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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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{

Vedclass · Accepted Answer

only $I$ is true

Vedclass · Answer

(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,

Similar Questions

If $\int_{0}^{1} \tan ^{-1} x \, dx = p$,then the value of $\int_{0}^{1} \tan ^{-1}\left(\frac{1-x}{1+x}\right) \, dx$ is

If $\int_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x \, dx}{(1+e^{\sin x})(1+\sin ^4 x)} = \alpha \pi + \beta \log _e(3+2 \sqrt{2})$,where $\alpha, \beta$ are integers,then $\alpha^2+\beta^2$ equals.....................

The value of $I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}} dx$ is

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{1+\sin^{2} x}{1+\pi^{\sin x}}\right) \, dx$ is

$\int_{0}^{100} e^{x-[x]} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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,