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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