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Câu hỏi: Nếu $\int\limits_{0}^{\pi }{f\left( x \right)\sin x \text{d}x=20}$, $\int\limits_{0}^{\pi }{x{f}'\left( x \right)\sin x \text{d}x}=5$ thì $\int\limits_{0}^{{{\pi }^{2}}}{f\left( \sqrt{x} \right)\cos \left( \sqrt{x} \right)\text{d}x}$ bằng
A. $-50$.
B. $-30$.
C. 15.
D. 25.
A. $-50$.
B. $-30$.
C. 15.
D. 25.
Từ $\int\limits_{0}^{\pi }{f\left( x \right)\sin x\text{d}x=20}$ và $\int\limits_{0}^{\pi }{x{f}'\left( x \right)\sin x\text{d}x=5}$ ta được
$25 = \int\limits_{0}^{\pi }{f\left( x \right)\sin x \text{d}x} + \int\limits_{0}^{\pi }{x{f}'\left( x \right)\sin x \text{d}x } = \int\limits_{0}^{\pi }{\left( f\left( x \right) + x{f}'\left( x \right) \right)\sin x \text{d}x} = \int\limits_{0}^{\pi }{{{\left( xf\left( x \right) \right)}^{\prime }}\sin x \text{d}x} $
$ = \int\limits_{0}^{\pi }{\sin x \text{d}\left( xf\left( x \right) \right)} = \left( xf\left( x \right).\sin x \right) \underset{0}{\overset{\pi }{\mathop{\left| {} \right.}}} - \int\limits_{0}^{\pi }{xf\left( x \right)\cos x \text{d}x} = - \int\limits_{0}^{\pi }{xf\left( x \right)\cos x \text{d}x}$.
Suy ra $\int\limits_{0}^{\pi }{xf\left( x \right)\cos x \text{d}x}=-25$.
Xét $I = \int\limits_{0}^{{{\pi }^{2}}}{f\left( \sqrt{x} \right)\cos \left( \sqrt{x} \right)\text{d}x}$. Đặt $t=\sqrt{x}$ suy ra ${{t}^{2}}=x$ nên $\text{d}x=2t\text{d}t$.
Đổi cận: với $x=0\Rightarrow t=0$, với $x={{\pi }^{2}} \Rightarrow t=\pi $.
Khi đó, $I = \int\limits_{0}^{\pi }{2t.f\left( t \right)\cos t \text{d}}t=2.\int\limits_{0}^{\pi }{x.f\left( x \right)\cos x \text{d}}x = 2.\left( -25 \right) = -50$.
$25 = \int\limits_{0}^{\pi }{f\left( x \right)\sin x \text{d}x} + \int\limits_{0}^{\pi }{x{f}'\left( x \right)\sin x \text{d}x } = \int\limits_{0}^{\pi }{\left( f\left( x \right) + x{f}'\left( x \right) \right)\sin x \text{d}x} = \int\limits_{0}^{\pi }{{{\left( xf\left( x \right) \right)}^{\prime }}\sin x \text{d}x} $
$ = \int\limits_{0}^{\pi }{\sin x \text{d}\left( xf\left( x \right) \right)} = \left( xf\left( x \right).\sin x \right) \underset{0}{\overset{\pi }{\mathop{\left| {} \right.}}} - \int\limits_{0}^{\pi }{xf\left( x \right)\cos x \text{d}x} = - \int\limits_{0}^{\pi }{xf\left( x \right)\cos x \text{d}x}$.
Suy ra $\int\limits_{0}^{\pi }{xf\left( x \right)\cos x \text{d}x}=-25$.
Xét $I = \int\limits_{0}^{{{\pi }^{2}}}{f\left( \sqrt{x} \right)\cos \left( \sqrt{x} \right)\text{d}x}$. Đặt $t=\sqrt{x}$ suy ra ${{t}^{2}}=x$ nên $\text{d}x=2t\text{d}t$.
Đổi cận: với $x=0\Rightarrow t=0$, với $x={{\pi }^{2}} \Rightarrow t=\pi $.
Khi đó, $I = \int\limits_{0}^{\pi }{2t.f\left( t \right)\cos t \text{d}}t=2.\int\limits_{0}^{\pi }{x.f\left( x \right)\cos x \text{d}}x = 2.\left( -25 \right) = -50$.
Đáp án A.