MathematicsQ1–1 of 1 questions
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1
MathematicsEasyMCQGSEB · 2008
वक्र $y = \tan x$,$x = 0$,$x = \frac{\pi}{4}$ और $X$-अक्ष द्वारा परिबद्ध क्षेत्र का क्षेत्रफल . . . . . . वर्ग इकाई है।
A
$\log 2$
B
$\frac{3}{2} \log 2$
C
$\frac{1}{2} \log 2$
D
$2 \log 2$
Solution
(C) वक्र $y = f(x)$,$X$-अक्ष और रेखाओं $x = a$ तथा $x = b$ द्वारा परिबद्ध क्षेत्र का क्षेत्रफल $A$ निम्नलिखित समाकल द्वारा दिया जाता है:
$A = \int_{a}^{b} |y| \, dx$
यहाँ,$y = \tan x$,$a = 0$,और $b = \frac{\pi}{4}$ है।
चूँकि अंतराल $[0, \frac{\pi}{4}]$ में $\tan x \geq 0$ है,इसलिए:
$A = \int_{0}^{\frac{\pi}{4}} \tan x \, dx$
$\tan x$ का समाकल $\ln|\sec x|$ होता है:
$A = [\ln|\sec x|]_{0}^{\frac{\pi}{4}}$
$A = \ln|\sec(\frac{\pi}{4})| - \ln|\sec(0)|$
$A = \ln(\sqrt{2}) - \ln(1)$
$A = \ln(2^{1/2}) - 0$
$A = \frac{1}{2} \ln 2$
अतः,सही विकल्प $C$ है।
$A = \int_{a}^{b} |y| \, dx$
यहाँ,$y = \tan x$,$a = 0$,और $b = \frac{\pi}{4}$ है।
चूँकि अंतराल $[0, \frac{\pi}{4}]$ में $\tan x \geq 0$ है,इसलिए:
$A = \int_{0}^{\frac{\pi}{4}} \tan x \, dx$
$\tan x$ का समाकल $\ln|\sec x|$ होता है:
$A = [\ln|\sec x|]_{0}^{\frac{\pi}{4}}$
$A = \ln|\sec(\frac{\pi}{4})| - \ln|\sec(0)|$
$A = \ln(\sqrt{2}) - \ln(1)$
$A = \ln(2^{1/2}) - 0$
$A = \frac{1}{2} \ln 2$
अतः,सही विकल्प $C$ है।
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