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Chapter 5: Problem 7

In Exercises \(7-12,\) evaluate the integral. $$\int_{-2}^{1} 5 d x$$

Short Answer

Expert verified
The solution to the given integral is 15.

Step by step solution

01

Identify the integral

The integral given is \( \int_{-2}^{1} 5 dx \), which is a constant function, as 5 does not vary with x.
02

Calculate the integral

The integral of a constant C over interval from a to b is C times (b-a). So, here it becomes 5 times (1- (-2)) which equals to 5 times 3.
03

Compute the result

Therefore, the result becomes 5*3 = 15.

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Key Concepts

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

Understanding Constant Functions in Integration
A constant function is a simple yet important concept in calculus. For integrals, when you see something like \( \int 5 \, dx \), the number 5 is a constant. This means it doesn't change whatever the value of \( x \) is. It's always 5, simple as that.

Knowing that you're dealing with a constant function simplifies the process greatly. There are no variables or complicated expressions to integrate. Constant functions are straightforward:
  • You just multiply the constant by the length of the interval you're integrating over.
This idea becomes clearer as you get comfortable with definite integrals.
Basic Integration Methods
Integration is like finding the sum of areas under curves. But if you're dealing with a constant function, it's even easier鈥攖he 'curve' is just a horizontal line. With integration, one of the core ideas is to find an antiderivative, or something that represents the original function along with what's under it.

In our example of \( \int_{-2}^{1} 5 \, dx \), we're using a pretty straightforward integration method:
  • Since we're working with a constant, you directly multiply this constant by the difference between the upper and lower bounds of the interval. So, in layman's terms: just multiply 5 by the result of \( 1 - (-2) \).
That simplifies the whole process and leads us straight to the power of number tricks and properties you learned in basic math.
Evaluation of Definite Integrals
Evaluating integrals is about calculating a specific value over an interval. This tells you how much 'total area' is under the function from one point to another along the x-axis. With definite integrals, however, you only care about the results from the specific bounds given.

For \( \int_{-2}^{1} 5 \, dx \), the steps are simple:
  • First, recognize the bounds: from \(-2\) to \(1\).
  • Then, compute the difference: \(1 - (-2)\), which equals 3.
  • Finally, multiply this difference by the constant: \(5 \times 3\).
You'll get 15, which represents the entire value of the integral in this interval.

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Most popular questions from this chapter

In Exercises \(23-26\) use a calculator program to find the Simpson's Rule approximations with \(n=50\) and \(n=100 .\) $$\int_{0}^{\pi / 2} \sin \left(x^{2}\right) d x$$

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True or False The average value of a function \(f\) on \([ a , b ]\) always lies between \(f ( a )\) and \(f ( b ) .\) Justify your answer.

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