Problem 91 - Calculus: Early Transcendentals

Question

If \( a \) and \( b \) are positive numbers, show that $$ \int^1_0 x^a(1 - x)^b \,dx = \int^1_0 x^b(1 - x)^a \,dx $$

Step-by-Step Solution

Answer

Both integrals are equal by symmetry through substitution.

Step 1: Understanding the Integral

We start with the given integral expressions \( \int^1_0 x^a(1 - x)^b \,dx \) and \( \int^1_0 x^b(1 - x)^a \,dx \). Algebraically, they both appear similar except for the switch in exponents of \( x \) and \( 1-x \). Hence, it suggests a potential symmetry.

Step 2: Symmetry with Substitution

Consider the substitution \( x = 1 - t \). Then \( dx = -dt \), and when \( x = 0, t = 1 \), and when \( x = 1, t = 0 \). This substitution results in reversing the limits: \[ \int_1^0 (1-t)^a t^b (-dt) = \int_0^1 t^b (1-t)^a \, dt \] This shows that after changing variables, \( \int^1_0 x^a(1 - x)^b \,dx \) transforms into \( \int^1_0 (1-x)^a x^b \,dx \), which is equivalent to \( \int^1_0 x^b (1-x)^a \,dx \).

Step 3: Conclusion from Symmetry

Since the transformations confirm the equality of the integrals in both variable positions due to the symmetry using the substitution \( x = 1 - t \), we conclude that the originally given integrals are equal.

Key Concepts

Definite Integral

The concept of a definite integral is a fundamental aspect in calculus. It is used to compute the accumulation of quantities, often interpreted as the area under a curve over a specified interval. In our case, the integral forms are given as \( \int^1_0 x^a(1 - x)^b \,dx \) and \( \int^1_0 x^b(1 - x)^a \,dx \). These represent areas under the curve \( y = x^a(1-x)^b \) and \( y = x^b(1-x)^a \),respectively, over the interval \([0, 1] \).
In definite integrals:

• The endpoints \(a\) and \(b\) are the limits of integration.
• The function \(f(x)\) (inside the integral) is the integrand.
• Integration is performed with respect to \(x\), which is the variable of integration.

Definite integrals can be seen as the net signed area between the integrand curve and the \(x\)-axis, from \(x = a\) to \(x = b\). This area can help identify factors or transformations, as seen in the symmetry of this exercise.

Symmetry

Symmetry in integrals can often simplify complex expressions or reveal hidden relationships. In the given problem, the symmetry is the key factor to understanding how the two integrals are equal.
When looking at \( \int^1_0 x^a(1 - x)^b \,dx \) and \( \int^1_0 x^b(1 - x)^a \,dx \), observe the pattern in terms of the powers of \(x\) and \(1-x\). Notice how simply swapping the exponents results in a symmetric form.
This kind of symmetry is often evident in scenarios involving:

• Geometric Symmetry: Involves symmetry in shapes and graphs.
• Algebraic Symmetry: Exhibited when expressions are invariant under certain transformations, like reversing limits through substitution.

By applying a substitution (as seen in the next section), we effectively flip the expression while not altering the absolute value of the integral, showcasing this integral symmetry.

Variable Substitution

Variable substitution, also known as change of variables, is an essential technique in calculus. It's particularly handy for solving integrals that otherwise seem complex or unmanageable. In our exercise,the substitution \(x = 1 - t\) was used to demonstrate the equivalence of the two integrals. Here's how it works:
1. **Choose a Substitution:** We substituted \(x = 1-t\). This changes the integrand's form.2. **Adjust Differential:** The differential \(dx\) becomes \(-dt\) when substituting \(dt\) for \(dx\).3. **Update Limits:** The limits of integration reverse, from \(x = 0\) to \(x = 1\), converting to \(t = 1\) to \(t = 0\).4. **Simplify and Solve:** The new integral becomes \(\int^1_0 t^b (1-t)^a \, dt\), paralleling the original problem’s counterpart.
By switching variables, we maintain the integral's value while highlighting its inherent symmetry. Variable substitution is a powerful tool not only for solving integrals but also for revealing their deeper characteristics via transformation.