Problem 54 - Biocalculus Calculus for the Life Sciences

Question

Evaluate \(\int_{0}^{1} x \sqrt{1-x^{4}} d x\) by making a substitution and interpreting the resulting integral in terms of an area.

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{\pi}{8}\).

Step 1: Identify a Suitable Substitution

To evaluate the integral, we need to perform a substitution that simplifies the expression under the integral. The presence of \(1-x^4\) inside the square root suggests a trigonometric substitution. Let \(x = \sin^{1/4}(u)\), meaning \(x^4 = \sin u\). Therefore, \(dx = \frac{1}{4\sin^{3/4}(u)\cos(u)} du\).

Step 2: Change Limits of Integration

When \(x = 0\), \(u = \arcsin(x^4) = 0\). When \(x = 1\), \(u = \arcsin(1) = \frac{\pi}{2}\). Thus, the integral becomes \(\int_{0}^{\frac{\pi}{2}} \sin^{1/4}(u) \sqrt{1-\sin(u)} \frac{1}{4\sin^{3/4}(u)\cos(u)} du\).

Step 3: Simplify the Integral

Substitute the expressions into the integral: \(\int_{0}^{\frac{\pi}{2}} \frac{1}{4\cos(u)} \sqrt{1-\sin(u)} du\). This simplifies further because \(\sqrt{1-\sin(u)} = \cos(u)\). So, the integral becomes \(\frac{1}{4} \int_{0}^{\frac{\pi}{2}} du\).

Step 4: Evaluate the Integral

The integral simplifies to \(\frac{1}{4} \times \left[ u \right]_{0}^{\frac{\pi}{2}} = \frac{1}{4} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{8}\). This integral evaluates to \(\frac{\pi}{8}\).

Step 5: Interpret as Area

The integral evaluated represents the area under the function \(x \sqrt{1-x^4}\) from \(x = 0\) to \(x = 1\). The substitution simplifies the problem to evaluating a simple region in the coordinate plane with respect to \(u\).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in calculus to simplify integrals containing expressions like \( \sqrt{1-x^2} \) or \( \sqrt{1+x^2} \), which resemble the Pythagorean identities. By choosing a trigonometric identity, such as \( x = \sin(\theta) \) for expressions involving \( \sqrt{1-x^2} \), we can transform the integral into a more manageable form.

In the given problem, we encounter \( \sqrt{1-x^4} \). Here, a creative choice of substitution is \( x = \sin^{1/4}(u) \). This implies that \( x^4 = \sin u \). As a result, the differential \( dx \) is derived using chain rule relationships, thus becoming \( dx = \frac{1}{4\sin^{3/4}(u)\cos(u)} du \). This substitution not only simplifies the radical but also changes the variable of integration to \( u \), transforming the original complex expression into a function of a single, simpler trigonometric function.

Key points of trigonometric substitution include:

• Choosing the correct trigonometric identity based on the integral's form.
• Altering the differential \( dx \) to match the new variable \( du \).
• Rewriting the limits of integration to correspond to the substituted variable.

Definite Integral

A definite integral represents the exact calculation of the area under a curve from one point to another along the x-axis. The integral \( \int_{0}^{1} x \sqrt{1-x^4} \, dx \) computes this area between \( x = 0 \) and \( x = 1 \).

After performing the trigonometric substitution, the bounds of the integral need to change accordingly. When \( x = 0 \), the new bound is determined as \( u = \arcsin(0) = 0 \). Similarly, when \( x = 1 \), \( u = \arcsin(1) = \frac{\pi}{2} \). These new limits of integration from 0 to \( \frac{\pi}{2} \) enable evaluation of the new, simplified integral \( \int_{0}^{\frac{\pi}{2}} \frac{1}{4\cos(u)} \sqrt{1-\sin(u)} \, du \).

The steps in handling definite integrals include:

• Changing the limits of integration to reflect the substitution.
• Reducing the complexity of the integral using identities or substitutions.
• Accurately computing the resultant integral to find the value representing an area.

Area Interpretation

The concept of a definite integral can be interpreted geometrically as the area under a curve. In our exercise, the integral \( \int_{0}^{1} x \sqrt{1-x^4} \, dx \) was simplified to \( \frac{1}{4} \int_{0}^{\frac{\pi}{2}} du \). This calculation gives us \( \frac{\pi}{8} \), which essentially represents the area under the original curve from \( x = 0 \) to \( x = 1 \).

Utilizing trigonometric substitution, the original problem is reinterpreted into an easier integral that describes a simpler geometric region. This is possible because the sine function's properties allow for area simplification under specific conditions, leading to manageable arithmetic or geometric interpretations.

Application of area interpretation:

• Simplifies visual understanding of complex functions through geometric areas.
• Relates calculus computations to real-world measurements like area or volume.
• Reveals the underlying significance of integrals beyond abstract numbers.