The value of the definite integral int_{0}^{1 | vedclass

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(A) Let $I = \int_{0}^{1} e^{e^x}(1 + x e^x) dx$. Consider the substitution $u = x e^x$. Then $du = (1 \cdot e^x + x \cdot e^x) dx = e^x(1 + x) dx$. T

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$e^e$

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(A) Let $I = \int_{0}^{1} e^{e^x}(1 + x e^x) dx$. Consider the substitution $u = x e^x$. Then $du = (1 \cdot e^x + x \cdot e^x) dx = e^x(1 + x) dx$. This does not simplify directly. Let us rewrite the integral as $I = \int_{0}^{1} e^{e^x} dx + \int_{0}^{1} x e^x e^{e^x} dx$. Consider the second part: $\int_{0}^{1} x e^x e^{e^x} dx$. Let $v = e^x$. Then $dv = e^x dx$. When $x=0, v=1$. When $x=1, v=e$. The integral becomes $\int_{1}^{e} \ln(v) e^v dv$. This is not standard. Let us re-examine the original expression: $\int_{0}^{1} e^{e^x} dx + \int_{0}^{1} x e^x e^{e^x} dx$. Actually,notice that $\frac{d}{dx}(x e^{e^x}) = 1 \cdot e^{e^x} + x \cdot e^{e^x} \cdot e^x = e^{e^x} + x e^x e^{e^x} = e^{e^x}(1 + x e^x)$. Thus,the integral is simply $\left[ x e^{e^x} \right]_{0}^{1}$. Evaluating at the limits: $(1 \cdot e^{e^1}) - (0 \cdot e^{e^0}) = e^e - 0 = e^e$.

The value of the definite integral $\int_{0}^{1} e^{e^x}(1 + x e^x) dx$ is equal to

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