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Properties of integrals Suppose \(\int_{0}^{3} f(x) d x=2\) \(\int_{3}^{6} f(x) d x=-5,\) and \(\int_{3}^{6} g(x) d x=1 .\) Evaluate the following integrals. a. \(\int_{0}^{3} 5 f(x) d x\) b. \(\int_{3}^{6}(-3 g(x)) d x\) c. \(\int_{3}^{6}(3 f(x)-g(x)) d x\) d. \(\int_{6}^{3}(f(x)+2 g(x)) d x\)

Short Answer

Expert verified
Question: Evaluate the following integrals: a. \(\int_{0}^{3} 5f(x) dx\) b. \(\int_{3}^{6} (-3g(x)) dx\) c. \(\int_{3}^{6} (3f(x) - g(x)) dx\) d. \(\int_{6}^{3} (f(x) + 2g(x)) dx\) Given that: \(\int_{0}^{3} f(x) dx = 2\), \(\int_{3}^{6} f(x) dx = -5\), and \(\int_{3}^{6} g(x) dx = 1\). Answer: a. \(10\) b. \(-3\) c. \(-16\) d. \(3\)

Step by step solution

01

Apply the constant multiplication property

We have \(\int_{0}^{3} 5f(x) dx\). Using the property that the integral of a constant times a function is the constant times the integral of the function, we get: \(\int_{0}^{3} 5f(x) dx = 5\int_{0}^{3} f(x) dx\)
02

Evaluate the integral

We know that \(\int_{0}^{3} f(x) dx = 2\). Therefore, the integral is: \(5\int_{0}^{3} f(x) dx = 5 \cdot 2 = 10\) b.
03

Apply the constant multiplication property

We have \(\int_{3}^{6} (-3g(x)) dx\). Again, using the constant multiplication property: \(\int_{3}^{6} (-3g(x)) dx = -3\int_{3}^{6} g(x) dx\)
04

Evaluate the integral

We know that \(\int_{3}^{6} g(x) dx = 1\). Therefore, the integral is: \(-3\int_{3}^{6} g(x) dx = -3 \cdot 1 = -3\) c.
05

Apply the addition and subtraction property

We have \(\int_{3}^{6} (3f(x) - g(x)) dx\). Using the addition and subtraction property: \(\int_{3}^{6} (3f(x) - g(x)) dx = \int_{3}^{6} 3f(x) dx - \int_{3}^{6} g(x) dx\)
06

Use the constant multiplication property

We have \(\int_{3}^{6} 3f(x) dx\). Applying the constant multiplication property: \(\int_{3}^{6} 3f(x) dx = 3\int_{3}^{6} f(x) dx\)
07

Evaluate the integrals

We know that \(\int_{3}^{6} f(x) dx = -5\) and \(\int_{3}^{6} g(x) dx = 1\). Therefore, the integral is: \(3\int_{3}^{6} f(x) dx - \int_{3}^{6} g(x) dx = 3 \cdot (-5) - 1 = -15 - 1 = -16\) d.
08

Apply the reverse order property

We have \(\int_{6}^{3} (f(x) + 2g(x)) dx\). Using the reverse order property: \(\int_{6}^{3} (f(x) + 2g(x)) dx = -\int_{3}^{6} (f(x) + 2g(x)) dx\)
09

Apply the addition and subtraction property

Using the addition and subtraction property: \(-\int_{3}^{6} (f(x) + 2g(x)) dx = -(\int_{3}^{6} f(x) dx + 2\int_{3}^{6} g(x) dx)\)
10

