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: Split the first integral.
We have \[ \int_{-a}^{a} f(x)\,dx = \int_{-a}^{0} f(x)\,dx+\int_{0}^{a} f(x)\,dx \] Therefore, \[ \int_{-a}^{a} f(x)\,dx-\int_{0}^{a} f(-x)\,dx \] becomes \[ \int_{-a}^{0} f(x)\,dx+\int_{0}^{a} f(x)\,dx-\int_{0}^{a} f(-x)\,dx \]

Step 2: Transform the integral from \(-a\) to \(0\).
In \[ \int_{-a}^{0} f(x)\,dx, \] put \[ x=-t \] Then, \[ dx=-dt \] When \[ x=-a,\quad t=a \] and when \[ x=0,\quad t=0 \] So, \[ \int_{-a}^{0} f(x)\,dx = \int_{a}^{0} f(-t)(-dt) \] \[ = \int_{0}^{a} f(-t)\,dt \] Changing the dummy variable \(t\) to \(x\), \[ \int_{-a}^{0} f(x)\,dx = \int_{0}^{a} f(-x)\,dx \]

Step 3: Substitute this result.
Now, \[ \int_{-a}^{a} f(x)\,dx-\int_{0}^{a} f(-x)\,dx \] becomes \[ \int_{0}^{a} f(-x)\,dx+\int_{0}^{a} f(x)\,dx-\int_{0}^{a} f(-x)\,dx \] The terms \[ \int_{0}^{a} f(-x)\,dx \] cancel each other.
Hence, \[ \int_{-a}^{a} f(x)\,dx-\int_{0}^{a} f(-x)\,dx = \int_{0}^{a} f(x)\,dx \]

Step 4: Convert \(\int_{0}^{a} f(x)\,dx\) into the required option form.
Using the property \[ \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx \] Therefore, \[ \int_{-a}^{a} f(x)\,dx-\int_{0}^{a} f(-x)\,dx = \int_{0}^{a} f(a-x)\,dx \]

Step 5: Final conclusion.
Hence, \[ \boxed{\int_{0}^{a} f(a-x)\,dx} \]