int_{0}^{sqrt{3}} (x+4)^2 e^{x^2} dx + int_{s | vedclass

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(B) We are given the expression $I = \int_{0}^{\sqrt{3}} (x+4)^2 e^{x^2} dx + \int_{\sqrt{3}}^{0} (x-4)^2 e^{x^2} dx$. Using the property $\int_{b}^{a

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$8(e^3 - 1)$

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(B) We are given the expression $I = \int_{0}^{\sqrt{3}} (x+4)^2 e^{x^2} dx + \int_{\sqrt{3}}^{0} (x-4)^2 e^{x^2} dx$. Using the property $\int_{b}^{a} f(x) dx = -\int_{a}^{b} f(x) dx$,we can rewrite the second integral as $-\int_{0}^{\sqrt{3}} (x-4)^2 e^{x^2} dx$. Thus,$I = \int_{0}^{\sqrt{3}} e^{x^2} \{(x+4)^2 - (x-4)^2\} dx$. Expanding the terms inside the bracket: $(x+4)^2 - (x-4)^2 = (x^2 + 8x + 16) - (x^2 - 8x + 16) = 16x$. So,$I = \int_{0}^{\sqrt{3}} e^{x^2} (16x) dx$. Let $u = x^2$,then $du = 2x dx$,which implies $8 du = 16x dx$. When $x=0, u=0$. When $x=\sqrt{3}, u=3$. $I = \int_{0}^{3} 8 e^u du = 8 [e^u]_{0}^{3} = 8(e^3 - e^0) = 8(e^3 - 1)$.

$\int_{0}^{\sqrt{3}} (x+4)^2 e^{x^2} dx + \int_{\sqrt{3}}^{0} (x-4)^2 e^{x^2} dx$ is equal to

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