e-x-2-dx-

Question Number 6240 by sanusihammed last updated on 19/Jun/16

$$\int{e}^{{x}^{\mathrm{2}} } \:\:{dx}\: \\ $$

Commented by FilupSmith last updated on 20/Jun/16

I=∫e^x^2 dx I^2 =∫e^x^2 dx∫e^y^2 dx =∫∫e^(x^2 +y^2 ) dxdy Take Jacobian r^2 =x^2 +y^2 0<θ≤2π =∫∫re^r^2 drdθ =∫re^r^2 dr∫dθ u=r^2 ⇒ du=2rdr =(1/2)∫e^u du∫dθ I^2 =(1/2)e^r^2 θ+C I=(1/( (√2)))(√(e^r^2 θ+C)) C=constant

$${I}=\int{e}^{{x}^{\mathrm{2}} } {dx} \\ $$$${I}^{\mathrm{2}} =\int{e}^{{x}^{\mathrm{2}} } {dx}\int{e}^{{y}^{\mathrm{2}} } {dx} \\ $$$$=\int\int{e}^{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } {dxdy} \\ $$$$\mathrm{Take}\:\mathrm{Jacobian} \\ $$$${r}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$\mathrm{0}<\theta\leqslant\mathrm{2}\pi \\ $$$$=\int\int{re}^{{r}^{\mathrm{2}} } {drd}\theta \\ $$$$=\int{re}^{{r}^{\mathrm{2}} } {dr}\int{d}\theta \\ $$$${u}={r}^{\mathrm{2}} \:\Rightarrow\:{du}=\mathrm{2}{rdr} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{{u}} {du}\int{d}\theta \\ $$$${I}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{e}^{{r}^{\mathrm{2}} } \theta+{C} \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{{e}^{{r}^{\mathrm{2}} } \theta+{C}} \\ $$$${C}=\mathrm{constant} \\ $$

Commented by FilupSmith last updated on 20/Jun/16

I=(1/( (√2)))(√(e^r^2 θ+C)) I=(1/( (√2)))(√(e^(x^2 +y^2 ) θ+C)) x and y are equivilent variables I=(1/( (√2)))(√(e^(2x^2 ) θ+C))

$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{{e}^{{r}^{\mathrm{2}} } \theta+{C}} \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{{e}^{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \theta+{C}} \\ $$$${x}\:{and}\:{y}\:{are}\:{equivilent}\:{variables} \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{{e}^{\mathrm{2}{x}^{\mathrm{2}} } \theta+{C}} \\ $$

Commented by sanusihammed last updated on 20/Jun/16

$${Thanks}\:{i}\:{really}\:{appreciate} \\ $$

Commented by prakash jain last updated on 20/Jun/16

$${if}\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi\:\mathrm{then}\:\mathrm{isn}'\mathrm{t} \\ $$$$\int\theta\mathrm{d}\theta=\mathrm{2}\pi? \\ $$

Commented by FilupSmith last updated on 20/Jun/16

I′m actually not sure considering this is an indefinate intgral

$$\mathrm{I}'\mathrm{m}\:\mathrm{actually}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{considering}\:\mathrm{this} \\ $$$$\mathrm{is}\:\mathrm{an}\:\mathrm{indefinate}\:\mathrm{intgral} \\ $$

Commented by nburiburu last updated on 25/Jun/16

$$\theta={atan}\left(\frac{{y}}{{x}}\right)={atan}\left(\mathrm{1}\right)=\:\pi/\mathrm{4}\:{then}? \\ $$

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