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Question Number 41651 by rahul 19 last updated on 10/Aug/18

∫( 1+2x+3x^2 +4x^3 +.........) dx , (0<∣x∣<1)

$$\int\left(\:\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +.........\right)\:{dx}\:,\:\:\: \\ $$ $$\left(\mathrm{0}<\mid{x}\mid<\mathrm{1}\right) \\ $$

Commented bymaxmathsup by imad last updated on 10/Aug/18

∫ (1+2x+3x^2 +4x^3 +...)dx = ∫( Σ_(n=1) ^∞ nx^(n−1) )dx =Σ_(n=1) ^∞ n ∫ x^(n−1) dx =Σ_(n=1) ^∞ x^n =(1/(1−x)) −1 +c = (x/(x−1)) +c with ∣x∣<1 .

$$\int\:\left(\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +...\right){dx}\:=\:\int\left(\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \right){dx} \\ $$ $$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{n}\:\int\:\:{x}^{{n}−\mathrm{1}} {dx}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:−\mathrm{1}\:\:+{c}\:=\:\frac{{x}}{{x}−\mathrm{1}}\:+{c}\:\:\:\:{with}\:\mid{x}\mid<\mathrm{1}\:. \\ $$

Answered by alex041103 last updated on 10/Aug/18

1+2x+3x^2 +...=Σ_(n=0) ^∞ (n+1)x^n ⇒∫( 1+2x+3x^2 +4x^3 +.........) dx = =∫Σ_(n=0) ^∞ (n+1)x^n dx=Σ_(n=0) ^∞ ∫(n+1)x^n dx= =C+Σ_(n=0) ^∞ ((n+1)/(n+1))x^(n+1) =C+Σ_(n=0) ^∞ x^(n+1) = =C+(Σ_(n=0) ^∞ x^n ) −1=(Σ_(n=0) ^∞ x^n )+C_1 For ∣x∣<1: Σ_(n=0) ^∞ x^n =(1/(1−x)) ⇒∫( 1+2x+3x^2 +4x^3 +.........) dx =(1/(1−x))+C

$$\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +...=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left({n}+\mathrm{1}\right){x}^{{n}} \\ $$ $$\Rightarrow\int\left(\:\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +.........\right)\:{dx}\:= \\ $$ $$=\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left({n}+\mathrm{1}\right){x}^{{n}} \:{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int\left({n}+\mathrm{1}\right){x}^{{n}} {dx}= \\ $$ $$={C}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}+\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} ={C}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}+\mathrm{1}} = \\ $$ $$={C}+\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \right)\:−\mathrm{1}=\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \right)+{C}_{\mathrm{1}} \\ $$ $${For}\:\mid{x}\mid<\mathrm{1}:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$ $$\Rightarrow\int\left(\:\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +.........\right)\:{dx}\:=\frac{\mathrm{1}}{\mathrm{1}−{x}}+{C} \\ $$

Commented byrahul 19 last updated on 10/Aug/18

thanks sir!

$$\mathrm{thanks}\:\mathrm{sir}! \\ $$

Answered by rahul 19 last updated on 10/Aug/18

∫ (d/dx)(x+x^2 +x^3 +.........)dx ⇒ x+x^2 +x^3 +.......... ⇒(x/(1−x)). What′s wrong in this?

$$\int\:\frac{\mathrm{d}}{\mathrm{d}{x}}\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +.........\right){dx}\: \\ $$ $$\Rightarrow\:{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +.......... \\ $$ $$\Rightarrow\frac{{x}}{\mathrm{1}−{x}}. \\ $$ $${W}\mathrm{hat}'\mathrm{s}\:\mathrm{wrong}\:\mathrm{in}\:\mathrm{this}? \\ $$

Commented byalex041103 last updated on 10/Aug/18

Nothing is wrong. If you look more carefully yoi will see that: (x/(1−x))=−((1−x −1)/(1−x))=−(((1−x)/(1−x))−(1/(1−x)))= =(1/(1+x))−1=(1/(1+x))+C As you can see the integration constant is sth very imortant

$${Nothing}\:{is}\:{wrong}.\:{If}\:{you}\:{look}\:{more} \\ $$ $${carefully}\:{yoi}\:{will}\:{see}\:{that}: \\ $$ $$\frac{{x}}{\mathrm{1}−{x}}=−\frac{\mathrm{1}−{x}\:−\mathrm{1}}{\mathrm{1}−{x}}=−\left(\frac{\mathrm{1}−{x}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)= \\ $$ $$=\frac{\mathrm{1}}{\mathrm{1}+{x}}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{1}+{x}}+{C} \\ $$ $${As}\:{you}\:{can}\:{see}\:{the}\:{integration}\:{constant} \\ $$ $${is}\:{sth}\:{very}\:{imortant} \\ $$

Commented byrahul 19 last updated on 10/Aug/18

ohh, yes u r right sir!

$$\mathrm{ohh},\:\mathrm{yes}\:\mathrm{u}\:\mathrm{r}\:\mathrm{right}\:\mathrm{sir}! \\ $$

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