
What is the area bounded by $y = \tan x$, $y = 0$ and $x = \dfrac{\pi }{4}$?
A. $\ln 2$ square units
B. $\dfrac{{\ln 2}}{2}$ square units
C. $2\left( {\ln 2} \right)$ square units
D. None of these
Answer
162.6k+ views
Hint: In this question, we are given the equation of the curve and the lines. We have to find the area bounded by them. Firstly, plot the graph of the following. Then, for the area integrate the equation of the curve with respect to $dx$ from $x = 0$ to $x = \dfrac{\pi }{4}$. Solve it further using integration, trigonometric and logarithm formulas.
Formula Used: Integration formula β
$\int {\tan xdx = \ln \left| {\sec x} \right| + c} $
Trigonometric table values β
$\sec \dfrac{\pi }{4} = \sqrt 2 $, $\sec {0^ \circ } = 1$
Logarithm formula β
$\log \sqrt[n]{x} = \dfrac{1}{n}\log x$
Complete step by step Solution:
Given that,
Equation of given curve is $y = \tan x$ and the line is $y = 0$, $x = \dfrac{\pi }{4}$
A graph of the given equations is attached below (figure 1);

Figure 1: A graph containing the given equation of the curve y=tan x and the line x=π/4
Now, as you can see the area of the bounded region lies between $x = 0$ to $x = \dfrac{\pi }{4}$
So, integrate $y = \tan x$ with respect to $dx$ from $0$ to $\dfrac{\pi }{4}$
Therefore, area will be $A = \int\limits_0^{\dfrac{\pi }{4}} {\tan xdx} $
As we know that, integration of the function $\tan x$ is $\ln \left| {\sec x} \right| + c$
So, area $A = \left[ {\ln \left| {\sec x} \right|} \right]_0^{\dfrac{\pi }{4}}$
Now, resolving the limits we get $A = \ln \left| {\sec \dfrac{\pi }{4}} \right| - \ln \left| {\sec 0} \right|$
Using trigonometric table,
$A = \ln \sqrt 2 - \ln 1$
$A = \ln \sqrt 2 - 0$
Also, written as $A = \dfrac{{\ln 2}}{2}$ (as $\log \sqrt[n]{x} = \dfrac{1}{n}\log x$)
Therefore, the correct option is (B).
Note: When we calculate the area between the curves we should also plot the graph correctly to calculate the area bounded by the curve and line. Here we have used the trigonometric tables of values of the function at a specific angle, logarithmic properties, and integration formulas so we must remember these values to calculate this kind of question.
Formula Used: Integration formula β
$\int {\tan xdx = \ln \left| {\sec x} \right| + c} $
Trigonometric table values β
$\sec \dfrac{\pi }{4} = \sqrt 2 $, $\sec {0^ \circ } = 1$
Logarithm formula β
$\log \sqrt[n]{x} = \dfrac{1}{n}\log x$
Complete step by step Solution:
Given that,
Equation of given curve is $y = \tan x$ and the line is $y = 0$, $x = \dfrac{\pi }{4}$
A graph of the given equations is attached below (figure 1);

Figure 1: A graph containing the given equation of the curve y=tan x and the line x=π/4
Now, as you can see the area of the bounded region lies between $x = 0$ to $x = \dfrac{\pi }{4}$
So, integrate $y = \tan x$ with respect to $dx$ from $0$ to $\dfrac{\pi }{4}$
Therefore, area will be $A = \int\limits_0^{\dfrac{\pi }{4}} {\tan xdx} $
As we know that, integration of the function $\tan x$ is $\ln \left| {\sec x} \right| + c$
So, area $A = \left[ {\ln \left| {\sec x} \right|} \right]_0^{\dfrac{\pi }{4}}$
Now, resolving the limits we get $A = \ln \left| {\sec \dfrac{\pi }{4}} \right| - \ln \left| {\sec 0} \right|$
Using trigonometric table,
$A = \ln \sqrt 2 - \ln 1$
$A = \ln \sqrt 2 - 0$
Also, written as $A = \dfrac{{\ln 2}}{2}$ (as $\log \sqrt[n]{x} = \dfrac{1}{n}\log x$)
Therefore, the correct option is (B).
Note: When we calculate the area between the curves we should also plot the graph correctly to calculate the area bounded by the curve and line. Here we have used the trigonometric tables of values of the function at a specific angle, logarithmic properties, and integration formulas so we must remember these values to calculate this kind of question.
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