Let `overset(a)underset(0)int f(x)dx=lambda` and `overset(a)underset(0)int f(2a-x)dx=mu`. Then,
`overset(2a)underset(0)int f(x) dx` equal to

Options

A. `lambda + mu`
B. `lambda - mu`
C. `2lambda - mu`
D. `lambda 2 mu`

Answer

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

1. \(\int_0^a f(x) \, dx = \lambda\)
2. \(\int_0^a f(2a - x) \, dx = \mu\)

### Step-by-Step Solution:

**Step 1: Break down the integral from 0 to 2a**

We can express the integral from 0 to \(2a\) as the sum of two integrals:
\[ \int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(x) \, dx \]

**Step 2: Change of variable for the second integral**

For the second integral, we will perform a change of variable. Let \(x = 2a - u\). Then when \(x = a\), \(u = a\) and when \(x = 2a\), \(u = 0\). Thus, we have:
\[ \int_a^{2a} f(x) \, dx = \int_a^0 f(2a - u) \, (-du) = \int_0^a f(2a - u) \, du \]
This simplifies to:
\[ \int_a^{2a} f(x) \, dx = \int_0^a f(2a - x) \, dx \]

**Step 3: Substitute known values**

From the problem, we know:
\[ \int_0^a f(x) \, dx = \lambda \]
and
\[ \int_0^a f(2a - x) \, dx = \mu \]
Thus, we can substitute these values into our expression:
\[ \int_0^{2a} f(x) \, dx = \lambda + \mu \]

**Final Result:**

Therefore, the value of \(\int_0^{2a} f(x) \, dx\) is:
\[ \int_0^{2a} f(x) \, dx = \lambda + \mu \]

### Conclusion:

The final answer is \(\lambda + \mu\).

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