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Question Number 119659 by ZiYangLee last updated on 26/Oct/20

$${\int }_0^1\left[(4x-1)f(12x^2-6x)\right]dx=?$$

Answered by bemath last updated on 26/Oct/20

$$letu=12x^2-6x\to du=24x-6dx$$$$\frac{1}{6}du=(4x-1)dx$$$$with\{\begin{array}{l}x=0\to u=0\\ x=1\to u=6\end{array}$$$$\underset{0}{\stackrel{6}{\int }}\frac{1}{6}f\left(u\right)du=\frac{1}{6}{\left[F\left(u\right)\right]}_0^6$$$$=\frac{1}{6}[F\left(6\right)-F\left(0\right)]$$$$whereF\left(u\right)antiderivativeoffunctionf\left(u\right)$$

Commented by ZiYangLee last updated on 26/Oct/20

$$\mathrm{thanks}$$

Answered by mathmax by abdo last updated on 26/Oct/20

$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}{12\mathrm{x}}^2-6\mathrm{x}=\mathrm{u}\Rightarrow (24\mathrm{x}-6)\mathrm{dx}=\mathrm{du}\Rightarrow$$$$6(4\mathrm{x}-1)\mathrm{dx}=\mathrm{du}\Rightarrow$$$${\int }_0^1(4\mathrm{x}-1)\mathrm{f}({12\mathrm{x}}^2-6\mathrm{x})\mathrm{dx}=\frac{1}{6}{\int }_0^6\mathrm{f}\left(\mathrm{u}\right)\mathrm{du}$$$${=}_{\mathrm{by}\mathrm{parts}}\frac{1}{6}(\mathrm{F}\left(6\right)-\mathrm{F}\left(0\right))\mathrm{with}\mathrm{F}\mathrm{is}\mathrm{a}\mathrm{primitive}\mathrm{of}\mathrm{f}$$

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