_0^ 3 ( x + 4 ) ^2 e^ x^2 dx + _ 3 ^0 ( x - 4 ) ^2 e^ x^2 dx is e… - Vidyadip

MCQ

$\int\limits_0^{\sqrt 3 } {{{\left( {x + 4} \right)}^2}{e^{{x^2}}}dx + \int\limits_{\sqrt 3 }^0 {{{\left( {x - 4} \right)}^2}{e^{{x^2}}}dx} } $ is equal to

A
$8e^3$
✓
$8(e^3 -1)$
C
$\sqrt 3 \left( {{e^4} - 1} \right)$
D
$\sqrt 3 \left( {{e^8} - 1} \right)$

✓

Answer

Correct option: B.

$8(e^3 -1)$

b
$\int_{0}^{\sqrt{3}} e^{x^{2}}\left\{(x+4)^{2}-(x-4)^{2}\right\} d x$

$\left.=\int_{0}^{\sqrt{3}} \mathrm{e}^{\mathrm{x}^{2}} 16 \mathrm{x} \mathrm{\,dx}=8 \mathrm{e}^{\mathrm{x}^{2}}\right)_{0}^{\sqrt{3}}=8\left(\mathrm{e}^{3}-1\right)$

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