Mathematics

Number of real solution of the given equation for $x$, $\int x^{2}\ e^{x}dx=0$

SOLUTION
$\int_{}^{} {{x^2}{e^x}dx = 0}$
$\Rightarrow {x^2}{e^x} - \int_{}^{} {2x{e^x}dx} = 0$
$\Rightarrow {x^2}{e^x} - 2\left[ {x{e^x} - \int_{}^{} {{e^x}dx} } \right] = 0$
$\Rightarrow {x^2}{e^x} - 2x{e^x} + 2{e^x} = 0$
$\Rightarrow {e^x}\left( {{x^2} - 2x + 2} \right) = 0$
$\Rightarrow \left( {{x^2} - 2x + 2} \right) = 0$
Now, $D = {\left( { - 2} \right)^2} - 4 \times 1 \times 2 = 4 - 8 = - 4 < 0$
So, it has no real roots

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One Word Medium Published on 17th 09, 2020
Questions 203525
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