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Câu hỏi: Cho ${\int{f(2x)dx}={{x}^{2}}-3x+C}$. Mệnh đề nào dưới đây đúng?
A. ${\int{f\left( x-2 \right)dx}=\dfrac{{{x}^{2}}}{4}-\dfrac{5x}{2}+{{C}_{1}}}$.
B. ${\int{f\left( x-2 \right)dx}=\dfrac{{{x}^{2}}}{2}-5x+{{C}_{1}}}$.
C. ${\int{f\left( x-2 \right)dx}=\dfrac{{{x}^{2}}}{2}+x+{{C}_{1}}}$.
D. ${\int{f\left( x-2 \right)dx}=\dfrac{{{x}^{2}}}{2}-x+{{C}_{1}}}$.
A. ${\int{f\left( x-2 \right)dx}=\dfrac{{{x}^{2}}}{4}-\dfrac{5x}{2}+{{C}_{1}}}$.
B. ${\int{f\left( x-2 \right)dx}=\dfrac{{{x}^{2}}}{2}-5x+{{C}_{1}}}$.
C. ${\int{f\left( x-2 \right)dx}=\dfrac{{{x}^{2}}}{2}+x+{{C}_{1}}}$.
D. ${\int{f\left( x-2 \right)dx}=\dfrac{{{x}^{2}}}{2}-x+{{C}_{1}}}$.
Ta đặt $t=2x\Rightarrow x=\dfrac{t}{2}\Rightarrow dx=\dfrac{1}{2}dt$
Suy ra $\int{f}(2x)dx=\dfrac{1}{2}\int{f}(t)dt=\dfrac{{{t}^{2}}}{4}-\dfrac{3t}{2}+C\Rightarrow \int{f}(t)dt=\dfrac{{{t}^{2}}}{2}-3t+C$
$\Rightarrow \int{f}(x-2)dx=\int{f}(x-2)d(x-2)=\dfrac{{{(x-2)}^{2}}}{2}-3(x-2)+C=\dfrac{{{x}^{2}}}{2}-5x+{{C}_{1}}$
Vậy $\int{f}(x-2)dx=\dfrac{{{x}^{2}}}{2}-5x+{{C}_{1}}$
Suy ra $\int{f}(2x)dx=\dfrac{1}{2}\int{f}(t)dt=\dfrac{{{t}^{2}}}{4}-\dfrac{3t}{2}+C\Rightarrow \int{f}(t)dt=\dfrac{{{t}^{2}}}{2}-3t+C$
$\Rightarrow \int{f}(x-2)dx=\int{f}(x-2)d(x-2)=\dfrac{{{(x-2)}^{2}}}{2}-3(x-2)+C=\dfrac{{{x}^{2}}}{2}-5x+{{C}_{1}}$
Vậy $\int{f}(x-2)dx=\dfrac{{{x}^{2}}}{2}-5x+{{C}_{1}}$
Đáp án B.