Let J = int_0^1 frac{x}{1+x^8} dx. Consider t | vedclass

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(A) To evaluate $J = \int_0^1 \frac{x}{1+x^8} dx$,we analyze the assertions.For assertion $I$:Since $1+x^8 < 2$ for all $x \in (0, 1)$,we have $\frac{

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only $I$ is true

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(A) To evaluate $J = \int_0^1 \frac{x}{1+x^8} dx$,we analyze the assertions.For assertion $I$:Since $1+x^8  \frac{1}{2}$.Multiplying by $x > 0$,we get $\frac{x}{1+x^8} > \frac{x}{2}$.Integrating both sides from $0$ to $1$:$J = \int_0^1 \frac{x}{1+x^8} dx > \int_0^1 \frac{x}{2} dx = \left[ \frac{x^2}{4} ight]_0^1 = \frac{1}{4}$.Thus,assertion $I$ is true.For assertion $II$:Let $u = x^4$,then $du = 4x^3 dx$,which is not directly helpful. Instead,consider the integral $J = \int_0^1 \frac{x}{1+x^8} dx$. Let $u = x^4$,then $du = 4x^3 dx$. This doesn't simplify easily. Let's use the substitution $u = x^4$,$du = 4x^3 dx$. Then $J = \int_0^1 \frac{x}{1+(x^4)^2} dx$. Let $u = x^4$,$du = 4x^3 dx$. This is not standard. Actually,let $u = x^4$,then $du = 4x^3 dx$. The integral is $J = \frac{1}{4} \int_0^1 \frac{4x^3}{x^2(1+x^8)} dx$. Alternatively,note that for $x \in (0, 1)$,$x^8 Thus $\frac{1}{1+x^8} > \frac{1}{1+x^4}$.Then $J = \int_0^1 \frac{x}{1+x^8} dx > \int_0^1 \frac{x}{1+x^4} dx$.Let $u = x^2$,$du = 2x dx$. Then $\int_0^1 \frac{x}{1+x^4} dx = \frac{1}{2} \int_0^1 \frac{du}{1+u^2} = \frac{1}{2} [	an^{-1}(u)]_0^1 = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}$.Since $J > \frac{\pi}{8}$,assertion $II$ is false.Therefore,only $I$ is true.

Let $J = \int_0^1 \frac{x}{1+x^8} dx$. Consider the following assertions:
$I$. $J > \frac{1}{4}$
$II$. $J < \frac{\pi}{8}$
Then,

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Let $J = \int_0^1 \frac{x}{1+x^8} dx$. Consider the following assertions:$I$. $J > \frac{1}{4}$$II$. $J < \frac{\pi}{8}$Then,