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 evaluate the expression \( \Sigma_{r=1}^{10} A_r \), where \( A_n = \int_0^{n\pi} |\sin x| \, dx \).

### Step-by-step Solution:

1. **Understanding the Function \( A_n \)**:
We start with the definition:
\[ A_n = \int_0^{n\pi} |\sin x| \, dx \]
The function \( |\sin x| \) is periodic with a period of \( 2\pi \). Therefore, we can calculate \( A_n \) by breaking the integral into sections of \( 2\pi \).

2. **Calculating \( A_1 \)**:
For \( n=1 \):
\[ A_1 = \int_0^{\pi} |\sin x| \, dx \]
Since \( \sin x \) is non-negative in the interval \( [0, \pi] \):
\[ A_1 = \int_0^{\pi} \sin x \, dx \]
Evaluating this integral:
\[ A_1 = \left[-\cos x\right]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 2 \]

3. **Calculating \( A_2 \)**:
For \( n=2 \):
\[ A_2 = \int_0^{2\pi} |\sin x| \, dx \]
The integral from \( 0 \) to \( 2\pi \) can be split into two intervals:
\[ A_2 = \int_0^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} -\sin x \, dx \]
The first integral is \( 2 \) (as calculated before), and the second integral evaluates to:
\[ \int_{\pi}^{2\pi} -\sin x \, dx = -\left[-\cos x\right]_{\pi}^{2\pi} = -(-1 - 1) = 2 \]
Thus:
\[ A_2 = 2 + 2 = 4 \]

4. **Calculating \( A_3 \)**:
For \( n=3 \):
\[ A_3 = \int_0^{3\pi} |\sin x| \, dx \]
This can be calculated as:
\[ A_3 = \int_0^{2\pi} |\sin x| \, dx + \int_{2\pi}^{3\pi} |\sin x| \, dx \]
We already know \( A_2 = 4 \) and the integral from \( 2\pi \) to \( 3\pi \) is the same as from \( 0 \) to \( \pi \):
\[ A_3 = 4 + 2 = 6 \]

5. **Generalizing \( A_n \)**:
From the calculations, we can see a pattern:
\[ A_n = 2n \]
This holds because each complete cycle of \( 2\pi \) contributes an area of \( 4 \) (two areas of \( 2 \)).

6. **Calculating \( \Sigma_{r=1}^{10} A_r \)**:
Now we can compute:
\[ \Sigma_{r=1}^{10} A_r = A_1 + A_2 + A_3 + \ldots + A_{10} = 2(1 + 2 + 3 + \ldots + 10) \]
The sum of the first \( n \) natural numbers is given by:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]
For \( n=10 \):
\[ 1 + 2 + 3 + \ldots + 10 = \frac{10 \cdot 11}{2} = 55 \]
Therefore:
\[ \Sigma_{r=1}^{10} A_r = 2 \cdot 55 = 110 \]

### Final Answer:
\[ \Sigma_{r=1}^{10} A_r = 110 \]