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

Q. How will you define integrating   constant C ? In how many ways can you  define C ?

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

Commented by vishalbhardwaj last updated on 03/Dec/19

what is C stand for ?

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

Commented by vishalbhardwaj last updated on 03/Dec/19

define c, its basic mean ??

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

Commented by MJS last updated on 03/Dec/19

(d/dx)[F(x)+C_1 ]=(d/dx)[F(x)+C_2 ]=f(x) ∀C_1 , C_2 ∈R  ⇔  ∫f(x)dx=F(x)+C with C∈R  or what do you mean?

$$\frac{{d}}{{dx}}\left[{F}\left({x}\right)+{C}_{\mathrm{1}} \right]=\frac{{d}}{{dx}}\left[{F}\left({x}\right)+{C}_{\mathrm{2}} \right]={f}\left({x}\right)\:\forall{C}_{\mathrm{1}} ,\:{C}_{\mathrm{2}} \in\mathbb{R} \\ $$$$\Leftrightarrow \\ $$$$\int{f}\left({x}\right){dx}={F}\left({x}\right)+{C}\:\mathrm{with}\:{C}\in\mathbb{R} \\ $$$$\mathrm{or}\:\mathrm{what}\:\mathrm{do}\:\mathrm{you}\:\mathrm{mean}? \\ $$

Commented by MJS last updated on 03/Dec/19

C is a constant factor. the integral is never  unique because while derivating you lose  the constant factor  (d/dx)[x^2 −3x+5]=(d/dx)[x^2 −3x−125890]=  =(d/dx)[x^2 −3x+C]=2x−3  ⇒  ∫2x−3 dx=x^2 −3x+C  C is any value

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

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