The value of int_0^1 (1 + e^{-x^2}) ,dx is: | vedclass

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(D) We are given the integral $I = \int_0^1 (1 + e^{-x^2}) \,dx$. By the linearity property of definite integrals,we can split this into two parts: $I

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) We are given the integral $I = \int_0^1 (1 + e^{-x^2}) \,dx$. By the linearity property of definite integrals,we can split this into two parts: $I = \int_0^1 1 \,dx + \int_0^1 e^{-x^2} \,dx$. The first part is $\int_0^1 1 \,dx = [x]_0^1 = 1 - 0 = 1$. The second part is $\int_0^1 e^{-x^2} \,dx$. The function $e^{-x^2}$ does not have an elementary antiderivative. Therefore,the integral $\int_0^1 e^{-x^2} \,dx$ cannot be expressed in terms of simple algebraic or transcendental functions. Thus,the value of the integral is $1 + \int_0^1 e^{-x^2} \,dx$. Comparing this with the given options,none of the options $A, B,$ or $C$ match this result. Therefore,the correct option is $D$.

The value of $\int_0^1 (1 + e^{-x^2}) \,dx$ is:

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