Question

Graph the curve \(y=\frac{x}{1+x}\) over the interval \([0,5] .\) Then, find the area of the region bounded by the curve, the \(x\) -axis, and the line \(x=4\).

Short answer

The area is \( 4 - \ln 5 \).

Step-by-step solution

Understand the Function

The function given is \( y = \frac{x}{1+x} \). This is a rational function, with a numerator \( x \) and a denominator \( 1+x \). The domain of the function is all real numbers except where the denominator is zero. However, since \( 1+x eq 0 \) for any \( x \in [0,5] \), this function is defined over the entire interval specified.

Graph the Function Over the Interval

To graph \( y = \frac{x}{1+x} \) over \([0, 5]\), evaluate the function at several points within the interval: \( x = 0, 1, 2, 3, 4, 5 \).- At \( x = 0 \), \( y = \frac{0}{1} = 0 \).- At \( x = 1 \), \( y = \frac{1}{2} = 0.5 \).- At \( x = 2 \), \( y = \frac{2}{3} \approx 0.67 \).- At \( x = 3 \), \( y = \frac{3}{4} = 0.75 \).- At \( x = 4 \), \( y = \frac{4}{5} = 0.8 \).- At \( x = 5 \), \( y = \frac{5}{6} \approx 0.83 \).Plot these points and draw a smooth curve through them. The graph should be increasing and approaching 1 as \( x \to \infty \).

Set Up the Integral for the Area

The area under the curve and above the x-axis from \( x = 0 \) to \( x = 4 \) is given by the definite integral of the function over this interval. Set up the integral:\[ \text{Area} = \int_{0}^{4} \frac{x}{1+x} \, dx \].

Simplify and Calculate the Integral

To integrate \( \frac{x}{1+x} \, dx \), use substitution. Let \( u = 1+x \), then \( du = dx \), and \( x = u-1 \).The limits of integration change as well:- When \( x = 0 \), \( u = 1 \).- When \( x = 4 \), \( u = 5 \).The integral becomes:\[ \int_{1}^{5} \frac{u-1}{u} \, du = \int_{1}^{5} \left( 1 - \frac{1}{u} \right) \, du \].Separate the integrals:\[ \int_{1}^{5} \, du - \int_{1}^{5} \frac{1}{u} \, du = \left[ u \right]_{1}^{5} - \left[ \ln |u| \right]_{1}^{5} \].Evaluate the integrals:\[ \left[ u \right]_{1}^{5} = 5 - 1 = 4 \],\[ \left[ \ln |u| \right]_{1}^{5} = \ln 5 - \ln 1 = \ln 5 \].Therefore, the area is \( 4 - \ln 5 \).

Conclusion

The area of the region bounded by the curve \( y = \frac{x}{1+x} \), the x-axis, and the line \( x = 4 \) is \( 4 - \ln 5 \). This represents the exact area under the curve over the interval from \( x = 0 \) to \( x = 4 \).

Key concepts

Rational functions

Rational functions are expressions of the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. They are called 'rational' because they are ratios of two polynomial functions. These functions can exhibit a variety of behaviors such as asymptotes, intercepts, and limits. In our example, the function \( y = \frac{x}{1+x} \) has a linear numerator and a linear denominator. This simplicity makes it easy to evaluate and graph.

One key aspect to consider with rational functions is their domain, which is typically all real numbers except where the denominator becomes zero and makes the function undefined. However, in this exercise, the denominator \( 1 + x \) never equals zero in the interval \([0, 5]\), simplifying our work. Rational functions can have vertical asymptotes where the denominator is zero, but in this case, the function behaves well over our given interval.

Definite integrals

Definite integrals are used to calculate the area under a curve over a specific interval on the x-axis. This is an essential concept in calculus that helps us understand the accumulated quantity, such as area, volume, and other physical properties. The integral is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.

In our problem, we calculate the area under the curve \( y = \frac{x}{1+x} \) from \( x = 0 \) to \( x = 4 \), which is given by \( \int_{0}^{4} \frac{x}{1+x} \, dx \). Calculating this integral involves using substitution, a common technique for simplifying integrals. Here, setting \( u = 1 + x \) allows us to transform the integral into a more manageable form. The definite integral provides the exact area, which we calculated as \( 4 - \ln 5 \).

Graphing functions

Graphing functions is a fundamental skill in calculus that involves plotting a function's behavior over an interval to visually understand its nature. For \( y = \frac{x}{1+x} \), the graph shows how \( y \) changes as \( x \) increases.

To create an accurate graph, it's helpful to calculate the function values at key points within the interval, such as \( x = 0, 1, 2, 3, 4, \) and \( 5 \). These calculated values provide a series of points that help in sketching the curve. As we note, the function increases smoothly and approaches the value 1 as \( x \) grows larger, though it never actually reaches 1 due to its asymptotic nature.

This technique of plotting specific points and connecting them with a smooth curve is a reliable method to visualize how the function behaves, which is critical for interpreting other concepts like the definite integral.

Area under a curve

Finding the area under a curve involves integrating the function over a specified interval. This is graphically represented as the region between the curve and the x-axis. It's a major application of integration in calculus.

In the case of \( y = \frac{x}{1+x} \), the area under the curve from \( x = 0 \) to \( x = 4 \) is the focus. We looked at how this area is bounded: by the curve itself, the x-axis, and the vertical line \( x = 4 \). By integrating the function over this interval, we effectively calculated the size of this region, obtaining an exact measure of \( 4 - \ln 5 \).

Understanding this process helps in various fields, from scientific computations to real-world applications where determining an area with precision is necessary. The integration showcases the elegance of calculus in handling complex shapes and spaces efficiently.