Question

Area of the region bounded by the curve \(y = tan x\), the x- axis and the line \(x = \frac{\pi}{3}\) is

• A. \(\log \frac{1}{2}\)
• B. 0
• C. log 2
• D. - log 2

Answer 1

The Correct Option is C

The region lies between \(x = 0\) and \(x = \frac{\pi}{3}\).

We can find the area by integrating the absolute value of the function \(y = \tan x\) within this interval.

Thus, the area (A) is given by:

\[ A = \int_{0}^{\pi/3} | \tan x | \, dx \]

To evaluate this integral, we need to break it up into two parts due to the nature of the tangent function.

\[ A = \int_{0}^{\pi/3} \tan x \, dx - \int_{0}^{\pi/3} (-\tan x) \, dx \]

Simplifying this expression, we have:

\[ A = \int_{0}^{\pi/3} \tan x \, dx + \int_{0}^{\pi/3} \tan x \, dx \]

Combining the integrals, we get:

\[ A = 2 \int_{0}^{\pi/3} \tan x \, dx \]

Using the integral property, we have:

\[ A = 2 \left[\log | \sec x | \right]_{0}^{\pi/3} \]

\[ A = 2 \left[\log(|\sec(\frac{\pi}{3})|) - \log(|\sec 0|)\right] \]

\[ A = 2 [\log(2) - \log(1)] \]

\[ A = 2 \log\left(\frac{2}{1}\right) \]

\[ A =  \log(2) \]

Therefore, the area of the region bounded by the curve \(y = \tan x\), the x-axis, and the line \(x = \frac{\pi}{3}\) is \( \log(2)\), which corresponds to option (C) \( \log(2)\).

Answer 2

We are given the curve \( y = \tan x \), the x-axis, and the vertical line \( x = \frac{\pi}{3} \).
Step 1: Set up the definite integral
The required area is: \[ \int_{0}^{\frac{\pi}{3}} \tan x \, dx \]

Step 2: Use the identity
\[ \int \tan x \, dx = -\log|\cos x| + C \] So, \[ \int_{0}^{\frac{\pi}{3}} \tan x \, dx = \left[ -\log|\cos x| \right]_0^{\frac{\pi}{3}} = -\log\left(\cos\left(\frac{\pi}{3}\right)\right) + \log\left(\cos(0)\right) \] 

Step 3: Simplify values
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \cos(0) = 1 \] \[ = -\log\left(\frac{1}{2}\right) + \log(1) = -\log\left(\frac{1}{2}\right) \] 

Step 4: Final simplification
\[ -\log\left(\frac{1}{2}\right) = \log 2 \] 

Final Answer: \(\log 2\)

Answer 3

The region is bounded by the curve \(y = \tan x\), the x-axis (\(y = 0\)), and the vertical line \(x = \frac{\pi}{3}\). 

The tangent function \(y = \tan x\) intersects the x-axis at \(x=0\) within the relevant domain. For \(x\) in the interval \(\left[0, \frac{\pi}{3}\right]\), \(\tan x \ge 0\). Therefore, the curve lies above the x-axis in this interval.

The area \(A\) of the region is given by the definite integral of the function \(y = \tan x\) from the lower limit \(x=0\) to the upper limit \(x = \frac{\pi}{3}\).

\[ A = \int_{0}^{\pi/3} \tan x \, dx \]

We know that the integral of \(\tan x\) is \(-\ln|\cos x|\) or \(\ln|\sec x|\). Let's use \(-\ln|\cos x|\).

\[ A = [-\ln|\cos x|]_{0}^{\pi/3} \]

Now, we evaluate the antiderivative at the upper and lower limits:

\[ A = \left(-\ln\left|\cos\left(\frac{\pi}{3}\right)\right|\right) - \left(-\ln|\cos(0)|\right) \]

We know that \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\) and \(\cos(0) = 1\).

Substitute these values:

\[ A = -\ln\left|\frac{1}{2}\right| - (-\ln|1|) \]

Since \(\frac{1}{2} > 0\) and \(1 > 0\), we can remove the absolute value signs:

\[ A = -\ln\left(\frac{1}{2}\right) + \ln(1) \]

We know that \(\ln(1) = 0\) and \(\ln\left(\frac{1}{a}\right) = -\ln(a)\).

\[ A = -(-\ln(2)) + 0 \]

\[ A = \ln(2) \]

Assuming 'log' represents the natural logarithm ('ln') as is common in calculus contexts involving trigonometric functions:

The area is log 2.

Comparing this with the given options, the correct option is:

log 2