{l}u = f\left( x \right)\\dv = xdx\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = f'\left( x \right)dx\\v = \dfrac{{{x^2} - 1}}{2}\end{array} \right.$.

$\begin{array}{l} \Rightarrow \int\limits_0^1 {xf\left( x \right)dx}  = \left. {\dfrac{{{x^2} - 1}}{2}f\left( x \right)} \right|_0^1 - \int\limits_0^1 {\dfrac{{{x^2} - 1}}{2}f'\left( x \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}f\left( 0 \right) - \dfrac{1}{2}\int\limits_0^1 {\left( {{x^2} - 1} \right).x{e^{{x^2}}}dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}.\dfrac{1}{2} - \dfrac{1}{2}I = \dfrac{1}{4} - \dfrac{1}{2}I\end{array}$

Đặt $t = {x^2} \Rightarrow dt = 2xdx$.

Đổi cận: $\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 1 \Rightarrow t = 1\end{array} \right.$.

Khi đó ta có:

$\begin{array}{l}I = \dfrac{1}{2}\int\limits_0^1 {\left( {t - 1} \right){e^t}dt}  = \dfrac{1}{2}\left( {\int\limits_0^1 {t{e^t}dt}  - \int\limits_0^1 {{e^t}dt} } \right)\\\,\,\,\, = \dfrac{1}{2}\left( {\left. {t.{e^t}} \right|_0^1 - \int\limits_0^1 {{e^t}dt}  - \int\limits_0^1 {{e^t}dt} } \right)\\\,\,\,\, = \dfrac{1}{2}\left( {e - \left. {2{e^t}} \right|_0^1} \right) = \dfrac{1}{2}\left( {e - 2e + 2} \right) = \dfrac{{2 - e}}{2}\end{array}$

Vậy $\int\limits_0^1 {xf\left( x \right)dx}  = \dfrac{1}{4} - \dfrac{1}{2}I = \dfrac{1}{4} - \dfrac{{2 - e}}{4} = \dfrac{{e - 1}}{4}$.

Ta có: $\int\limits_1^2 {\left( {x + \dfrac{2}{x}} \right)\ln xdx}  = \int\limits_1^2 {x\ln xdx}  + 2\int\limits_1^2 {\dfrac{{\ln xdx}}{x}}  = {I_1} + 2{I_2}$.

Xét ${I_1} = \int\limits_1^2 {x\ln xdx}$.

Đặt $\left\{ \begin{array}{l}u = \ln x\\dv = xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{x}dx\\v = \dfrac{{{x^2}}}{2}\end{array} \right.$

$\begin{array}{l} \Rightarrow {I_1} = \left. {\dfrac{{{x^2}}}{2}\ln x} \right|_1^2 - \int\limits_1^2 {\dfrac{{{x^2}}}{2}.\dfrac{1}{x}dx} \\\,\,\,\,\,\,\,\,\,\,\,\, = 2\ln 2 - \dfrac{1}{2}\int\limits_1^2 {xdx} \\\,\,\,\,\,\,\,\,\,\,\,\, = 2\ln 2 - \dfrac{1}{2}.\left. {\dfrac{{{x^2}}}{2}} \right|_1^2\\\,\,\,\,\,\,\,\,\,\,\,\, = 2\ln 2 - \dfrac{1}{2}.\left( {2 - \dfrac{1}{2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = 2\ln 2 - \dfrac{3}{4}\end{array}$

Xét ${I_2} = \int\limits_1^2 {\dfrac{{\ln xdx}}{x}}$.

Đặt $t = \ln x \Rightarrow dt = \dfrac{{dx}}{x}$.

Đổi cận: $\left\{ \begin{array}{l}x = 1 \Rightarrow t = \ln 1 = 0\\x = 2 \Rightarrow t = \ln 2\end{array} \right.$.

$\Rightarrow {I_2} = \int\limits_0^{\ln 2} {tdt}  = \left. {\dfrac{{{t^2}}}{2}} \right|_0^{\ln 2} = \dfrac{{{{\ln }^2}2}}{2}$.

$\Rightarrow I = {I_1} + 2{I_2} = 2\ln 2 - \dfrac{3}{4} + {\ln ^2}2 =  - \dfrac{3}{4} + 2\ln 2 + {\ln ^2}2$.

$\Rightarrow a =  - 3,\,\,b = 2,\,\,c = 1$.

Vậy $a + b + c =  - 3 + 2 + 1 = 0$.

Xét $I = \int\limits_0^1 {xf'\left( x \right)dx}$.

