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Chapter 5: Problem 11

In Exercises \(7-12,\) evaluate the integral. $$\int_{-2.1}^{3.4} 0.5 d s$$

Short Answer

Expert verified
The integral evaluates to \(2.75\).

Step by step solution

01

Identify the type of integral

The integral we are dealing with is a definite integral over a constant function. The basic theorem that states that the integral of a constant over the interval \([a, b]\) is equal to that constant multiplied by the size of the interval can be put to use here.
02

Calculate the size of the interval

Subtract the lower limit of integration from the upper limit: \(3.4 - (-2.1) = 5.5\). So, the size of the interval is \(5.5\).
03

Evaluate the integral

Finally, the value of the integral is equal to the constant (0.5 in this case) multiplied by the size of the interval. So the value of the integral will be \(0.5 * 5.5 = 2.75\).

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Key Concepts

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

Constant Function
When you encounter a constant function in calculus, you are dealing with a function where the output value is the same regardless of the input. In our exercise, the function is represented by the constant 0.5.

Since the function doesn't change, it simplifies the integration process significantly. Instead of worrying about curves or areas under varying graphs, you focus on simple multiplication. This constant output is a building block in calculus and provides a straightforward path for solving integrals involving constant functions.
Evaluation of Integrals
Evaluating an integral involves calculating the area under a curve defined by a function over a specified interval. In our example of a definite integral \[ \int_{-2.1}^{3.4} 0.5 \, ds \]it specifically tells us to find the area under the curve of the constant function 0.5 from -2.1 to 3.4.

Here鈥檚 how we evaluate it:
  • First, determine the size of the interval, which we find by subtracting the lower limit from the upper limit: \[ 3.4 - (-2.1) = 5.5 \]
  • Then, multiply the constant value by this interval length: \[0.5 \times 5.5 = 2.75 \]
This result, 2.75, tells you the exact quantity that represents the area under the constant curve within the specified range.
Properties of Definite Integrals
Understanding the properties of definite integrals can make solving problems faster and clearer. Here are a few essential properties:

  • **Linearity:** If you have \[\int_{a}^{b} c f(x) \, dx,\]where c is a constant, you can pull out the constant: \[c \times \int_{a}^{b} f(x) \, dx\]
  • **Additivity:** The integral from a to b can be split into the sum of two integrals: \[\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \]
  • **Interval Length:** For a constant function, the length of the interval is crucial. As seen in our exercise, calculate this interval carefully to resolve accurate area under the curve. \[\text{If } f(x) = c, \text{ then } \int_{a}^{b} f(x) \, dx = c \times (b-a)\]
These properties help break down more complex integral problems into manageable pieces, allowing you to leverage basic numerical principles to find solutions.

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Most popular questions from this chapter

Multiple Choice If an MRAM sum with four rectangles of equal width is used to approximate the area enclosed between the \(x\) -axis and the graph of \(y=4 x-x^{2},\) the approximation is (A) 10 (B) 10.5 (C) \(10 . \overline{6}\) (D) 10.75 (E) 11

In Exercises \(49-54,\) use NINT to solve the problem. Find the average value of \(\sqrt{\cos x}\) on the interval \([-1,1]\)

Rectangular Approximation Methods Prove or disprove the following statement: MRAM \(_{n}\) is always the average of LRAM \(_{n}\) and \(\operatorname{RRAM}_{n}\).

In Exercises \(49-54,\) use NINT to solve the problem. Evaluate \(\int_{0}^{10} \frac{1}{3+2 \sin x} d x\)

Distance Traveled by a Projectile An object is shot straight upward from sea level with an initial velocity of 400 ft/sec. (a) Assuming gravity is the only force acting on the object, give an upper estimate for its velocity after 5 sec have elapsed. Use \(g=32 \mathrm{ft} / \mathrm{sec}^{2}\) for the gravitational constant. (b) Find a lower estimate for the height attained after 5 sec.
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