/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Suppose that \(\int_{1}^{2} f(x)... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that \(\int_{1}^{2} f(x) d x=5 .\) Find $$ \begin{array}{ll}{\text { a. } \int_{1}^{2} f(u) d u} & {\text { b. } \int_{1}^{2} \sqrt{3} f(z) d z} \\ {\text { c. } \int_{2}^{1} f(t) d t} & {\text { d. } \int_{1}^{2}[-f(x)] d x}\end{array} $$

Short Answer

Expert verified
a. 5, b. \(5\sqrt{3}\), c. -5, d. -5.

Step by step solution

01

Understanding direct substitution for a.

Given \( \int_{1}^{2} f(x) \, dx = 5 \), we are asked to find \( \int_{1}^{2} f(u) \, du \). Since the variable being integrated (\( x \) or \( u \)) does not change the value of the integral as long as the limits and function are the same, we have \( \int_{1}^{2} f(u) \, du = 5 \).
02

Applying scaling property for b.

We have \( \int_{1}^{2} \sqrt{3} f(z) \, dz \). By the property of integrals \( \int_{a}^{b} c f(x) \, dx = c \int_{a}^{b} f(x) \, dx \), where \( c \) is a constant, we get \( \sqrt{3} \int_{1}^{2} f(z) \, dz = \sqrt{3} \times 5 = 5\sqrt{3} \).
03

Reversing limits for c.

To find \( \int_{2}^{1} f(t) \, dt \), we use the property \( \int_{a}^{b} f(x) \, dx = - \int_{b}^{a} f(x) \, dx \). Therefore, \( \int_{2}^{1} f(t) \, dt = - \int_{1}^{2} f(t) \, dt = -5 \).
04

Applying negative to the integrand for d.

We need \( \int_{1}^{2} [-f(x)] \, dx \). By linearity, this becomes \( - \int_{1}^{2} f(x) \, dx = -5 \).

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Key Concepts

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

Properties of Integrals
Definite integrals possess several fundamental properties that make them versatile and broadly applicable in mathematics. One key property is that the integral of a function over a certain interval is independent of the variable of integration. This means that if you have a definite integral represented as \( \int_{a}^{b} f(x) \, dx \), this will yield the same result as \( \int_{a}^{b} f(u) \, du \), provided the function \( f \) and the limits \( a \) and \( b \) are unchanged.

  • Changing the variable inside the integral does not affect its solution.
  • The integral depends solely on the function and the limits.
These properties allow integrals to be manipulated into forms that are easier to calculate or that reveal certain properties of a function. Understanding these properties is crucial for solving complex integral calculations efficiently.
Reversal of Limits in Integration
A fascinating aspect of definite integrals is the effect of interchanging the limits of integration. Normally, when integrating a function \( f(x) \) from \( a \) to \( b \), such as \( \int_{a}^{b} f(x) \, dx \), you obtain a certain value. However, if you reverse the limits to \( \int_{b}^{a} f(x) \, dx \), the result becomes the negative of the original integral.

  • For any continuous function \( f(x) \), \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \).
  • This property is useful when rearranging integrals for symmetry or simplifying expressions.
Understanding this property helps when the limits of integration are manipulated, ensuring calculations remain accurate even if the direction of integration changes.
Linearity of Integrals
The linearity property of integrals plays a critical role in simplifying computations. This property consists fundamentally of two components: the ability to factor constants out of an integral and the additive nature of integrals.

  • If you need to integrate \( c \times f(x) \) over \( [a, b] \), you can take the constant \( c \) outside: \( \int_{a}^{b} c \times f(x) \, dx = c \int_{a}^{b} f(x) \, dx \).
  • The integral of a sum of functions equals the sum of their integrals: \( \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \).
This property allows for easier handling of complex functions by breaking them into simpler parts or moving constants outside the integral, thus streamlining computation.
Scaling Property of Integrals
The scaling property, a subset of the linearity of integrals, highlights how integrals respond to constants that multiply a function. Specifically, when a constant multiplies the integrand, this constant can be factored out and multiplied by the integral's final result. This is expressed mathematically as \( \int_{a}^{b} c \times f(x) \, dx = c \int_{a}^{b} f(x) \, dx \), where \( c \) is a constant scalar.

  • The constant is a scalar and does not depend on the variable of integration.
  • Scaling simplifies computation by factoring the constant out, reducing the complexity of the problem within the integral.
The scaling property is particularly useful in practical applications where functions are accompanied by constants, such as in physics and engineering problems. Understanding this allows for efficient computation without sacrificing accuracy.

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Most popular questions from this chapter

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