Problem 78
Question
Evaluate \( \displaystyle \int^1_0 x \sqrt{1 - x^4} \,dx \) by making a substitution and interpreting the resulting integral in terms of an area.
Step-by-Step Solution
Verified Answer
The integral evaluates to \( \frac{\pi}{8} \).
1Step 1: Identify a Suitable Substitution
To simplify the integral \( \int_0^1 x \sqrt{1 - x^4} \, dx \), look for a substitution that involves the composition in the integrand. Choose \( u = x^2 \), which means \( du = 2x \, dx \), hence \( x \, dx = \frac{1}{2} du \).
2Step 2: Change the Limits of Integration
With the substitution \( u = x^2 \), adjust the limits accordingly. When \( x = 0 \), \( u = 0^2 = 0 \), and when \( x = 1 \), \( u = 1^2 = 1 \). The new limits for \( u \) are 0 to 1.
3Step 3: Substitute and Simplify the Integral
Substitute \( u = x^2 \) and \( x \, dx = \frac{1}{2} du \) into the integral. The integral becomes \( \frac{1}{2} \int_0^1 \sqrt{1 - u^2} \, du \). This is a standard integral that represents the area of a quarter-circle.
4Step 4: Recognize the Standard Integral
The integral \( \int \sqrt{1 - u^2} \, du \) is known to represent the area under a quarter of a unit circle when evaluated from 0 to 1. The area of a unit circle is \( \pi \), and one quarter of it is \( \frac{\pi}{4} \).
5Step 5: Calculate the Final Result
Multiply the result by the constant factor that was factored out earlier: \( \frac{1}{2} \times \frac{\pi}{4} = \frac{\pi}{8} \). Thus, the original integral evaluates to \( \frac{\pi}{8} \).
Key Concepts
Substitution MethodDefinite IntegralsGeometric Interpretation of Integrals
Substitution Method
The substitution method is a powerful technique for evaluating integrals, especially when dealing with complex compositions. It involves changing variables to simplify an integral, making it easier to solve. In this problem, we are given the integral \( \int_0^1 x \sqrt{1 - x^4} \, dx \).
The first step is to identify a substitution that simplifies the integrand. Here, we notice that the expression \( 1 - x^4 \) contains a square, suggesting a trigonometric or power substitution could work. We let \( u = x^2 \), resulting in \( du = 2x \, dx \), and consequently, \( x \, dx = \frac{1}{2} du \).
Substitution not only changes the variable but also the bounds of integration. When \( x = 0 \), \( u = 0 \), and when \( x = 1 \), \( u = 1 \). Hence, our integral becomes \( \frac{1}{2} \int_0^1 \sqrt{1 - u^2} \, du \), which simplifies the original expression and makes the calculation manageable.
The first step is to identify a substitution that simplifies the integrand. Here, we notice that the expression \( 1 - x^4 \) contains a square, suggesting a trigonometric or power substitution could work. We let \( u = x^2 \), resulting in \( du = 2x \, dx \), and consequently, \( x \, dx = \frac{1}{2} du \).
Substitution not only changes the variable but also the bounds of integration. When \( x = 0 \), \( u = 0 \), and when \( x = 1 \), \( u = 1 \). Hence, our integral becomes \( \frac{1}{2} \int_0^1 \sqrt{1 - u^2} \, du \), which simplifies the original expression and makes the calculation manageable.
- Substitution is most effective if it simplifies functions within the integral.
- It requires altering the differential and limits of integration accordingly.
- This process turns a complex integral into a more recognizable form, often easier to evaluate.
Definite Integrals
The concept of definite integrals is fundamental in calculus. A definite integral of a function over a given interval \([a, b]\) is represented as \( \int_a^b f(x) \, dx \). It essentially calculates the "net area" under the curve from \( x = a \) to \( x = b \).
In our problem, after substitution, we deal with the definite integral \( \frac{1}{2} \int_0^1 \sqrt{1 - u^2} \, du \). The limits \( 0 \) and \( 1 \) are directly linked to our variable substitution where \( u \) ranges from \( 0 \) to \( 1 \) as \( x \) changes accordingly.
Definite integrals have several key characteristics:
In our problem, after substitution, we deal with the definite integral \( \frac{1}{2} \int_0^1 \sqrt{1 - u^2} \, du \). The limits \( 0 \) and \( 1 \) are directly linked to our variable substitution where \( u \) ranges from \( 0 \) to \( 1 \) as \( x \) changes accordingly.
Definite integrals have several key characteristics:
- They provide the exact "signed" area under the curve, not just a function's antiderivative.
- The evaluated result depends on the limits of integration.
- Applications include areas under curves, displacement, and accumulative quantities in physics and engineering.
Geometric Interpretation of Integrals
The geometric interpretation of integrals can often simplify their evaluation and deepen understanding. Integrals can represent the area of geometric shapes. In this case, \( \int \sqrt{1 - u^2} \, du \) is known to represent the area under a quarter of a unit circle.
A unit circle's equation is \( x^2 + y^2 = 1 \). Thus, the integral \( \int \sqrt{1 - u^2} \, du \) covers the upper half-quarter circle from \( u = 0 \) to \( u = 1 \).
Recognizing this as a quarter of a circle simplifies evaluating our integral to simply finding this known area. The entire area of a unit circle is \( \pi \), making a quarter-circle's area \( \frac{\pi}{4} \).
A unit circle's equation is \( x^2 + y^2 = 1 \). Thus, the integral \( \int \sqrt{1 - u^2} \, du \) covers the upper half-quarter circle from \( u = 0 \) to \( u = 1 \).
Recognizing this as a quarter of a circle simplifies evaluating our integral to simply finding this known area. The entire area of a unit circle is \( \pi \), making a quarter-circle's area \( \frac{\pi}{4} \).
- Integrals can represent areas, contributing to physical interpretations of mathematics.
- Common geometric functions in calculus often trace back to simple, recognizable shapes like circles.
- Identifying shapes can speed up the evaluation process without requiring full computation of the integral.
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