Problem 13 - Calculus Single Variable

Question

Suppose that \(\int_{2}^{1} f(x) d x=0\) and \(\int_{2}^{1} g(x) d x=0 .\) Evaluate \(\int_{2}^{1}(f(x)-3 g(x)+5) d x\).

Step-by-Step Solution

Answer

The integral evaluates to \(-5\).

Step 1: Understanding the Given Integrals

We are given that \(\int_{2}^{1} f(x)\, dx = 0\) and \(\int_{2}^{1} g(x)\, dx = 0\). This means that both functions \(f(x)\) and \(g(x)\) have a net area of zero between the limits 2 and 1.

Step 2: Apply Properties of Integrals

The integral \(\int_{a}^{b} (f(x) + g(x))\, dx \) leads to \(\int_{a}^{b} f(x)\, dx + \int_{a}^{b} g(x)\, dx\). We can apply this property to \( (f(x) - 3g(x) + 5) \) to break it into separate integrals: \(\int_{2}^{1} f(x)\, dx - \int_{2}^{1} 3g(x)\, dx + \int_{2}^{1} 5\, dx\).

Step 3: Evaluate Separate Integrals

From the given, we know \(\int_{2}^{1} f(x)\, dx = 0\) and \(\int_{2}^{1} g(x)\, dx = 0\). The property of integrals with constants allows us to write \(\int_{2}^{1} 3g(x)\, dx = 3\int_{2}^{1} g(x)\, dx = 3 \times 0 = 0\). This leaves us with \(\int_{2}^{1} 5\, dx\).

Step 4: Integrate the Constant Function

The integral of a constant \(5\) over an interval \([a, b]\) is given by \(5(b-a)\). Thus, \(\int_{2}^{1} 5\, dx = 5(1 - 2) = -5\).

Step 5: Add the Evaluated Integrals Together

Combine the results of the integrals: \(0 - 0 - 5 = -5\). This means the value of \(\int_{2}^{1} (f(x) - 3g(x) + 5)\, dx\) is \(-5\).

Key Concepts

Understanding Properties of Integrals

When working with integrals, especially definite integrals, it's important to recognize some of their core properties. These properties help in simplifying complex integrals into manageable parts. One fundamental property is the ability to split the integral of a sum or difference of functions into separate integrals.

For example, if you have an integral of the form \[ \int_{a}^{b} ( f(x) + g(x) ) \, dx, \]this can be broken into \[ \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx. \]This property allows for evaluating each function's integral individually and then combining the results.

• It holds true for differences as well as sums.
• Let's say if you have \[ \int_{a}^{b} ( f(x) - 3g(x) + 5 ) \, dx, \] it becomes \[ \int_{a}^{b} f(x) \, dx - \int_{a}^{b} 3g(x) \, dx + \int_{a}^{b} 5 \, dx. \]
• This helps in simplifying problems, especially when parts of the integral are known.

Understanding this property is crucial because it provides the flexibility needed to effectively evaluate definite integrals by simplifying the given functions into their component parts.

Constant Function Integration Simplified

Integrating a constant function is one of the simplest tasks in calculus. A constant function is simply a function that takes the same value, regardless of the input; for example, in this case, the constant is 5. The integral of a constant function over an interval \([a, b]\) is given by the formula:

\[ \int_{a}^{b} c \, dx = c(b-a). \]

• Think about it as calculating the area of a rectangle. The base of the rectangle is \( b-a \), and the height is the value of the constant, \( c \).
• For our case, to integrate 5 from 2 to 1, it’s \[ \int_{2}^{1} 5 \, dx = 5(1 - 2) = -5. \]
• This simple formula helps quickly determine the area under a constant function without complex calculations.

The simplicity of integrating constant functions makes it a fundamental skill in calculus, building the foundation for understanding more complex integrals.

Calculating Net Area Under a Curve

The concept of a net area under a curve is fundamentally tied to definite integrals. When you evaluate a definite integral from a lower limit to an upper limit of a function, the resulting number is the net area between the curve and the x-axis over that interval.

• If the curve is above the x-axis over the interval, the area contributes positively to the net area.
• If the curve lies below the x-axis, it contributes negatively, which can lead to the net area being zero, or even negative, depending on the heights and widths of the sections of the curve.

In the given problem, both integrals of \( f(x) \) and \( g(x) \) have a net area of zero from 2 to 1. This means:

• The parts of the function that are above the x-axis perfectly balance out the parts below, resulting in no net area.
• Net area could involve regions both above and below the x-axis, leading to positive and negative areas offsetting each other.

Understanding the net area is essential in applications where balancing positive and negative contributions determine outcomes, such as in physics or probability. It gives a combined measure of how much function values rise above or fall below the x-axis within a given interval.