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Question Number 109546 by bemath last updated on 24/Aug/20

$$Iff\left(x\right)continuein[1,30]and$$$$\underset{6}{\stackrel{30}{\int }}f\left(x\right)dx=30,then\underset{1}{\stackrel{9}{\int }}f(3y+3)dy=\mathrm{\_}\mathrm{\_}$$

Commented by kaivan.ahmadi last updated on 24/Aug/20

$$t=3y+3\Rightarrow dt=3dy\Rightarrow dy=\frac{1}{3}dt$$$$y=1\Rightarrow t=6$$$$y=9\Rightarrow t=30$$$$\frac{1}{3}{\int }_6^{30}f\left(t\right)dt=\frac{1}{3}\times 30=10$$

Commented by bemath last updated on 24/Aug/20

$$cooll..thanks$$

Answered by 1549442205PVT last updated on 24/Aug/20

$$\mathrm{Put}\mathrm{x}=3\mathrm{y}+3\Rightarrow \mathrm{dx}=3\mathrm{dy}.\mathrm{Then}$$$${\int }_1^9\mathrm{f}(3\mathrm{y}+3)\mathrm{dy}=\frac{1}{3}{\int }_6^{30}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\frac{1}{3}\times 30=10$$

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