Problem 31 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^{1}_{0} x \bigl( \sqrt[3]{x} + \sqrt[4]{x} \bigr) \,dx \)

Step-by-Step Solution

Answer

The value of the integral is \( \frac{55}{63} \).

Step 1: Expand the Integral

The given integral is \( \int^{1}_{0} x \bigl( \sqrt[3]{x} + \sqrt[4]{x} \bigr) \,dx \). Start by distributing \( x \) inside the parenthesis. This gives us \( \int^{1}_{0} \left( x^{\frac{4}{3}} + x^{\frac{5}{4}} \right) \, dx \), which we can separate into two integrals: \( \int^{1}_{0} x^{\frac{4}{3}} \, dx + \int^{1}_{0} x^{\frac{5}{4}} \, dx \).

Step 2: Integrate the First Term

Now, evaluate the first integral \( \int^{1}_{0} x^{\frac{4}{3}} \, dx \). Use the power rule of integration: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). This gives \( \frac{x^{\frac{4}{3}+1}}{\frac{4}{3}+1} \) evaluated from 0 to 1. Simplify to \( \frac{x^{\frac{7}{3}}}{\frac{7}{3}} = \frac{3}{7}x^{\frac{7}{3}} \).

Step 3: Evaluate the First Integral

Substitute the limits into \( \frac{3}{7}x^{\frac{7}{3}} \) from 0 to 1. Thus, \( \frac{3}{7}(1)^{\frac{7}{3}} - \frac{3}{7}(0)^{\frac{7}{3}} = \frac{3}{7} - 0 = \frac{3}{7} \).

Step 4: Integrate the Second Term

Next, evaluate the second integral \( \int^{1}_{0} x^{\frac{5}{4}} \, dx \). Again use the power rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). This gives \( \frac{x^{\frac{5}{4}+1}}{\frac{5}{4}+1} \) evaluated from 0 to 1, which simplifies to \( \frac{x^{\frac{9}{4}}}{\frac{9}{4}} = \frac{4}{9}x^{\frac{9}{4}} \).

Step 5: Evaluate the Second Integral

Substitute the limits into \( \frac{4}{9}x^{\frac{9}{4}} \) from 0 to 1. Thus, \( \frac{4}{9}(1)^{\frac{9}{4}} - \frac{4}{9}(0)^{\frac{9}{4}} = \frac{4}{9} - 0 = \frac{4}{9} \).

Step 6: Combine the Results

Sum the results of the two integrals obtained: \( \frac{3}{7} + \frac{4}{9} \). Find a common denominator to add these fractions. The common denominator of 7 and 9 is 63, so convert the fractions: \( \frac{3}{7} = \frac{27}{63} \) and \( \frac{4}{9} = \frac{28}{63} \). Adding these, \( \frac{27}{63} + \frac{28}{63} = \frac{55}{63} \).

Key Concepts

Power Rule

The power rule is one of the foundations of calculus and is especially handy when working with polynomials or functions raised to powers. It states that when you have an integral of the form \( \int x^n \, dx \), the solution can be found using \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. This rule simplifies the process of finding antiderivatives. It's important to note that this rule applies when \( n eq -1 \).
In the given exercise, applying the power rule was straightforward as we worked with terms \( x^{\frac{4}{3}} \) and \( x^{\frac{5}{4}} \).

• For \( x^{\frac{4}{3}} \), increasing the exponent by one gives \( x^{\frac{7}{3}} \), and dividing by the new exponent \( \frac{7}{3} \) results in \( \frac{3}{7}x^{\frac{7}{3}} \).
• Similarly, for \( x^{\frac{5}{4}} \), the process results in \( \frac{4}{9}x^{\frac{9}{4}} \).

Understanding the power rule helps simplify complex-looking integrals into more manageable parts, which we can solve confidently.

Definite Integral

A definite integral represents the area under a curve between two specific points on the x-axis. It's different from an indefinite integral since it does not include the constant of integration \( C \). Instead, you evaluate it over an interval \([a, b]\).
In the provided problem, the definite integral \( \int_{0}^{1} \) was used, meaning we calculate the net area under the curve from \( x = 0 \) to \( x = 1 \).

• Applying the power rule gave us expressions like \( \frac{3}{7}x^{\frac{7}{3}} \) and \( \frac{4}{9}x^{\frac{9}{4}} \). We then substituted the limits \( x = 1 \) and \( x = 0 \) into these expressions.
• For both cases, when substituting \( x = 0 \), the contribution to the area is zero due to the term becoming zero, simplifying the calculations.

By evaluating at these endpoints, the calculation yields an exact numerical result, compared to just a formula when evaluating an indefinite integral.

Fraction Addition

Adding fractions might seem mundane, but it's crucial here to combine results of the definite integrals. Each fraction \( \frac{3}{7} \) and \( \frac{4}{9} \) represented areas, which needed to be added to find the final integral value.
To add fractions:

• Find a common denominator. For 7 and 9, that number is 63.
• Convert each fraction: \( \frac{3}{7} = \frac{27}{63} \) and \( \frac{4}{9} = \frac{28}{63} \).
• Add the converted fractions: \( \frac{27}{63} + \frac{28}{63} = \frac{55}{63} \).

This step ensured accuracy when combining the results of the integrals, since decimals could lead to rounding errors. It’s always a good idea to work with fractions to maintain precision until the final answer is reached. Thus, understanding fraction addition enhances our confidence in solving problems involving integrals.