Problem 44 - Calculus and its applications

Question

Find each integral. $$ \int\left(2 e^{6 x}-\frac{3}{x}+\sqrt[3]{x^{4}}\right) d x, x>0 $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{3} e^{6x} - 3 \ln x + \frac{3}{7} x^{\frac{7}{3}} + C \).

Step 1: Break Down the Integral

The given integral is \( \int (2e^{6x} - \frac{3}{x} + \sqrt[3]{x^4}) \, dx \). This can be broken down as the sum of three separate integrals. Let's write it as:\[ \int 2e^{6x} \, dx - \int \frac{3}{x} \, dx + \int \sqrt[3]{x^4} \, dx \]

Step 2: Integrate \(2e^{6x}\)

To integrate \(2e^{6x}\), we apply the rule for integrating exponential functions. The rule states \( \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C \). Thus,\[\int 2e^{6x} \, dx = 2 \cdot \frac{1}{6} e^{6x} = \frac{1}{3} e^{6x}\]

Step 3: Integrate \(\frac{3}{x}\)

This integral involves a simple logarithmic rule \( \int \frac{1}{x} \, dx = \ln |x| + C \). Applying the constant multiple rule, we have:\[\int \frac{3}{x} \, dx = 3 \cdot \ln |x| = 3 \ln x\] (assuming \(x > 0\))

Step 4: Integrate \(\sqrt[3]{x^4}\)

Rewrite \(\sqrt[3]{x^4}\) as \(x^{\frac{4}{3}}\). Now use the power rule for integration \( \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \) where \(n = \frac{4}{3}\).\[\int x^{\frac{4}{3}} \, dx = \frac{x^{\frac{4}{3} + 1}}{\frac{4}{3} + 1} = \frac{x^{\frac{7}{3}}}{\frac{7}{3}} = \frac{3}{7} x^{\frac{7}{3}}\]

Step 5: Combine All Parts

Combine all the results from the previous integrals:\[\int (2e^{6x} - \frac{3}{x} + \sqrt[3]{x^4}) \, dx = \frac{1}{3} e^{6x} - 3 \ln x + \frac{3}{7} x^{\frac{7}{3}} + C\]

Key Concepts

Exponential Functions

Exponential functions are a crucial component in calculus, often appearing in differential and integral equations. They have the general form of \(e^{kx}\), where \(e\) is Euler's number, approximately equal to 2.71828, and \(k\) is a constant.
A key feature of exponential functions is their rate of growth. They grow, or decay, at a rate proportional to their current value. This makes them highly applicable in scenarios involving continuous growth, such as population models and compound interest.
In integral calculus, when integrating functions of the form \(e^{kx}\), we use the rule:

• \( \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C \)

In the exercise, the expression \(2e^{6x}\) follows this rule, making it straightforward to integrate it to \(\frac{1}{3}e^{6x}\). Understanding how to manipulate and integrate exponential functions is fundamental in solving calculus problems efficiently.

Logarithmic Integration

Logarithmic integration deals specifically with the integration of the expression \(\frac{1}{x}\), which yields the natural logarithm. The natural logarithm, denoted as \(\ln x\), is the inverse operation of the exponential function with base \(e\).
In integral calculus, we have an important rule:

• \( \int \frac{1}{x} \, dx = \ln |x| + C \)

This rule stems from the fact that the rate of change of \(\ln x\) is \(\frac{1}{x}\). In our exercise, this principle is applied to \(\frac{3}{x}\), resulting in the integral \(3\ln x\), assuming \(x > 0\). Logarithmic integration is widely used in solving differential equations and calculating areas under curves involving rational functions.

Power Rule for Integration

The power rule for integration is a fundamental tool in calculus that simplifies the process of finding antiderivatives for power functions. A power function has the form \(x^n\), where \(n\) is a real number.
The power rule states:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) where \(n eq -1\)

This powerful rule allows for the direct integration of polynomials by incrementing the exponent by one and dividing by the new exponent.
Within the provided exercise, the term \(\sqrt[3]{x^4}\) is rewritten as the power form \(x^{\frac{4}{3}}\). Applying the power rule offers the integral \(\frac{3}{7}x^{\frac{7}{3}}\). Mastery of the power rule is crucial for tackling a wide range of integrals encountered in both basic and advanced calculus problems.