Question

Use a symbolic integration utility to evaluate the integral. $$ \int_{1}^{e} x^{9} \ln x d x $$

Short answer

The evaluated integral is \( e^{10} - e^9 \).

Step-by-step solution

Choose u and dv

Firstly, choose \( u = x^9 \) and \( dv = \ln(x) dx \) since the derivative of \( x^9 \) and antiderivative of \( \ln(x) \) dx are simpler.

Calculate du and v

Next, calculate the differential of \( u \) and the integral of \( dv \). \( du = 9x^8 dx \) and \( v = x \ln(x) - x \).

Apply integration by parts formula

Now apply the integration by parts formula. \(\int u dv = uv - \int v du\). Substituting, we get \(\int_{1}^{e} x^9 \ln(x) dx = [x^9 * (x \ln(x) - x)]_{1}^{e} - \int_{1}^{e} (x^9) (x \ln(x) - x) * 9x^8 dx\). This simplifies to \([x^{10} (\ln(x) - 1)]_{1}^{e} - \int_{1}^{e} 9x^{17} (\ln(x) - 1) dx\).

Simplify integration

Reduce the integral to simpler parts: logistics of \( 9x^{17} (\ln(x) - 1) \), \( x^{18} \) is a simple function to integrate and \(\ln(x) \) dx is now dealt with. The new integral becomes \(-9 \int_{1}^{e} x^{17} dx + 9 \int_{1}^{e} x^{18} \ln(x) dx\). The second integral still has \(\ln(x)\), therefore repeat Step 1: let \( u = x^{18} \) and \( dv = \ln(x) dx \). Repeated application of integration by parts yields to zero because of decreasing powers. The terms with \(\ln(x)\) will eventually disappear since the derivative of any power of \( x \) approaches zero.

Evaluate integral and insert the bounds of integration

Finally, evaluate the integrals and insert the bounds of integration to get the answer. The process yields \( e^{10} - e^9 - 0 + e^{18} - e^{18} = e^{10} - e^9 \).

Key concepts

Symbolic Integration

Symbolic integration refers to finding an exact antiderivative of a given function using algebraic methods. Unlike numerical integration, which approximates the integral's value with numbers, symbolic integration provides a formula that represents the integral's value for any point within the domain of the function.

For example, the integral \[ \int x^{9} \ln(x) dx \] involves the function \( x^{9} \ln(x) \), and symbolic integration would yield an expression that exactly represents its antiderivative. Tools like computer algebra systems can perform this task efficiently, but understanding the underlying principles is essential for a student's mathematical development.

One common technique used in symbolic integration is 'integration by parts', which is especially useful when dealing with products of functions, such as a power of \( x \) multiplied by the natural logarithm of \( x \) in our exercise.

Definite Integral Evaluation

Evaluating a definite integral means calculating the exact area under the curve of a function, between two specified points. This is different from an indefinite integral, which does not have specified bounds and represents a family of functions (antiderivatives).

The evaluation often involves finding the antiderivative and then applying the Fundamental Theorem of Calculus, which states that if \( F(x) \) is an antiderivative of \( f(x) \), the definite integral of \( f(x) \) from \( a \) to \( b \) is \( F(b) - F(a) \). In our exercise, after applying integration by parts, you get an expression that can be evaluated at the upper and lower bounds \( e \) and \( 1 \), to find the area under the curve between these two points.

Integration Techniques

There are several techniques for integration that can be applied according to the function's characteristics such as 'integration by substitution', 'integration by parts', 'partial fraction decomposition', and others. In the example at hand, integration by parts is particularly useful. This method is based on the product rule for differentiation and is generally used when the integral involves a product of functions.

The chosen technique often simplifies the integration process and breaks complicated functions into simpler parts that can be more easily integrated. For instance, in our example, integration by parts is applied to break down the product of \( x^{9} \) and \( \ln(x) \). This technique involves identifying parts of the integral as \( u \) and \( dv \), and then using the formula \[ \int u dv = uv - \int v du \] to find the integral. Such techniques are essential for efficiently and effectively solving complex integrals.

Natural Logarithm Properties

The natural logarithm, denoted as \( \ln(x) \), has properties that are beneficial when dealing with integrals. Some of these properties include the natural log of 1 being 0 (\( \ln(1) = 0 \)), and the derivative of \( \ln(x) \) with respect to \( x \) being \( 1/x \).

These properties are particularly useful in the exercise where the natural logarithm is part of the integrand. Knowing that the logarithm of 1 is 0 simplifies the evaluation after applying the limits of integration. Additionally, the fact that the derivative of \( \ln(x) \) is \( 1/x \) is used in determining the differential \( du \) when applying integration by parts. Such properties contribute to the overall strategy for integrating functions involving the natural logarithm.