यदि f(x) = int frac{dx}{x^2+2} और f(sqrt{2}) | vedclass

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

Question

(C) दिया गया है $f(x) = \int \frac{dx}{x^2+(\sqrt{2})^2}$.मानक समाकलन सूत्र $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$ का उपयो

Vedclass · Accepted Answer

$\frac{-\pi}{4 \sqrt{2}}$

Vedclass · Answer

(C) दिया गया है $f(x) = \int \frac{dx}{x^2+(\sqrt{2})^2}$.मानक समाकलन सूत्र $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$ का उपयोग करने पर:$f(x) = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{x}{\sqrt{2}}) + C$.चूँकि $f(\sqrt{2}) = 0$ है,$x = \sqrt{2}$ प्रतिस्थापित करने पर:$0 = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{\sqrt{2}}{\sqrt{2}}) + C$$0 = \frac{1}{\sqrt{2}} 	an^{-1}(1) + C$$0 = \frac{1}{\sqrt{2}} (\frac{\pi}{4}) + C$$C = -\frac{\pi}{4\sqrt{2}}$.अब,$f(0)$ ज्ञात करते हैं:$f(0) = \frac{1}{\sqrt{2}} 	an^{-1}(\frac{0}{\sqrt{2}}) - \frac{\pi}{4\sqrt{2}}$$f(0) = \frac{1}{\sqrt{2}} (0) - \frac{\pi}{4\sqrt{2}} = -\frac{\pi}{4\sqrt{2}}$.

यदि $f(x) = \int \frac{dx}{x^2+2}$ और $f(\sqrt{2}) = 0$ है,तो $f(0) =$

Similar Questions

$\int(1-\cos x) \operatorname{cosec}^2 x \, dx$ का मान ज्ञात कीजिए।

फलन का समाकलन कीजिए: $\frac{1}{\sqrt{9-25 x^{2}}}$

$\int \frac{\cos 2x}{\cos x} dx = $

$\int \frac{dx}{\tan x + \cot x} = $

$\int \frac{dx}{\cos x + \sqrt{3} \sin x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module