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

$\begin{array}{l} \Rightarrow I = \left. {xf\left( x \right)} \right|_0^1 - \int\limits_0^1 {f\left( x \right)dx} \\ \Rightarrow I = f\left( 1 \right) - \int\limits_0^1 {f\left( x \right)dx} \\ \Rightarrow \dfrac{1}{{36}} =  - \dfrac{1}{{18}} - \int\limits_0^1 {f\left( x \right)dx} \\ \Rightarrow \int\limits_0^1 {f\left( x \right)dx}  =  - \dfrac{1}{{18}} - \dfrac{1}{{36}} =  - \dfrac{1}{{12}}.\end{array}$

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

$\begin{array}{l} \Rightarrow \int\limits_0^1 {xf'\left( x \right)dx}  = \left. {xf\left( x \right)} \right|_0^1 - \int\limits_0^1 {f\left( x \right)dx} \\ \Rightarrow \int\limits_0^1 {xf'\left( x \right)dx}  = f\left( 1 \right) - \int\limits_0^1 {f\left( x \right)dx} \\ \Rightarrow \int\limits_0^1 {xf'\left( x \right)dx}  = 4 - 3 = 1.\end{array}$

Công thức tính tích phân từng phần $\int\limits_a^b {udv}  = \left. {uv} \right|_a^b - \int\limits_a^b {vdu}$

Đặt $\left\{ \begin{array}{l}u = x\\{\rm{d}}v = \cos x\,{\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = {\rm{d}}x\\v = \sin x\end{array} \right.,$ khi đó $\int\limits_0^{\dfrac{\pi }{2}} {x\cos x\,{\rm{d}}x}  = \left. {x.\sin x} \right|_0^{\dfrac{\pi }{2}} - \int\limits_0^{\dfrac{\pi }{2}} {\sin x\,{\rm{d}}x}$

$= \dfrac{\pi }{2} + \left. {\cos x} \right|_0^{\dfrac{\pi }{2}} = \dfrac{\pi }{2} + \cos \dfrac{\pi }{2} - \cos 0 = \dfrac{\pi }{2} - 1.$

Suy ra $\dfrac{\pi }{m} - \int\limits_0^{\dfrac{\pi }{2}} {x\cos x\,{\rm{d}}x}  = \dfrac{\pi }{m} - \dfrac{\pi }{2} + 1 = 1 \Rightarrow m = 2.$

Do đó $9{m^2} - 6 = {9.2^2} - 6 = 30$.

Đặt $\left\{ \begin{array}{l}u = \dfrac{1}{{x + 1}}\\{\rm{d}}v = f'\left( x \right){\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u =  - \dfrac{{{\rm{d}}x}}{{{{\left( {x + 1} \right)}^2}}}\\v = f\left( x \right)\end{array} \right.,$ khi đó $\int\limits_0^1 {\dfrac{{f'\left( x \right)}}{{x + 1}}{\rm{d}}x}  = \left. {\dfrac{{f\left( x \right)}}{{x + 1}}} \right|_0^1 + \int\limits_0^1 {\dfrac{{f\left( x \right)}}{{{{\left( {x + 1} \right)}^2}}}{\rm{d}}x}$

Suy ra $1 = \left. {\dfrac{{f\left( x \right)}}{{x + 1}}} \right|_0^1 + I \Leftrightarrow I = 1 - \left[ {\dfrac{{f\left( 1 \right)}}{2} - f\left( 0 \right)} \right] = 1 - \dfrac{1}{2}\left[ {f\left( 1 \right) - 2f\left( 0 \right)} \right] = 1 - \dfrac{1}{2}.2 = 0.$

$\int\limits_{0}^{\pi }{\left( 3x+2 \right){{\cos }^{2}}xdx}=\frac{1}{2}\int\limits_{0}^{\pi }{\left( 3x+2 \right)\left( 1+\cos 2x \right)dx}=\frac{1}{2}\left[ \int\limits_{0}^{\pi }{\left( 3x+2 \right)dx}+\int\limits_{0}^{\pi }{\left( 3x+2 \right)\cos 2xdx} \right]=\frac{1}{2}\left( {{I}_{1}}+{{I}_{2}} \right)$

Tính ${{I}_{1}}?$

${{I}_{1}}=\int\limits_{0}^{\pi }{\left( 3x+2 \right)dx}=\left. \left( \frac{3{{x}^{2}}}{2}+2x \right) \right|_{0}^{\pi }=\frac{3}{2}{{\pi }^{2}}+2\pi$

Tính ${{I}_{2}}?$

${{I}_{2}}=\int\limits_{0}^{\pi }{\left( 3x+2 \right)\cos 2xdx}$

Đặt $\left\{ \begin{array}{l}u = 3x + 2\\dv = \cos 2xdx\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = 3dx\\v = \frac{1}{2}\sin 2x\end{array} \right.$

