Problem 46 - Calculus for Biology and Medicine

Question

In Problems , use an area formula from geometry to find the value of each integral by interpreting it as the (signed) area under the graph of an appropriately chosen function. $$ \int_{0}^{2}|x-1| d x $$

Step-by-Step Solution

Answer

The integral \(\int_{0}^{2}|x-1| dx\) equals 1.

Step 1: Understanding the Absolute Value Function

The function \(|x-1|\) can be split into two linear pieces based on the value of \(x\): \(x-1\) when \(x \geq 1\), and \(-(x-1)\) when \(x < 1\). The graph forms a "V" shape, with a vertex at \((1,0)\).

Step 2: Setting Up the Graph

The graph of \(|x-1|\) is a straight line with a positive slope from \(x=0\) to \(x=1\) (\(|x-1| = 1-x \,\text{for}\, x < 1\)), and a straight line with a positive slope from \(x=1\) to \(x=2\) (\(|x-1| = x-1 \, \text{for}\, x \geq 1\)).

Step 3: Dividing the Integral

Break down the integral using the point \(x=1\), where the function changes:\[\int_{0}^{2} |x - 1| \, dx = \int_{0}^{1} (1-x) \, dx + \int_{1}^{2} (x-1) \, dx\]

Step 4: Calculating the First Integral

Calculate the area from \(x=0\) to \(x=1\):\[\int_0^1 (1-x) \, dx\]This integral represents the area of a right triangle with base and height both equal to 1. The area is \(\frac{1}{2} \times 1 \times 1 = \frac{1}{2}\).

Step 5: Calculating the Second Integral

Calculate the area from \(x=1\) to \(x=2\):\[\int_1^2 (x-1) \, dx\]This integral also represents the area of a right triangle with base and height both equal to 1, so the area is \(\frac{1}{2} \times 1 \times 1 = \frac{1}{2}\).

Step 6: Summing the Areas

Sum the areas of the two triangles obtained from the integrals:\[\frac{1}{2} + \frac{1}{2} = 1\]

Step 7: Conclusion

The value of the integral \(\int_{0}^{2}|x-1| \, dx\) is the total area under the graph of \(|x-1|\), which is 1.

Key Concepts

Absolute Value Function

The absolute value function, particularly \(|x-1|\), is an essential concept in math. It takes any real number input and returns its distance from zero on the number line, without considering its direction. This means the function will always yield a non-negative result. For example, when the input is 3 or -3, the function returns 3.More specifically, the function we are working with here, \(|x-1|\), presents a peculiar behavior. It splits into two linear pieces depending on the value of \(x\): \(x-1\) or \(-(x-1)\). For values where \(x\) is less than 1, the expression \(-(x-1)\) works as \(|x-1| = 1-x\). Conversely, for \(x\) equal to or greater than 1, it simplifies to \(x-1\).Visually, this creates a distinctive 'V' shape on a graph, as it combines the slopes from both segments, with a vertex located at \((1,0)\). Understanding and identifying these characteristics allow us to manipulate and interpret absolute value functions effectively in various mathematical applications.

Area Under a Curve

When working with integrals in calculus, we often refer to the "area under a curve." In simple terms, this means the total area that lies between the curve of a function and the x-axis on a coordinate plane.In the case of the integral mentioned \[ \int_{0}^{2} |x-1| \, dx \], the curve is represented by the graph of \(|x-1|\). The integral calculates the total "signed area" from \(x=0\) to \(x=2\). This implies:

• From \(x=0\) to \(x=1\), we see a right triangle formed with the x-axis. It has a base and a height both of length 1.
• Similarly, from \(x=1\) to \(x=2\), another right triangle emerges, again having base and height of 1.

The signed areas found here are \(rac{1}{2}\) for each triangle, resulting in a total area of \(1\) when added together. Thus, the integral represents an efficient method of summing these separate areas to determine the overall area covered by the function's graph.

Integral Calculus

Integral calculus is a branch of mathematics that involves summation processes, primarily to find areas, volumes, central points, and more. At its core, it involves the operations of integration, which allow us to accumulate quantities.There are two main types of integrals in calculus:

• Definite Integrals: These calculate the exact area under a curve within given limits on the x-axis, as seen in \[ \int_{0}^{2} |x-1| \, dx \].
• Indefinite Integrals: These do not have set limits and represent a family of functions, essentially serving as the antiderivative of a given function.

Integration works by breaking a complex region into easier, manageable pieces (like rectangles or triangles). It then systematically sums these pieces to ascertain the region's total size or measure. In our problem, we divided the integral at the point where the function transitioned from one linear segment to another to calculate simpler areas and sum them for a result of \(1\).Understanding integral calculus forms the backbone for advanced studies in mathematics and physics. It aids in modeling and solving real-world problems involving continuous data.