Problem 49 - College Algebra and Calculus: An Applied Approach

Question

Find the area of the region bounded by the graphs of the equations. Then use a graphing utility to graph the region and verify your answer. $$ y=x^{2} \ln x, y=0, x=1, x=e $$

Step-by-Step Solution

Answer

The area of the bounded region is \( \frac{8e^3 + 1}{27}\).

Step 1: Set up the integral

The area \(A\) between the curve \(y=x^{2} \ln x\), the x-axis (y = 0), and the lines \(x=1\) and \(x=e\) can be found by evaluating the definite integral of \(x^{2} \ln x\) from 1 to \(e\). For this, the integral is set up as:\[A = \int^{e}_{1} x^{2} \ln x\:dx\]

Step 2: Use integration by parts

Integration by parts is needed to solve the integral, a method where one function is chosen to be differentiated and the other to be integrated when you have the product of two functions. The integration by parts formula is: \(\int u\:dv = uv - \int v\: du \). For our integral, we can let \(u = \ln x\) and \(dv = x^2\: dx\). Which results in \(du = \frac{1}{x}\: dx\) and \(v = \frac{1}{3} x^3\). Applying the integration by parts formula:\[\int^{e}_{1} x^{2} \ln x\:dx = [ \frac{1}{3} x^3 \ln x ]_{1}^{e} - \int^{e}_{1} \frac{1}{3} x^3 \frac{1}{x} dx\]\[\int^{e}_{1} x^{2} \ln x\:dx = [ \frac{1}{3} x^3 \ln x ]_{1}^{e} - \int^{e}_{1} \frac{1}{3} x^2 dx\]

Step 3: Evalute the Integral

Now, we can evaluate our integral. The integral of \( \frac{1}{3} x^2 \) from 1 to \( e \) is \( \frac{1}{9} \left[e^3 - 1\right]\). Then, evaluate the rest: \[A = [ \frac{1}{3} x^3 \ln x ]_{1}^{e} - \frac{1}{9} \left[e^3 - 1\right]\]\[A = \left[\frac{e^3}{3} \ln(e) - \frac{1}{3}\cdot 1\cdot \ln(1)\right] - \frac{1}{9} \left[e^3 - 1\right]\]\[A = \left[\frac{e^3}{3} - 0\right] - \frac{1}{9} \left[e^3 - 1\right]\]\[A = \frac{e^3}{3} - \frac{e^3 - 1}{9}\]

Step 4: Simplify

Finally, simplify to get the answer:\[A = \left[\frac{9e^3 - e^3 + 1}{27}\right]\]\[A = \left[\frac{8e^3 + 1}{27}\right]\]

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that helps to find the accumulation of quantities. In simpler terms, it can be used to compute areas under curves, among other things. This is represented by the notation \( \int_{a}^{b} f(x)\,dx \), where \( a \) and \( b \) are the bounds of integration, and \( f(x) \) is a function whose integral we're calculating.

• The lower bound, \( a \), and the upper bound, \( b \), are limits that specify the region of interest on the x-axis.
• By evaluating \( \int_{a}^{b} f(x)\,dx \), we can find the exact area under the curve \( f(x) \) between \( x = a \) and \( x = b \).

In the exercise example, the definite integral \( \int_{1}^{e} x^2 \ln x\,dx \) is solved to find the area between the curve and the x-axis, from x = 1 to x = \( e \). Integration by parts is used to handle such functions where a product exists, allowing us to break down the integral into simpler parts.

Area Bounded by Curves

The area bounded by curves refers to the space enclosed between two or more curves on a graph. This is a typical problem in calculus, where we might want to know the exact size of such enclosed regions.

• To find this area, we typically set up a definite integral over the desired interval.
• For curves \( y = f(x) \) and \( y = g(x) \), the area between them from \( x = a \) to \( x = b \) is calculated as \( \int_{a}^{b} [f(x) - g(x)]\,dx \).

In the exercise, we are tasked to find the area enclosed by \( y = x^2 \ln x \) and the x-axis, between \( x = 1 \) and \( x = e \). This is achieved by evaluating the definite integral of \( x^2 \ln x \), considering \( y=0 \) as one of the boundary lines, essentially forming the lower bound for the enclosed region.

Natural Logarithm Function

The natural logarithm function, denoted \( \ln x \), is a logarithmic function that is widely used in calculus. It represents the logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828.

• The function \( \ln x \) is the inverse of the exponential function \( e^x \).
• It is key in solving problems involving growth and decay, as well as in integration and differentiation tasks.

In the problem at hand, \( \ln x \) appears in the term \( x^2 \ln x \), which we need to integrate to find the area under the curve. The natural logarithm function influences how the curve behaves, hence understanding its characteristics helps in setting up and solving the integral. Integration by parts is particularly useful in this context, as it allows us to handle the product of \( x^2 \) and \( \ln x \), resulting in a solution to the area problem described in our initial exercise.