If I = int_{0}^{1} frac{dx}{1+x^{pi / 2}}, th | vedclass

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(A) We are given the integral $I = \int_{0}^{1} \frac{dx}{1+x^{\pi / 2}}$.Since $1 < \pi / 2 < 2$, for $x \in (0, 1)$, we have $x^2 < x^{\pi / 2} < x^

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$\log_{e} 2 < 1 < \pi / 4$

Vedclass · Answer

(A) We are given the integral $I = \int_{0}^{1} \frac{dx}{1+x^{\pi / 2}}$.Since $1 Adding $1$ to all sides, we get $1+x^2 Taking the reciprocal reverses the inequality: $\frac{1}{1+x^2} > \frac{1}{1+x^{\pi / 2}} > \frac{1}{1+x}$.Integrating from $0$ to $1$ with respect to $x$:$\int_{0}^{1} \frac{dx}{1+x^2} > \int_{0}^{1} \frac{dx}{1+x^{\pi / 2}} > \int_{0}^{1} \frac{dx}{1+x}$.Evaluating the integrals:$[ an^{-1}(x)]_{0}^{1} > I > [\log_{e}(1+x)]_{0}^{1}$.$\frac{\pi}{4} > I > \log_{e}(2)$.Thus, the correct relation is $\log_{e} 2 < I < \pi / 4$.

If $I = \int_{0}^{1} \frac{dx}{1+x^{\pi / 2}}$, then

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