The smallest interval [a, b] such that int_0^ | vedclass

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

Question

(A) Let $I = \int_0^1 \frac{dx}{\sqrt{1 + x^4}}$.For $x \in [0, 1]$,we have $0 \le x^4 \le 1$.Adding $1$ to all sides,we get $1 \le 1 + x^4 \le 2$.Tak

Vedclass · Accepted Answer

$[\frac{1}{\sqrt{2}}, 1]$

Vedclass · Answer

(A) Let $I = \int_0^1 \frac{dx}{\sqrt{1 + x^4}}$.For $x \in [0, 1]$,we have $0 \le x^4 \le 1$.Adding $1$ to all sides,we get $1 \le 1 + x^4 \le 2$.Taking the square root,we get $1 \le \sqrt{1 + x^4} \le \sqrt{2}$.Taking the reciprocal,the inequality reverses: $\frac{1}{\sqrt{2}} \le \frac{1}{\sqrt{1 + x^4}} \le 1$.Integrating from $0$ to $1$ with respect to $x$,we get $\int_0^1 \frac{1}{\sqrt{2}} dx \le \int_0^1 \frac{dx}{\sqrt{1 + x^4}} \le \int_0^1 1 dx$.This simplifies to $\frac{1}{\sqrt{2}}(1 - 0) \le I \le 1(1 - 0)$,which gives $\frac{1}{\sqrt{2}} \le I \le 1$.Thus,the smallest interval is $[\frac{1}{\sqrt{2}}, 1]$.

The smallest interval $[a, b]$ such that $\int_0^1 \frac{dx}{\sqrt{1 + x^4}} \in [a, b]$ is given by

Similar Questions

Let $\alpha > 0$. If $\int \limits _0^\alpha \frac{ x }{\sqrt{ x +\alpha}-\sqrt{ x }} dx =\frac{16+20 \sqrt{2}}{15}$,then $\alpha$ is equal to :

The value of the definite integral $\int_{0}^{1} e^{e^x}(1 + x e^x) dx$ is equal to

If $[2,6]$ is divided into four intervals of equal length,then the approximate value of $\int_2^6 \frac{1}{x^2-x} dx$ using Simpson's rule is

$\int_{0}^{\frac{\pi}{4}} (\tan^n x + \tan^{n-2} x) d(x - [x])$ is : (where $[.]$ denotes the greatest integer function)

$\int_{1}^{3} (x - 1)(x - 2)(x - 3) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module