If `A_(n)=int_(0)^(npi)|sinx|dx, AA n in N`, then `Sigma_(r=1)^(10)A_(r)` is equal to

Options

A. 100
B. 110
C. 55
D. 105

Answer

To solve the problem, we need to calculate the sum \( \Sigma_{r=1}^{10} A_r \) where \( A_n = \int_0^{n\pi} |\sin x| \, dx \).

### Step 1: Understanding \( A_n \)
We start with the expression for \( A_n \):
\[ A_n = \int_0^{n\pi} |\sin x| \, dx \]
The function \( |\sin x| \) has a periodic behavior with a period of \( 2\pi \).

### Step 2: Calculate \( A_1 \)
For \( n = 1 \):
\[ A_1 = \int_0^{\pi} |\sin x| \, dx \]
Since \( \sin x \) is non-negative in the interval \( [0, \pi] \), we can write:
\[ A_1 = \int_0^{\pi} \sin x \, dx \]
Calculating this integral:
\[ A_1 = [-\cos x]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

### Step 3: Calculate \( A_2 \)
For \( n = 2 \):
\[ A_2 = \int_0^{2\pi} |\sin x| \, dx \]
This integral can be split into two parts:
\[ A_2 = \int_0^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} -\sin x \, dx \]
Calculating the second integral:
\[ \int_{\pi}^{2\pi} -\sin x \, dx = -[-\cos x]_{\pi}^{2\pi} = -(-\cos(2\pi) + \cos(\pi)) = -(-1 + 1) = 0 \]
Thus,
\[ A_2 = 2 + 2 = 4 \]

### Step 4: Generalizing \( A_n \)
For \( n \):
- Each complete cycle from \( 0 \) to \( 2\pi \) contributes an area of \( 4 \).
- For \( n \) complete cycles, the area is \( 2n \).

Thus, we can generalize:
\[ A_n = 2n \]

### Step 5: Calculate \( \Sigma_{r=1}^{10} A_r \)
Now, we compute the sum:
\[ \Sigma_{r=1}^{10} A_r = A_1 + A_2 + A_3 + \ldots + A_{10} = 2(1 + 2 + 3 + \ldots + 10) \]
Using the formula for the sum of the first \( n \) natural numbers:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \]
For \( n = 10 \):
\[ 1 + 2 + 3 + \ldots + 10 = \frac{10 \times 11}{2} = 55 \]
Thus,
\[ \Sigma_{r=1}^{10} A_r = 2 \times 55 = 110 \]

### Final Answer
The value of \( \Sigma_{r=1}^{10} A_r \) is \( 110 \).