Question

Evaluate the following integrals. $$\int_{1}^{e^{2}} \frac{(\ln x)^{5}}{x} d x$$

Short answer

Question: Evaluate the integral $$\int_{1}^{e^{2}} \frac{(\ln x)^{5}}{x} d x$$. Answer: The value of the integral is $$\frac{64}{6}$$.

Step-by-step solution

Define substitution variable

Let's define a substitution variable to simplify the integration process: $$u = \ln x$$ Now find the derivative of the variable with respect to x: $$\frac{d u}{d x} = \frac{1}{x}$$

Find the relationship between dx and du

We need to find the relationship between \(dx\) and \(du\). To do this, we will rewrite our previous expression in terms of \(dx\): $$ d u = \frac{1}{x} d x$$

Change the limits of integration

Now, we have to change the limits of integration according to the new variable \(u\). When \(x = 1\), we get: $$u = \ln x = \ln 1$$ $$u = 0$$ When \(x = e^{2}\), we get: $$u = \ln x = \ln (e^{2})$$ $$u = 2$$ Our new limits of integration are \(0\) and \(2\).

Rewrite the integral in terms of u

Now we can rewrite the integral in terms of \(u\). Replace \(\ln x\) with \(u\) and substitute \(dx\) from Step 2: $$\int_{1}^{e^{2}} \frac{(\ln x)^{5}}{x} d x = \int_{0}^{2} u^{5} d u$$

Find the antiderivative

To find the antiderivative of \(u^{5}\), we use the power rule for integration: $$\int u^{5} d u = \frac{u^{6}}{6} + C$$

Evaluate the antiderivative at the limits of integration

Now we will evaluate the antiderivative at the limits of integration and find the definite integral: $$\int_{0}^{2} u^{5} d u = \left[\frac{u^{6}}{6}\right]_{0}^{2}$$ $$= \frac{(2)^{6}}{6} - \frac{(0)^{6}}{6}$$ $$= \frac{64}{6}$$ The integral equals: $$\int_{1}^{e^{2}} \frac{(\ln x)^{5}}{x} d x = \frac{64}{6}$$

Key concepts

Definite Integral

In calculus, a definite integral is used to calculate the total accumulation of quantities, such as area under a curve, between two specified points. It provides a precise value that denotes this accumulated sum.
The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where:

• \( f(x) \) is the function to be integrated, also known as the integrand.
• \( a \) and \( b \) are the limits of integration, with \( a \) as the lower limit and \( b \) as the upper limit.

Instead of a general antiderivative, the definite integral results in a specific number by invoking the fundamental theorem of calculus.
This theorem connects differentiation with integration, allowing us to solve definite integrals using antiderivatives. If \( F(x) \) is an antiderivative of \( f(x) \), then the definite integral from \( a \) to \( b \) can be expressed as:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \] The definite integral provides a powerful tool to measure quantities across intervals with mathematical precision.

Integration by Substitution

Integration by substitution is a technique used to simplify the process of finding antiderivatives. It's particularly helpful when dealing with complex expressions. The basic idea is to change variables to make the integration more manageable.
The substitution method involves choosing a new variable, typically \( u \), which is a function of \( x \). This new variable simplifies the integrand, often into a form that's easier to integrate.\[\text{Let } u = g(x), \, \text{then} \, \frac{du}{dx} = g'(x), \, \text{hence} \, du = g'(x) \cdot dx\] Here, we rewrite \( dx \) in terms of \( du \), allowing us to transform the entire integral into a new expression in terms of \( u \). Once integration is complete, it's crucial to substitute back to the original variable to find the integral in its original context.

• Identify a part of the integrand to substitute.
• Find \( du \) in terms of \( dx \).
• Change the limits of integration if dealing with definite integrals.
• Substitute and integrate.
• Convert back to the original variable.

This technique is analogous to the chain rule in differentiation and is invaluable for solving integrals that are otherwise difficult to tackle.

Power Rule for Integration

The power rule for integration is a fundamental tool in calculus used to find antiderivatives for power functions. It's quite straightforward and one of the first integration rules students learn. It applies to functions of the form \( ax^n \) where \( n eq -1 \).
The basic formula for the power rule is:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\] Where \( C \) is the constant of integration that appears in indefinite integrals.
Here’s how it works:

• Increase the exponent by 1.
• Divide by the new exponent.
• Don't forget to add \( C \), only for indefinite integrals.

In definite integrals like \( \int_{a}^{b} x^n \, dx \), you will evaluate the antiderivative at the bounds \( a \) and \( b \), leading to:\[\left[ \frac{x^{n+1}}{n+1} \right]_a^b = \frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1}\] The power rule is not limited to simple polynomial strokes but is foundational for more complex integration techniques.