If `overset(a)underset(0) int f(2a-x)dx = m and overset(a)underset(0)int f(x) dx=n`, then `overset(2a)underset(0) int f(x) dx` is equal to

Options

A. `2m+n`
B. `m+2n`
C. `m-n`
D. `m+n`

Answer

To solve the problem, we need to evaluate the integral \( \int_0^{2a} f(x) \, dx \) given the following:

1. \( \int_0^a f(2a - x) \, dx = m \)
2. \( \int_0^a f(x) \, dx = n \)

### Step 1: Break down the integral \( \int_0^{2a} f(x) \, dx \)

We can split the integral from \( 0 \) to \( 2a \) into two parts:

\[ \int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(x) \, dx \]

### Step 2: Evaluate \( \int_a^{2a} f(x) \, dx \)

To evaluate \( \int_a^{2a} f(x) \, dx \), we can use a substitution. Let \( x = 2a - u \). Then, when \( x = a \), \( u = a \) and when \( x = 2a \), \( u = 0 \). The differential \( dx = -du \).

Thus, we have:

\[ \int_a^{2a} f(x) \, dx = \int_a^0 f(2a - u)(-du) = \int_0^a f(2a - u) \, du \]

### Step 3: Relate the integral to \( m \)

From the substitution, we find that:

\[ \int_a^{2a} f(x) \, dx = \int_0^a f(2a - u) \, du = m \]

### Step 4: Combine the results

Now we can combine our results:

\[ \int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(x) \, dx = n + m \]

### Conclusion

Thus, we find that:

\[ \int_0^{2a} f(x) \, dx = m + n \]

### Final Answer

The value of \( \int_0^{2a} f(x) \, dx \) is \( m + n \).

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