cho ∫f(4x)dx=$e^{2x}$ -x²+C. Khi đó ∫f(-x)dx bằng? Mong mn giúp mình vs ạ!!! 28/10/2021 Bởi Parker cho ∫f(4x)dx=$e^{2x}$ -x²+C. Khi đó ∫f(-x)dx bằng? Mong mn giúp mình vs ạ!!!
$\displaystyle\int f(4x) \, dx =e^{2x}-x^2+C\\ \Leftrightarrow -\displaystyle\int f(4x) \, d(4x) =-4e^{2x}+4x^2+C(*)\\ t=-4x \Rightarrow x=\dfrac{-t}{4}\\ (*)\Leftrightarrow -\displaystyle\int f(-t) \, d(-t) =-4e^{2.\tfrac{-t}{4}}+4\left(\dfrac{-t}{4}\right)^2+C\\ \Leftrightarrow \displaystyle\int f(-t) \, dt =-4e^{\tfrac{-t}{2}}+\dfrac{t^2}{4}+C\\ \Leftrightarrow \displaystyle\int f(-x) \, dx =-4e^{\tfrac{-x}{2}}+\dfrac{x^2}{4}+C$ Bình luận
$\displaystyle\int f(4x) \, dx =e^{2x}-x^2+C\\ \Leftrightarrow -\displaystyle\int f(4x) \, d(4x) =-4e^{2x}+4x^2+C(*)\\ t=-4x \Rightarrow x=\dfrac{-t}{4}\\ (*)\Leftrightarrow -\displaystyle\int f(-t) \, d(-t) =-4e^{2.\tfrac{-t}{4}}+4\left(\dfrac{-t}{4}\right)^2+C\\ \Leftrightarrow \displaystyle\int f(-t) \, dt =-4e^{\tfrac{-t}{2}}+\dfrac{t^2}{4}+C\\ \Leftrightarrow \displaystyle\int f(-x) \, dx =-4e^{\tfrac{-x}{2}}+\dfrac{x^2}{4}+C$