MathematicsQ1–1 of 1 questions
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1
MathematicsEasyMCQGSEB · 2008
The area of the region bounded by the curve $y = \tan x$,$x = 0$,$x = \frac{\pi}{4}$,and the $X$-axis is . . . . . . sq. units.
A
$\log 2$
B
$\frac{3}{2} \log 2$
C
$\frac{1}{2} \log 2$
D
$2 \log 2$
Solution
(C) The area $A$ of the region bounded by the curve $y = f(x)$,the $X$-axis,and the lines $x = a$ and $x = b$ is given by the integral:
$A = \int_{a}^{b} |y| \, dx$
Here,$y = \tan x$,$a = 0$,and $b = \frac{\pi}{4}$.
Since $\tan x \geq 0$ in the interval $[0, \frac{\pi}{4}]$,we have:
$A = \int_{0}^{\frac{\pi}{4}} \tan x \, dx$
The integral of $\tan x$ is $\ln|\sec x|$:
$A = [\ln|\sec x|]_{0}^{\frac{\pi}{4}}$
$A = \ln|\sec(\frac{\pi}{4})| - \ln|\sec(0)|$
$A = \ln(\sqrt{2}) - \ln(1)$
$A = \ln(2^{1/2}) - 0$
$A = \frac{1}{2} \ln 2$
Thus,the correct option is $C$.
$A = \int_{a}^{b} |y| \, dx$
Here,$y = \tan x$,$a = 0$,and $b = \frac{\pi}{4}$.
Since $\tan x \geq 0$ in the interval $[0, \frac{\pi}{4}]$,we have:
$A = \int_{0}^{\frac{\pi}{4}} \tan x \, dx$
The integral of $\tan x$ is $\ln|\sec x|$:
$A = [\ln|\sec x|]_{0}^{\frac{\pi}{4}}$
$A = \ln|\sec(\frac{\pi}{4})| - \ln|\sec(0)|$
$A = \ln(\sqrt{2}) - \ln(1)$
$A = \ln(2^{1/2}) - 0$
$A = \frac{1}{2} \ln 2$
Thus,the correct option is $C$.
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