Prove the following equality: integral from a to b

Prove the following equality: $\int_a^b f(a+b-x) \, dx = \int_a^b f(x) \, dx$.

Proof using substitution:

Let $u = a + b - x$. Then $du = -dx$, so $dx = -du$; also $x = a+b-u$.

Change the limits of integration:

When $x = a$: $u = b$. When $x = b$: $u = a$.

Transform the integral:

\[\int_a^b f(a+b-x) \, dx = \int_b^a f(u) \cdot (-du) = -\int_b^a f(u) \, du = \int_a^b f(u) \, du\]

Since the variable of integration is a dummy variable, we can replace $u$ with $x$:

\[= \int_a^b f(x) \, dx\]

Therefore:

\[\boxed{\int_a^b f(a+b-x) \, dx = \int_a^b f(x) \, dx}\]

Geometric interpretation:

The transformation $u = a + b - x$ reflects the interval $[a, b]$ about its midpoint $\frac{a+b}{2}$. The equality says that integrating $f(a+b-x)$ from $a$ to $b$ is the same as integrating $f(x)$ from $a$ to $b$. This means $f(a+b-x)$ is the reflection of $f(x)$ about the vertical line $x = \frac{a+b}{2}$, and the areas under these curves over $[a, b]$ are equal.

Example application:

For $\int_0^{\pi} x\sin x \, dx$, use this property with $a = 0$, $b = \pi$:

\[\int_0^\pi x\sin x \, dx = \int_0^\pi (\pi - x)\sin x \, dx\]

Adding these:

\[2\int_0^\pi x\sin x \, dx = \int_0^\pi [x + (\pi-x)]\sin x \, dx = \pi\int_0^\pi \sin x \, dx\]

This allows us to evaluate:

\[\int_0^\pi x\sin x \, dx = \frac{\pi}{2}\int_0^\pi \sin x \, dx = \frac{\pi}{2}[-\cos x]_0^\pi = \frac{\pi}{2}[1-(-1)] = \pi\]

General principle:

This property is particularly useful when the integrand becomes simpler under the transformation $x \to a+b-x$, when we want to relate two integrals that are reflections of each other, or when we can add the original integral to its reflected version to obtain a simpler expression.