Let `f(x)=x^(2), g(x)="cos" x` and `alpha,beta (alpha lt beta)` be the roots of the equation `18x^(2)-19pi x+pi^(2)=0`. Then the area bounded by the curves `u="fog"(x)`, the ordinates `x=alpha, x=beta` and the X-asis is

Options

A. `(1)/(2)(pi-3)` sq units
B. `(pi)/(3)` sq units
C. `(pi)/(4)` sq units
D. None of these

Answer

To find the area bounded by the curves \( u = f(g(x)) \), the ordinates \( x = \alpha \), \( x = \beta \), and the x-axis, we will follow these steps:

### Step 1: Find the roots \( \alpha \) and \( \beta \) of the quadratic equation
The given quadratic equation is:
\[ 18x^2 - 19\pi x + \pi^2 = 0 \]
We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 18 \), \( b = -19\pi \), and \( c = \pi^2 \).

### Step 2: Substitute the values into the quadratic formula
Substituting the values:
\[ x = \frac{19\pi \pm \sqrt{(-19\pi)^2 - 4 \cdot 18 \cdot \pi^2}}{2 \cdot 18} \]
Calculating \( b^2 - 4ac \):
\[ (-19\pi)^2 = 361\pi^2 \quad \text{and} \quad 4ac = 72\pi^2 \]
Thus,
\[ b^2 - 4ac = 361\pi^2 - 72\pi^2 = 289\pi^2 \]
Now substituting back:
\[ x = \frac{19\pi \pm \sqrt{289\pi^2}}{36} \]
Since \( \sqrt{289\pi^2} = 17\pi \):
\[ x = \frac{19\pi \pm 17\pi}{36} \]

### Step 3: Calculate the roots \( \alpha \) and \( \beta \)
Calculating the roots:
\[ \alpha = \frac{19\pi - 17\pi}{36} = \frac{2\pi}{36} = \frac{\pi}{18} \]
\[ \beta = \frac{19\pi + 17\pi}{36} = \frac{36\pi}{36} = \pi \]

### Step 4: Define the function \( u = f(g(x)) \)
Given \( f(x) = x^2 \) and \( g(x) = \cos x \), we find:
\[ u = f(g(x)) = f(\cos x) = (\cos x)^2 \]

### Step 5: Set up the integral for the area
The area \( A \) bounded by the curve \( u \), the x-axis, and the lines \( x = \alpha \) and \( x = \beta \) is given by:
\[ A = \int_{\alpha}^{\beta} u \, dx = \int_{\frac{\pi}{18}}^{\pi} \cos^2 x \, dx \]

### Step 6: Use the identity for \( \cos^2 x \)
Using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \):
\[ A = \int_{\frac{\pi}{18}}^{\pi} \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2} \int_{\frac{\pi}{18}}^{\pi} (1 + \cos 2x) \, dx \]

### Step 7: Evaluate the integral
Calculating the integral:
\[ A = \frac{1}{2} \left[ \int_{\frac{\pi}{18}}^{\pi} 1 \, dx + \int_{\frac{\pi}{18}}^{\pi} \cos 2x \, dx \right] \]
The first integral:
\[ \int_{\frac{\pi}{18}}^{\pi} 1 \, dx = \left[ x \right]_{\frac{\pi}{18}}^{\pi} = \pi - \frac{\pi}{18} = \frac{17\pi}{18} \]
The second integral:
\[ \int \cos 2x \, dx = \frac{1}{2} \sin 2x \]
Evaluating it from \( \frac{\pi}{18} \) to \( \pi \):
\[ \left[ \frac{1}{2} \sin 2x \right]_{\frac{\pi}{18}}^{\pi} = \frac{1}{2} (\sin 2\pi - \sin \frac{\pi}{9}) = \frac{1}{2} (0 - \sin \frac{\pi}{9}) = -\frac{1}{2} \sin \frac{\pi}{9} \]

### Step 8: Combine results
Thus, the area becomes:
\[ A = \frac{1}{2} \left( \frac{17\pi}{18} - \frac{1}{2} \sin \frac{\pi}{9} \right) = \frac{17\pi}{36} - \frac{1}{4} \sin \frac{\pi}{9} \]

### Final Answer
The area bounded by the curves is:
\[ A = \frac{17\pi}{36} - \frac{1}{4} \sin \frac{\pi}{9} \]