The value of int_{0}^{pi /2} frac{e^{x^2}}{e^ | vedclass

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Question

(A) Let $I = \int_{0}^{\pi /2} \frac{e^{x^2}}{e^{x^2} + e^{(\pi /2 - x)^2}} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we

Vedclass · Accepted Answer

$\pi /4$

Vedclass · Answer

(A) Let $I = \int_{0}^{\pi /2} \frac{e^{x^2}}{e^{x^2} + e^{(\pi /2 - x)^2}} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{\pi /2} \frac{e^{(\pi /2 - x)^2}}{e^{(\pi /2 - x)^2} + e^{x^2}} dx$. Adding the two expressions for $I$: $2I = \int_{0}^{\pi /2} \frac{e^{x^2} + e^{(\pi /2 - x)^2}}{e^{x^2} + e^{(\pi /2 - x)^2}} dx$. $2I = \int_{0}^{\pi /2} 1 dx = [x]_{0}^{\pi /2} = \pi /2$. Therefore,$I = \pi /4$.

The value of $\int_{0}^{\pi /2} \frac{e^{x^2}}{e^{x^2} + e^{(\pi /2 - x)^2}} dx$ is

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