$\begin{array}{l} \Rightarrow {I_2} = \left. {\frac{1}{2}\left( {3x + 2} \right)\sin 2x} \right|_0^\pi  - \frac{3}{2}\int\limits_0^\pi  {\sin 2xdx} \\ = \left. {\frac{1}{2}\left( {3x + 2} \right)\sin 2x} \right|_0^\pi  + \frac{3}{4}\left. {\cos 2x} \right|_0^\pi \\ = \frac{3}{4}\left( {1 - 1} \right) = 0\end{array}$ 

Vậy $I=\frac{1}{2}\left( \frac{3}{2}{{\pi }^{2}}+2\pi  \right)=\frac{3}{4}{{\pi }^{2}}+\pi$

$\begin{array}{l}\int\limits_0^{\dfrac{\pi }{3}} {\dfrac{{{x^2}dx}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}}  = \int\limits_0^{\dfrac{\pi }{3}} {\dfrac{x}{{\cos x}}\dfrac{{x\cos xdx}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}}  = \int\limits_0^{\dfrac{\pi }{3}} {\dfrac{x}{{\cos x}}\dfrac{{d\left( {x\sin x + \cos x} \right)}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} \\ =  - \int\limits_0^{\dfrac{\pi }{3}} {\dfrac{x}{{\cos x}}d\left( {\dfrac{1}{{x\sin x + \cos x}}} \right)}  =  - \left. {\dfrac{x}{{\cos x}}.\dfrac{1}{{x\sin x + \cos x}}} \right|_0^{\dfrac{\pi }{3}} + \int\limits_0^{\dfrac{\pi }{3}} {\dfrac{1}{{x\sin x + \cos x}}d\left( {\dfrac{x}{{\cos x}}} \right)} \\ =  - \left. {\dfrac{x}{{\cos x\left( {x\sin x + \cos x} \right)}}} \right|_0^{\dfrac{\pi }{3}} + \int\limits_0^{\dfrac{\pi }{3}} {\dfrac{1}{{{{\cos }^2}x}}dx} \\ =  - \left. {\dfrac{x}{{\cos x\left( {x\sin x + \cos x} \right)}}} \right|_0^{\dfrac{\pi }{3}} + \left. {\tan x} \right|_0^{\dfrac{\pi }{3}} =  - \dfrac{{\dfrac{\pi }{3}}}{{\dfrac{1}{2}\left( {\dfrac{\pi }{3}.\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2}} \right)}} + \sqrt 3  = \dfrac{{ - 4\pi }}{{\pi \sqrt 3  + 3}} + \sqrt 3 \\ =  - \dfrac{{a\pi }}{{b + c\sqrt 3 }} + d\sqrt 3 \,\,\left( {a,b,c,d \in {Z^ + }} \right) \Rightarrow a = 4,b = 3,c = 1,d = 1 \Rightarrow a + b + c + d = 9\end{array}$

 Ta đặt $\left\{ \begin{array}{l}u = x\\{e^{2x}}dx = dv\end{array} \right. \Rightarrow \left\{ \begin{array}{l}dx = du\\v = \frac{1}{2}{e^{2x}}\end{array} \right.$

Khi đó $\int\limits_{0}^{100}{x.{{e}^{2x}}dx}=\left. \frac{1}{2}x.{{e}^{2x}} \right|_{0}^{100}-\frac{1}{2}\int\limits_{0}^{100}{{{e}^{2x}}dx}=\left. \frac{1}{2}x.{{e}^{2x}} \right|_{0}^{100}-\left. \frac{1}{4}{{e}^{2x}} \right|_{0}^{100}$$=\frac{1}{2}.100.{{e}^{200}}-\frac{1}{4}{{e}^{200}}+\frac{1}{4}=\frac{1}{4}\left( 199{{e}^{200}}+1 \right)$

Đặt $\left\{ \begin{array}{l}u = \ln x\\{\rm{d}}v = \dfrac{{{\rm{d}}x}}{{{x^2}}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = \dfrac{{{\rm{d}}x}}{x}\\v =  - \dfrac{1}{x}\end{array} \right.,$ khi đó $I = \left. { - \dfrac{{\ln x}}{x}} \right|_1^m + \int\limits_1^m {\dfrac{{{\rm{d}}x}}{{{x^2}}}}  =  - \dfrac{{\ln m}}{m} - \left. {\dfrac{1}{x}} \right|_1^m =  - \dfrac{{\ln m}}{m} - \dfrac{1}{m} + 1$

Mặt khác $I = \dfrac{1}{2} - \dfrac{1}{2}\ln 2\,\, \Rightarrow \,\,\dfrac{1}{2} - \dfrac{1}{2}\ln 2 =  - \dfrac{{\ln m}}{m} - \dfrac{1}{m} + 1 \Rightarrow m = 2 \in \left( {\dfrac{3}{2};\dfrac{5}{2}} \right).$