Question

Evaluate \[ \int_{-a}^{a} f(x)\,dx-\int_{0}^{a} f(-x)\,dx \]

• A. \(\displaystyle \int_{-a}^{a} f(a-x)\,dx\)
• B. \(\displaystyle \int_{-a}^{a} \{f(x)+f(a-x)\}\,dx\)
• C. \(\displaystyle \int_{0}^{a} \{f(x)+f(a-x)\}\,dx\)
• D. \(\displaystyle \int_{0}^{a} f(a-x)\,dx\)

Hint:

For definite integrals, remember the useful property: \[ \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx \] This property is very helpful in simplifying expressions involving limits \(0\) and \(a\).

Answer 1

The Correct Option is D

Step 1: Write the expression to simplify.
$ E=\int_{-a}^{a}f(x)dx-\int_0^a f(-x)dx $.
Step 2: Split the first integral at 0.
\[ \int_{-a}^{a}f(x)dx = \int_{-a}^{0}f(x)dx+\int_0^a f(x)dx \]
Step 3: Substitute $ x=-t $ in $ \int_{-a}^{0}f(x)dx $.
\[ \int_{-a}^{0}f(x)dx=\int_0^a f(-t)dt=\int_0^a f(-x)dx \]
Step 4: Cancel the $ f(-x) $ integrals.
\[ E=\int_0^a f(-x)dx+\int_0^a f(x)dx-\int_0^a f(-x)dx=\int_0^a f(x)dx \]
Step 5: Apply the reflection property.
$ \int_0^a f(x)dx=\int_0^a f(a-x)dx $ (substitute $ x\to a-x $).
Step 6: State the final result.
\[ E=\int_0^a f(a-x)dx \] \[ \boxed{\int_0^a f(a-x)\,dx} \]