∫_1 ^( 3) e^x^2 dx help me sir


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Question Number 142049 by mohammad17 last updated on 25/May/21

$\int_{1}^{3} e^{x^{2}} d x$ $h e l p m e s i r$
	Answered by Dwaipayan Shikari last updated on 26/May/21

	$\int e^{x^{2}} d x$ $= {x e}^{x^{2}} - 2 \int x^{2} e^{x^{2}} d x$ $= {x e}^{x^{2}} - \frac{2}{3} x^{3} e^{x^{2}} + \frac{2.2}{3} \int x^{5} e^{x^{2}} d x$ $= {x e}^{x^{2}} - \frac{2}{3} x^{3} e^{x^{2}} + \frac{2.2}{3.5} x^{5} e^{x^{2}} - \frac{2^{3}}{3.5.7} x^{7} e^{x^{2}} + . . . + C$ $\int_{1}^{3} e^{x^{2}} d x = e^{9} (\frac{3}{1} - \frac{2}{1.3} . 3^{3} + \frac{2^{2}}{1.3.5} 3^{5} - \frac{2^{3}}{1.3.5.7} 3^{7} + . . .) - e (\frac{1}{1} - \frac{2}{1.3} + \frac{2^{2}}{1.3.5} - . .)$
	Answered by 676597498 last updated on 26/May/21
	$I = ∫_(1 ) ^( 3) e^x^2 dx ⇒ I^2 = ∫_1 ^( 3) e^(x^2 +y^2 ) dxdy { ((r^2 =x^2 +y^2 )),( { ((x=rcosθ)),((y=rsinθ)) :}) :}⇒ I^2 =∫_1 ^( 3) e^r^2 rdrdθ { ((1 ≤ x ≤ 3)),((1 ≤ y ≤ 3)) :} ⇒ { ((x^2 ≤ 9)),((y^2 ≤ 9)) :}⇒ r^2 ≤ 18⇒ 0 ≤ r ≤ 3(√2) ⇒ I^2 = ∫_0 ^( 3(√2)) (∫_0 ^( 2π) dθ)re^r^2 dr ⇒ I^2 = 2π∫_0 ^( 3(√2)) re^r^2 dr = 2π[(e^r^2 /2)]_0 ^(3(√2)) ⇒ I^2 = 2π [(e^(18) /2)−1]=π(e^(18) −2) ⇒ ∫_1 ^( 3) e^x^2 dx = (√π).(√(e^(18) −2))$
	$I = \int_{1}^{3} e^{x^{2}} d x \Rightarrow I^{2} = \int_{1}^{3} e^{x^{2} + y^{2}} d x d y$ ${\begin{cases} r^{2} = x^{2} + y^{2} \\ {\begin{cases} x = r c o s θ \\ y = r s i n θ \end{cases} \end{cases} \Rightarrow I^{2} = \int_{1}^{3} e^{r^{2}} r d r d θ$ ${\begin{cases} 1 ⩽ x ⩽ 3 \\ 1 ⩽ y ⩽ 3 \end{cases} \Rightarrow {\begin{cases} x^{2} ⩽ 9 \\ y^{2} ⩽ 9 \end{cases} \Rightarrow r^{2} ⩽ 18 \Rightarrow 0 ⩽ r ⩽ 3 \sqrt{2}$ $\Rightarrow I^{2} = \int_{0}^{3 \sqrt{2}} (\int_{0}^{2 π} d θ) {r e}^{r^{2}} d r$ $\Rightarrow I^{2} = 2 π \int_{0}^{3 \sqrt{2}} {r e}^{r^{2}} d r = 2 π {[\frac{e^{r^{2}}}{2}]}_{0}^{3 \sqrt{2}}$ $\Rightarrow I^{2} = 2 π [\frac{e^{18}}{2} - 1] = π (e^{18} - 2)$ $\Rightarrow \int_{1}^{3} e^{x^{2}} d x = \sqrt{π} . \sqrt{e^{18} - 2}$
	Commented by Ar Brandon last updated on 27/May/21

	$I {don}^{'} t agree, Sir .$ $The area bounded by x = 3, y = 3 forms a square$ $whereas that bounded by 0 ⩽ r ⩽ 3 \sqrt{3}, 0 ⩽ θ ⩽ 2 π forms$ $a circle .$
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