Evaluate the integrals

We know that \(\int_{3}^{6} f(x) dx = -5\) and \(\int_{3}^{6} g(x) dx = 1\). Therefore, the integral is: \(-(\int_{3}^{6} f(x) dx + 2\int_{3}^{6} g(x) dx) = -(-5 + 2 \cdot 1) = -(-5 + 2) = 3\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Multiplication Property
The constant multiplication property of integrals is straightforward yet powerful. It tells us that if you multiply a function by a constant before integrating, you can actually factor the constant out of the integral.
This means that the process of integration becomes much simpler.For example, let's say you have the integral \( \int_{a}^{b} c \cdot f(x) \, dx \), where \( c \) is a constant.
  • According to the constant multiplication property, this is equivalent to \( c \cdot \int_{a}^{b} f(x) \, dx \).
  • This simplification allows us to evaluate the integral of \( f(x) \) and then multiply the result by the constant \( c \).
For instance, if you have \( \int_{0}^{3} 5f(x) \) and you know \( \int_{0}^{3} f(x) \, dx = 2 \), you simply multiply 2 by 5 to get 10.
This makes calculations both efficient and less prone to error.
Addition and Subtraction Property
The addition and subtraction property of integrals allows us to handle more complex expressions by splitting them into simpler, individual integrals. This property is useful when dealing with the sum or difference of two functions inside an integrand.
Suppose you have an integral of the form \( \int_{a}^{b} (f(x) + g(x)) \, dx \). According to this property, you can separate it as:
  • \( \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \).
  • Similarly, for subtraction, \( \int_{a}^{b} (f(x) - g(x)) \, dx = \int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx \).
This makes it much easier to work with each function separately before combining the results.
In our exercise, we dealt with \( \int_{3}^{6} (3f(x) - g(x)) \, dx \). By applying the property, we split it into \( \int_{3}^{6} 3f(x) \, dx - \int_{3}^{6} g(x) \, dx \), and performed the calculations based on known integrals for each function.
This decomposition approach simplifies the problem significantly and helps tackle even more complicated integrands effectively.
Reverse Order Property
The reverse order property is a useful feature of definite integrals that allows us to flip the limits of integration and change the sign of the integral.
This property can often simplify calculations or help make them more intuitive.Consider an integral \( \int_{a}^{b} f(x) \, dx \). According to the reverse order property, it can be expressed as:
  • \( -\int_{b}^{a} f(x) \, dx \).
This means if the upper limit is smaller than the lower limit, you can reverse them and multiply the result by -1.In our example with \( \int_{6}^{3} (f(x) + 2g(x)) \, dx \), applying the reverse order property allowed us to flip the limits:
\( -\int_{3}^{6} (f(x) + 2g(x)) \, dx \). By carrying out the rest of the operation using properties we already understood, we turned a potentially tricky negative area calculation into one that aligns with familiar integration properties.
This transformation is invaluable when dealing with negative integrals or when simplifying calculations.

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Most popular questions from this chapter

The population of a culture of bacteria has a growth rate given by \(p^{\prime}(t)=\frac{200}{(t+1)^{r}}\) bacteria per hour, for \(t \geq 0,\) where \(r > 1\) is a real number. In Chapter 6 it is shown that the increase in the population over the time interval \([0, t]\) is given by \(\int_{0}^{t} p^{\prime}(s) d s\). (Note that the growth rate decreases in time, reflecting competition for space and food.) a. Using the population model with \(r=2,\) what is the increase in the population over the time interval \(0 \leq t \leq 4 ?\) b. Using the population model with \(r=3,\) what is the increase in the population over the time interval \(0 \leq t \leq 6 ?\) c. Let \(\Delta P\) be the increase in the population over a fixed time interval \([0, T] .\) For fixed \(T,\) does \(\Delta P\) increase or decrease with the parameter \(r ?\) Explain. d. A lab technician measures an increase in the population of 350 bacteria over the 10 -hr period [0,10] . Estimate the value of \(r\) that best fits this data point. e. Looking ahead: Use the population model in part (b) to find the increase in population over the time interval \([0, T],\) for any \(T > 0 .\) If the culture is allowed to grow indefinitely \((T \rightarrow \infty)\) does the bacteria population increase without bound? Or does it approach a finite limit?

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