The value of the integral int_{0}^{1} e^{x^{2 | vedclass

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(D) We know that for $x \in [0, 1]$,$0 \leq x^2 \leq 1$.Since $e^x$ is an increasing function,we have $e^0 \leq e^{x^2} \leq e^1$,which implies $1 \le

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lies in the closed interval $[1, e]$

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(D) We know that for $x \in [0, 1]$,$0 \leq x^2 \leq 1$.Since $e^x$ is an increasing function,we have $e^0 \leq e^{x^2} \leq e^1$,which implies $1 \leq e^{x^2} \leq e$.Integrating the inequality over the interval $[0, 1]$:$\int_{0}^{1} 1 \, dx \leq \int_{0}^{1} e^{x^2} \, dx \leq \int_{0}^{1} e \, dx$.Calculating the integrals:$[x]_{0}^{1} \leq \int_{0}^{1} e^{x^2} \, dx \leq [ex]_{0}^{1}$.$1 \leq \int_{0}^{1} e^{x^2} \, dx \leq e$.Thus,the value of the integral lies in the closed interval $[1, e]$.

The value of the integral $\int_{0}^{1} e^{x^{2}} d x$ is:

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