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Evaluate: $\large \int \limits_0^1$$ x\;e^{x^2}\; dx$

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Substitute $ x^2=u $
Differentiating both the sides we get
$\Rightarrow  2x\;dx=du$
when $x=0,\:u=o$ and when $x=1,\:u=1$
$\Rightarrow\:\int^1_0 x\;e^{x^2}\; dx =$$\large\frac{1}{2} \large \int^1_0 $$ e^u du$
$\qquad = \large \frac{1}{2} $$[e^u]^1_0$
$\Rightarrow\:I = \large\frac{1}{2} $$(e^1-e^0) $
$\qquad = \large\frac{1}{2}$$(e-1) \approx 0.859$
answered Mar 21, 2014 by balaji
edited Mar 23, 2014 by rvidyagovindarajan_1

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