Let `f(x) and g(x)` be any two continuous function in the interval `[0, b]` and 'a' be any point between 0 and b. Which satisfy the following conditions : `f(x)=f(a-x), g(x)+g(a-x)=3, f(a+b-x)=f(x)`. Also `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx, int_(a)^(b)f(x)dx=int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`
If `int_(0)^(a//2)f(x)dx=p," then "int_(0)^(a)f(x)dx` is equal to

Options

A. 2p
B. 0
C. p
D. 3p

Answer

To solve the problem step by step, we will analyze the given conditions and use them to find the value of the integral \( \int_{0}^{a} f(x) \, dx \).

### Step 1: Understanding the Given Conditions
We have two continuous functions \( f(x) \) and \( g(x) \) defined on the interval \([0, b]\), and a point \( a \) between \( 0 \) and \( b \). The conditions provided are:
1. \( f(x) = f(a - x) \)
2. \( g(x) + g(a - x) = 3 \)
3. \( f(a + b - x) = f(x) \)

### Step 2: Analyze the Symmetry of \( f(x) \)
From the first condition \( f(x) = f(a - x) \), we can see that \( f(x) \) is symmetric about \( x = \frac{a}{2} \). This implies that:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
This means that the area under the curve from \( 0 \) to \( a \) is the same as from \( a \) to \( 0 \).

### Step 3: Split the Integral
We can split the integral \( \int_{0}^{a} f(x) \, dx \) into two parts:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{\frac{a}{2}} f(x) \, dx + \int_{\frac{a}{2}}^{a} f(x) \, dx \]

### Step 4: Change of Variable in the Second Integral
For the second integral, we can use the substitution \( x = a - t \) where \( dx = -dt \). When \( x = \frac{a}{2} \), \( t = \frac{a}{2} \) and when \( x = a \), \( t = 0 \). Thus:
\[ \int_{\frac{a}{2}}^{a} f(x) \, dx = \int_{0}^{\frac{a}{2}} f(a - t) \, (-dt) = \int_{0}^{\frac{a}{2}} f(t) \, dt \]
This means:
\[ \int_{\frac{a}{2}}^{a} f(x) \, dx = \int_{0}^{\frac{a}{2}} f(t) \, dt \]

### Step 5: Combine the Integrals
Now we can combine the two parts:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{\frac{a}{2}} f(x) \, dx + \int_{0}^{\frac{a}{2}} f(x) \, dx = 2 \int_{0}^{\frac{a}{2}} f(x) \, dx \]

### Step 6: Substitute the Given Value
We are given that \( \int_{0}^{\frac{a}{2}} f(x) \, dx = p \). Thus:
\[ \int_{0}^{a} f(x) \, dx = 2p \]

### Final Answer
Therefore, the value of \( \int_{0}^{a} f(x) \, dx \) is:
\[ \int_{0}^{a} f(x) \, dx = 2p \]