/*! 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 53 Are the statements true or false... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Are the statements true or false? Give an explanation for your answer. If \(F(x)=\int_{0}^{x} f(t) d t\) and \(G(x)=\int_{0}^{x} g(t) d t,\) then \(F(x)+G(x)=\int_{0}^{x}(f(t)+g(t)) d t\).

Short Answer

Expert verified
True, the equation holds by the property of integral sum.

Step by step solution

01

Recall the Property of Integrals

To solve the problem, we need to remember a key property of integrals: the integral of a sum is equal to the sum of the integrals. This can be expressed mathematically as: \[\int (a(t) + b(t)) \, dt = \int a(t) \, dt + \int b(t) \, dt\]where \(a(t)\) and \(b(t)\) are functions.
02

Evaluate the Left Hand Side

Given the definitions of \( F(x) \) and \( G(x) \), the left-hand side of the equation is:\[F(x) + G(x) = \int_{0}^{x} f(t) \, dt + \int_{0}^{x} g(t) \, dt\]This uses the property of integrals that states that the sum of integrals of functions is additive.
03

Evaluate the Right Hand Side

The right-hand side of the equation is given by:\[\int_{0}^{x} (f(t) + g(t)) \, dt\]This uses the integral of a sum property, which combines the functions \( f(t) \) and \( g(t) \) under one integral.
04

Compare Both Sides

Now compare both sides: \[F(x) + G(x) = \int_{0}^{x} f(t) \, dt + \int_{0}^{x} g(t) \, dt = \int_{0}^{x} (f(t) + g(t)) \, dt\]Since the equation holds true using the property that sums of functions within integrals can be separated or combined, the original statement is valid and true.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Properties of Integrals
When working with integrals, it's important to understand some fundamental properties that can help you simplify and manipulate expressions more easily. One of these key properties is the integral of a sum, which states that the integral of two functions added together is equal to the sum of their individual integrals. This can be neatly expressed as:
  • \( \int (a(t) + b(t)) \, dt = \int a(t) \, dt + \int b(t) \, dt \)
Essentially, what this tells you is that you can "distribute" the integral over addition. This property is very useful in breaking down complex integrals into more manageable parts, making calculations simpler and more intuitive.Another important thing to remember is linearity in integrals. This means that if you have constants multiplied by the functions, they can be factored out of the integral:
  • \( \int c \cdot f(t) \, dt = c \cdot \int f(t) \, dt \)
These basic properties are not just theoretical—they offer practical techniques for solving problems efficiently.
Definite Integrals
Definite integrals are a central concept in integral calculus that help you calculate the exact area under a curve between two points, typically denoted as \( a \) and \( b \). The notation \( \int_{a}^{b} f(t) \, dt \) refers to this particular type of integral. Here, \( a \) and \( b \) are the lower and upper limits of the integral, respectively. Such integrals allow you to evaluate the total accumulation of quantities, like distance, area, volume, etc., over a specific interval. Unlike indefinite integrals, definite integrals do not have a constant of integration, resulting in a specific numerical value once evaluated.
Another unique property of definite integrals is their ability to handle both positive and negative values. If the function dips below the x-axis within the interval \([a, b]\), the integral sums both positive and negative areas, reflecting net accumulation.
These features make definite integrals incredibly versatile in applications across physics, engineering, and beyond.
Sum of Integrals
The concept of the sum of integrals is crucial for simplifying expressions that feature multiple integral terms. In many cases, you might encounter a situation where you need to integrate a sum of functions over the same interval. Fortunately, thanks to the properties discussed earlier, you know that:
  • \( \int_{a}^{b} (f(t) + g(t)) \, dt = \int_{a}^{b} f(t) \, dt + \int_{a}^{b} g(t) \, dt \)
This property holds true for definite integrals, where the integration is carried out over specified limits. It allows you to separate or combine integrals to make calculations easier.
If the original problem involved finding \( F(x) + G(x) \) or \( \int_{0}^{x} (f(t) + g(t)) \, dt \), understanding that these two expressions are equivalent thanks to the sum of integrals property can save a lot of time.
By recognizing this equivalence, you can confidently approach problems involving sums of functions in integral form, knowing that splitting or combining these integrals will not alter the result.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Evaluate the definite integrals exactly \([as in \)\ln (3 \pi)]\(,\) using the Fundamental Theorem, and numerically \([\ln (3 \pi) \approx 2.243]\) $$\int_{2}^{5}\left(x^{3}-\pi x^{2}\right) d x$$

Evaluate the definite integrals exactly \([as in \)\ln (3 \pi)]\(,\) using the Fundamental Theorem, and numerically \([\ln (3 \pi) \approx 2.243]\) $$\int_{0}^{1} 2 e^{x} d x$$

Newton's law of gravity says that the gravitational force between two bodies is attractive and given by $$F=\frac{G M m}{r^{2}}$$, where \(G\) is the gravitational constant, \(m\) and \(M\) are the masses of the two bodies, and \(r\) is the distance between them. This is the famous inverse square law. For a falling body, we take \(M\) to be the mass of the earth and \(r\) to be the distance from the body to the center of the earth. So, actually, \(r\) changes as the body falls, but for anything we can easily observe (say, a ball dropped from the Tower of Pisa), it won't change significantly over the course of the motion. Hence, as an approximation, it is reasonable to assume that the force is constant. According to Newton's second law, acceleration is caused by a force and Force \(=\) Mass \(\times\) Acceleration. (a) Find the differential equation for the position, \(s\) of a moving body as a function of time. (b) Explain how the differential equation shows the acceleration of the body is independent of its mass.

Give an example of: Two different solutions to the differential equation $$\frac{d y}{d t}=t+3$$

Evaluate the definite integrals exactly \([as in \)\ln (3 \pi)]\(,\) using the Fundamental Theorem, and numerically \([\ln (3 \pi) \approx 2.243]\) $$\int_{0}^{\pi / 4} \sin x d x$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.