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Question Number 97660 by bobhans last updated on 09/Jun/20

Commented by bemath last updated on 09/Jun/20

$$\mathrm{nice}\mathrm{question}$$

Answered by bemath last updated on 09/Jun/20

$$\mathrm{let}\mathrm{a}=\underset{0}{\stackrel{2}{\int }}\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx};\mathrm{b}=\underset{0}{\stackrel{1}{\int }}\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{c}=\underset{0}{\stackrel{3}{\int }}\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}$$$$\iff f\left(x\right)={ax}^2+bx+c+2$$$$\iff f{}^{\prime }\left(x\right)=2ax+b$$$$f{}^{\prime }\left(1\right)=2a+b=2\underset{0}{\stackrel{2}{\int }}g\left(x\right)dx+\underset{0}{\stackrel{1}{\int }}g\left(x\right)dx$$$$$$

Commented by bemath last updated on 09/Jun/20

$$\mathrm{let}\mathrm{h}\left(\mathrm{x}\right)\mathrm{antiderivative}\mathrm{of}\mathrm{g}\left(\mathrm{x}\right)$$$$a=\underset{0}{\stackrel{2}{\int }}g\left(x\right)dx=h\left(2\right)-h\left(0\right)$$$$b=\underset{0}{\stackrel{1}{\int }}g\left(x\right)dx=h\left(1\right)-h\left(0\right)$$$$\iff 2a+b=2h\left(2\right)-2h\left(0\right)+h\left(1\right)-h\left(0\right)$$$$=2h\left(2\right)+h\left(1\right)-3h\left(0\right)$$

Answered by mathmax by abdo last updated on 09/Jun/20

$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=2\mathrm{x}{\int }_0^2\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}+{\int }_0^1\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(1\right)=2{\int }_0^2\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}+{\int }_0^1\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}$$

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