Question:
Prove that: \( \int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx \).
Prove that: \( \int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx \).
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Use substitution to transform integral limits; check variable changes carefully.
Updated On: Nov 10, 2025
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Solution and Explanation
Consider the integral \( \int_a^{2a} f(x) \, dx \). Substitute \( u = 2a - x \):
\[
x = a \Rightarrow u = 2a - a = a, \quad x = 2a \Rightarrow u = 0, \quad dx = -du.
\]
\[
\int_a^{2a} f(x) \, dx = \int_a^0 f(2a - u) (-du) = \int_0^a f(2a - u) \, du.
\]
Thus:
\[
\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(2a - x) \, dx.
\]
Proved.
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