Question

Properties of integrals Consider two functions \(f\) and \(g\) on [1,6] such that \(\int_{1}^{6} f(x) d x=10, \int_{1}^{6} g(x) d x=5, \int_{4}^{6} f(x) d x=5,\) and \(\int_{1}^{4} g(x) d x=2 .\) Evaluate the following integrals. a. \(\int_{1}^{4} 3 f(x) d x\) b. \(\int_{1}^{6}(f(x)-g(x)) d x\) c. \(\int_{1}^{4}(f(x)-g(x)) d x\) d. \(\int_{4}^{6}(g(x)-f(x)) d x\) e. \(\int_{4}^{6} 8 g(x) d x\) f. \(\int_{4}^{1} 2 f(x) d x\)

Short answer

In summary, we have found the values of the following integrals: a) \(\int_{1}^{4} 3f(x) dx = 15\) b) \(\int_{1}^{6}(f(x)-g(x)) dx = 5\) c) \(\int_{1}^{4}(f(x)-g(x)) dx = 3\) d) \(\int_{4}^{6}(g(x)-f(x)) dx = -2\) e) \(\int_{4}^{6} 8g(x) dx = 24\) f) \(\int_{4}^{1} 2f(x) dx = -10\)

Step-by-step solution

Problem a: \(\int_{1}^{4} 3f(x) dx\)

To find this integral, we can use the linearity property of integrals. If \(\int_{a}^{b} f(x) dx = k\), then \(\int_{a}^{b} cf(x) dx = ck\), where \(c\) is a constant. We know that \(\int_{1}^{6} f(x) dx = 10\). We are given the integral of \(f(x)\) from \(1\) to \(6\), but we need the integral from \(1\) to \(4\). We can find this by using the integral value from \(4\) to \(6\), which is given as \(\int_{4}^{6} f(x) dx = 5\). Using the additivity property of integrals, we have \(\int_{1}^{4} f(x) dx = \int_{1}^{6} f(x) dx - \int_{4}^{6} f(x) dx = 10 - 5 = 5\). Now we can find the required integral: \(\int_{1}^{4} 3f(x) dx = 3 \int_{1}^{4} f(x) dx = 3 \cdot 5 = 15\).

Problem b: \(\int_{1}^{6}(f(x)-g(x)) dx\)

We can use the linearity and additivity properties of integrals to solve this problem. We know that \(\int_{1}^{6} f(x) dx = 10\) and \(\int_{1}^{6} g(x) dx = 5\). From the additivity property, we get \(\int_{1}^{6}(f(x) - g(x)) dx = \int_{1}^{6} f(x) dx - \int_{1}^{6} g(x) dx = 10 - 5 = 5\).

Problem c: \(\int_{1}^{4}(f(x)-g(x)) dx\)

Using similar methodology as before, we first find the integral values from \(1\) to \(4\) for \(f(x)\) and \(g(x)\). For \(f(x)\), we have \(\int_{1}^{4} f(x) dx = 5\) (obtained in problem a). We were given \(\int_{1}^{4} g(x) dx = 2\). Now, using additivity property, we get \(\int_{1}^{4}(f(x) - g(x)) dx = \int_{1}^{4} f(x) dx - \int_{1}^{4} g(x) dx = 5 - 2 = 3\).

Problem d: \(\int_{4}^{6}(g(x)-f(x)) dx\)

Using a similar approach and the additivity property: \(\int_{4}^{6}(g(x) - f(x)) dx = \int_{4}^{6} g(x) dx - \int_{4}^{6} f(x) dx\) We know that \(\int_{1}^{6} g(x) dx = 5\) and \(\int_{1}^{4} g(x) dx = 2\). So, \(\int_{4}^{6} g(x) dx = 5 - 2 = 3\). We were given \(\int_{4}^{6} f(x) dx = 5\). Therefore, the integral becomes \(\int_{4}^{6}(g(x) - f(x)) dx = 3 - 5 = -2\).

Problem e: \(\int_{4}^{6} 8g(x) dx\)

Similarly, using the linearity property of integrals, we can find the solution. We had already found \(\int_{4}^{6} g(x) dx = 3\). Now, \(\int_{4}^{6} 8g(x) dx = 8 \int_{4}^{6} g(x) dx = 8 \cdot 3 = 24\).

Problem f: \(\int_{4}^{1} 2f(x) dx\)

To find this integral, we can use the property of reversal of limits. If \(\int_{a}^{b} f(x) dx = k\), then \(\int_{b}^{a} f(x) dx = -k\). We know that \(\int_{1}^{4} f(x) dx = 5\). Therefore, \(\int_{4}^{1} f(x) dx = -5\). Now, using the linearity property, we get \(\int_{4}^{1} 2f(x) dx = 2 \int_{4}^{1} f(x) dx = 2 \cdot (-5) = -10\).

Key concepts

Linearity Property of Integrals

Understanding the linearity property of integrals is essential when dealing with scalar multiplication within an integral. Imagine you are working with an integral of a function multiplied by a constant, such as \( c \cdot f(x) \). The linearity property tells us that you can take the constant out of the integral and multiply it with the integral value of the function. In mathematical terms, if you have \( \int_{a}^{b} c \cdot f(x) dx \), you can simplify it to \( c \cdot \int_{a}^{b} f(x) dx \).

This means, the integral of a constant multiplied by a function over an interval is the same as the constant multiplied by the integral of the function over that interval. For example, if \( \int_{1}^{6} f(x) dx = 10 \) and you want to calculate \( \int_{1}^{6} 3f(x) dx \), you can simply multiply the constant 3 by 10, giving you 30, which would be the value of the latter integral.

Additivity Property of Integrals

When we look at the additivity property of integrals, we're observing how integrals can be broken down or combined over different intervals. The property states that if you have two integrals with the same function but different intervals, you can add or subtract these integrals depending on how the intervals relate to each other.

For instance, if you have the integral of \( f(x) \) from 1 to 6, that is \( \int_{1}^{6} f(x) dx \) and you know the integral from 1 to 4, as well as from 4 to 6, you can find the integral from 1 to 6 by adding these two together: \( \int_{1}^{4} f(x) dx + \int_{4}^{6} f(x) dx = \int_{1}^{6} f(x) dx \). This property is particularly useful when you're dealing with functions that are easy to integrate over specific intervals but not over others, or when you have prior knowledge about the integral values over different ranges, as it allows you to combine or disassemble these values as needed.

Integral Evaluation

Lastly, evaluating an integral involves computing the definite integral of a function over a specific interval. To evaluate an integral, you may apply the Fundamental Theorem of Calculus if you have the antiderivative of the function, or rely on properties of integrals if given particular known values, as showcased in our example problems.

For example, to evaluate \( \int_{4}^{6} g(x) dx \) when you know \( \int_{1}^{6} g(x) dx = 5 \) and \( \int_{1}^{4} g(x) dx = 2 \), you subtract the known value of the integral from 1 to 4 from the integral from 1 to 6, resulting in the integral from 4 to 6. These steps simplify the sometimes challenging process of finding the antiderivative and calculating the difference at the endpoints of the interval. By understanding the concept of integral evaluation and utilizing these integral properties, you can simplify complex integral problems into more manageable parts and efficiently find their solutions.