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Câu hỏi: Cho $\int\limits_{{}}^{{}}{f\left( 4x \right)dx}={{e}^{2x}}-{{x}^{2}}+C.$ Khi đó $\int\limits_{{}}^{{}}{f\left( -x \right)dx}$ bằng
A. $\dfrac{{{e}^{2x}}}{4}+4{{x}^{2}}+C.$
B. $4{{e}^{\dfrac{x}{2}}}-\dfrac{1}{4}{{x}^{2}}+C$.
C. $-4{{e}^{\dfrac{x}{2}}}+\dfrac{1}{4}{{x}^{2}}+C.$
D. $-{{e}^{-\dfrac{x}{2}}}+{{\left( \dfrac{x}{4} \right)}^{2}}+C.$
A. $\dfrac{{{e}^{2x}}}{4}+4{{x}^{2}}+C.$
B. $4{{e}^{\dfrac{x}{2}}}-\dfrac{1}{4}{{x}^{2}}+C$.
C. $-4{{e}^{\dfrac{x}{2}}}+\dfrac{1}{4}{{x}^{2}}+C.$
D. $-{{e}^{-\dfrac{x}{2}}}+{{\left( \dfrac{x}{4} \right)}^{2}}+C.$
Ta có $\int\limits_{{}}^{{}}{f\left( 4x \right)dx}={{e}^{2x}}-{{x}^{2}}+C\Rightarrow f\left( 4x \right)=\left( {{e}^{2x}}-{{x}^{2}}+C \right)'=2{{e}^{2x}}-2x.$
Đặt $x=-\dfrac{1}{4}t$ suy ra $f\left( 4x \right)=f\left( -t \right)=2{{e}^{-\dfrac{1}{2}t}}+\dfrac{1}{2}t.$
Khi đó $f\left( -x \right)=2{{e}^{-\dfrac{1}{2}x}}+\dfrac{1}{2}x.$
Ta có $\int\limits_{{}}^{{}}{f\left( -x \right)dx}=\int\limits_{{}}^{{}}{\left( 2{{e}^{-\dfrac{1}{2}x}}+\dfrac{1}{2}x \right)dx}=-4{{e}^{-\dfrac{x}{2}}}+\dfrac{1}{4}{{x}^{2}}+C.$
Đặt $x=-\dfrac{1}{4}t$ suy ra $f\left( 4x \right)=f\left( -t \right)=2{{e}^{-\dfrac{1}{2}t}}+\dfrac{1}{2}t.$
Khi đó $f\left( -x \right)=2{{e}^{-\dfrac{1}{2}x}}+\dfrac{1}{2}x.$
Ta có $\int\limits_{{}}^{{}}{f\left( -x \right)dx}=\int\limits_{{}}^{{}}{\left( 2{{e}^{-\dfrac{1}{2}x}}+\dfrac{1}{2}x \right)dx}=-4{{e}^{-\dfrac{x}{2}}}+\dfrac{1}{4}{{x}^{2}}+C.$
Đáp án C.