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Question Number 54273 by rahul 19 last updated on 01/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19

$$\frac{df}{dx}=f\left(x\right)+{\int }_0^{1}f\left(x\right)dx$$$$\frac{df}{dx}=f\left(x\right)+a$$$$\frac{df}{f\left(x\right)+a}=dx$$$$\frac{d[f\left(x\right)+a]}{f\left(x\right)+a}=dx$$$$ln[f\left(x\right)+a]=x+c$$$$f\left(x\right)+a=e^{x+c}$$$$f\left(0\right)+a=e^{0+c}$$$$1+a=e^c$$$$f\left(x\right)+a=e^x\times e^c$$$$f\left(x\right)+a=e^x(1+a)$$$$f\left(x\right)+a=e^x+{ae}^x$$$$f\left(x\right)=(1+a)e^x-a$$$${\int }_0^{1}f\left(x\right)dx=(1+a){\int }_0^1{e}^{x}dx-a{\int }_0^{1}dx$$$$a=(1+a)(e-1)-a$$$$2a=e-1+ae-a$$$$3a-ae=e-1$$$$a=\frac{e-1}{3-e}$$$$f\left(x\right)=(1+a)e^x-a$$$$=(1+\frac{e-1}{3-e})e^x-\frac{e-1}{3-e}$$$$f\left(x\right)=\frac{2}{3-e}{e}^x-\frac{e-1}{3-e}$$$$\int f\left(x\right)dx=\frac{2}{3-e}{e}^x+\frac{1-e}{3-e}x+c$$$$sooptionbisanswer$$

Commented by rahul 19 last updated on 01/Feb/19

thank you sir!

Commented by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19

$$mostwelcome...$$

Commented by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19

$$isthereanyshorttricktogetanswer...then$$$$howthesetypeofquestioncanbesolvedinshorttime..$$

Commented by tanmay.chaudhury50@gmail.com last updated on 02/Feb/19

$$thankyou$$

Commented by rahul 19 last updated on 02/Feb/19

$$sir,i^{\prime }vepostedans.ofthoseintegrals.$$$$docheck!$$

Commented by rahul 19 last updated on 03/Feb/19

$$Q.No.=54248$$

Answered by ajfour last updated on 01/Feb/19

$$f{}^{″}\left(x\right)=f{}^{\prime }\left(x\right)$$$$\Rightarrow f{}^{\prime }\left(x\right)={Ae}^x$$$$f\left(x\right)={Ae}^x+b$$$$f\left(0\right)=A+b=1\Rightarrow b=1-A$$$$f{}^{\prime }\left(x\right)=f\left(x\right)+{\int }_0^{1}f\left(x\right)dx\Rightarrow$$$${Ae}^x={Ae}^x+b+A(e-1)+b$$$$butb=1-A$$$$\Rightarrow 2-2A+Ae-A=0$$$$\Rightarrow \mathit{A}=\frac{2}{3-e},hence$$$$\Rightarrow f\left(x\right)=\frac{2e^x}{3-e}+1-\frac{2}{3-e}$$$$f\left(x\right)=\frac{2e^x+1-e}{3-e}$$

Commented by ajfour last updated on 02/Feb/19

$$Imusthavemademistake,please$$$$help..$$

Commented by rahul 19 last updated on 02/Feb/19

$$Sir,howcanyousayf{}^{{}^{\prime }}\left(x\right)={Ae}^{x}isthe$$$$onlypossibility?$$

Commented by ajfour last updated on 02/Feb/19

$$f{}^{″}\left(x\right)=f{}^{\prime }\left(x\right)$$$$\Rightarrow \int \frac{f{}^{″}\left(x\right)dx}{f{}^{\prime }\left(x\right)}=\int dx$$$$\Rightarrow \ln f{}^{\prime }\left(x\right)=x+c$$$$\Rightarrow f{}^{\prime }\left(x\right)=e^x{e}^c={Ae}^x.$$

Commented by rahul 19 last updated on 03/Feb/19

$$thankyousir.$$

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