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Question Number 74891 by vishalbhardwaj last updated on 03/Dec/19

$$\mathrm{Q}.\mathrm{How}\mathrm{will}\mathrm{you}\mathrm{define}\mathrm{integrating}$$$$\mathrm{constant}\mathrm{C}?\mathrm{In}\mathrm{how}\mathrm{many}\mathrm{ways}\mathrm{can}\mathrm{you}$$$$\mathrm{define}\mathrm{C}?$$$$$$

Commented by vishalbhardwaj last updated on 03/Dec/19

$$\mathrm{what}\mathrm{is}\mathrm{C}\mathrm{stand}\mathrm{for}?$$

Commented by vishalbhardwaj last updated on 03/Dec/19

$$\mathrm{define}\mathrm{c},\mathrm{its}\mathrm{basic}\mathrm{mean}??$$

Commented by MJS last updated on 03/Dec/19

$$\frac{d}{dx}[F\left(x\right)+C_1]=\frac{d}{dx}[F\left(x\right)+C_2]=f\left(x\right)\mathrm{\forall }{C}_1,C_2\in \mathbb{R}$$$$\iff$$$$\int f\left(x\right)dx=F\left(x\right)+C\mathrm{with}C\in \mathbb{R}$$$$\mathrm{or}\mathrm{what}\mathrm{do}\mathrm{you}\mathrm{mean}?$$

Commented by MJS last updated on 03/Dec/19

$$C\mathrm{is}\mathrm{a}\mathrm{constant}\mathrm{factor}.\mathrm{the}\mathrm{integral}\mathrm{is}\mathrm{never}$$$$\mathrm{unique}\mathrm{because}\mathrm{while}\mathrm{derivating}\mathrm{you}\mathrm{lose}$$$$\mathrm{the}\mathrm{constant}\mathrm{factor}$$$$\frac{d}{dx}[x^2-3x+5]=\frac{d}{dx}[x^2-3x-125890]=$$$$=\frac{d}{dx}[x^2-3x+C]=2x-3$$$$\Rightarrow$$$$\int 2x-3dx=x^2-3x+C$$$$C\mathrm{is}\mathrm{any}\mathrm{value}$$

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