If f is continuous, show that
Z b
a
f(−x)dx =
Z −a
−b
f(x)dx .
When f(x) ≥ 0 and 0 ≤ a < b, draw a picture to interpret this equality as an equality of areas.

We are asked to prove the equality:

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

### Proof of the Equality

We begin with the left-hand side:

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

To change the limits of integration, we perform the substitution \( u = -x \). Then, \( du = -dx \), and the limits change as follows:
- When \( x = a \), \( u = -a \).
- When \( x = b \), \( u = -b \).

Thus, the integral becomes:

\[
\int_a^b f(-x) \, dx = \int_{-a}^{-b} f(u) (-du).
\]

Since \( du = -dx \), we can factor out the negative sign, giving:

\[
\int_{-a}^{-b} f(u) (-du) = \int_{-b}^{-a} f(u) \, du.
\]

Since \( u \) is a dummy variable (it represents the same thing as \( x \)), we can replace \( u \) with \( x \) again:

\[
\int_{-b}^{-a} f(u) \, du = \int_{-b}^{-a} f(x) \, dx.
\]

Thus, we have shown that:

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

Finally, we can rewrite the right-hand side of the original expression to match this result:

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

so the equality is proven.

### Graphical Interpretation

When \( f(x) \geq 0 \) and \( 0 \leq a < b \), the equality can be interpreted as an equality of areas under the curve.

- The integral \( \int_a^b f(-x) \, dx \) represents the area under the curve of \( f(-x) \), which is a reflection of the function \( f(x) \) across the y-axis, evaluated between \( a \) and \( b \).
- The integral \( \int_{-a}^{-b} f(x) \, dx \) represents the area under the curve of \( f(x) \), but between the limits \( -a \) and \( -b \).

The equality shows that the area under the curve on one side of the y-axis is the same as the area on the other side, reflecting the symmetry in the function when transformed by \( f(-x) \).

### Diagram:

The diagram would consist of two areas:
- One between \( a \) and \( b \), for the function \( f(-x) \), on the positive side of the x-axis.
- One between \( -b \) and \( -a \), for the function \( f(x) \), on the negative side of the x-axis.

The areas under the curves on both sides of the y-axis will be identical if the integral equality holds.

---

Do you want any further clarifications or a visual representation of the graph?

Here are some related questions:
1. How does the substitution \( u = -x \) work in definite integrals?
2. What is the significance of continuity in proving integral equalities?
3. Can you provide an example of a function \( f(x) \) and calculate both integrals?
4. How does symmetry play a role in interpreting these kinds of integrals?
5. What would happen if \( f(x) \) were not continuous or had discontinuities?

**Tip:** When performing a substitution in definite integrals, always adjust the limits of integration accordingly to match the new variable.

If f is continuous, show that ∫(from a to b) f(−x) dx = ∫(from -a to -b) f(x) dx. When f(x) ≥ 0 and 0 ≤ a < b, draw a picture to interpret this equality as an equality of areas.

Explore the detailed proof of the integral equality ∫(from a to b) f(−x) dx = ∫(from -a to -b) f(x) dx for continuous functions, utilizing the substitution method. Learn how symmetry in functions leads to an equality of areas under the curve. Suitable for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation