/*! 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 19 Evaluate the integrals using Par... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integrals using Part 1 of the Fundamental Theorem of Calculus. $$\int_{-\pi / 2}^{\pi / 2} \sin \theta d \theta$$

Short Answer

Expert verified
The integral evaluates to 0.

Step by step solution

01

Identify the function and its antiderivative

The integral given is \( \int_{-\pi / 2}^{\pi / 2} \sin \theta \, d\theta \). According to Part 1 of the Fundamental Theorem of Calculus, we need an antiderivative (indefinite integral) of \( \sin \theta \). The antiderivative of \( \sin \theta \) is \(-\cos \theta + C\), where \( C \) is a constant of integration.
02

Apply the Fundamental Theorem of Calculus Part 1

The Fundamental Theorem of Calculus Part 1 states that if \( F \) is an antiderivative of \( f \) on an interval \( [a, b] \), then \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \].Here, set \( F(\theta) = -\cos \theta \). We need to evaluate \( F(\theta) \) at the upper limit \( \pi/2 \) and the lower limit \( -\pi/2 \).
03

Evaluate the antiderivative at the upper limit

To evaluate at the upper limit, substitute \( \theta = \pi/2 \) into the antiderivative:\[ F\left(\frac{\pi}{2}\right) = -\cos\left(\frac{\pi}{2}\right) = -0 = 0. \]
04

Evaluate the antiderivative at the lower limit

To evaluate at the lower limit, substitute \( \theta = -\pi/2 \) into the antiderivative:\[ F\left(-\frac{\pi}{2}\right) = -\cos\left(-\frac{\pi}{2}\right) = -0 = 0. \]
05

Calculate the value of the definite integral

Now, subtract the value of the antiderivative at the lower limit from the value at the upper limit:\[ \int_{-\pi / 2}^{\pi / 2} \sin \theta \, d\theta = F\left(\frac{\pi}{2}\right) - F\left(-\frac{\pi}{2}\right) = 0 - 0 = 0. \]

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Key Concepts

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

Antiderivative
An antiderivative of a function is essentially the reverse of a derivative. If you have a function, and you know its derivative, the antiderivative is the function you originally started with, before differentiation took place.For example, the antiderivative of \( \sin \theta \) is \(-\cos \theta + C\). The \( C \) represents a constant of integration because derivatives of constants are zero, thus you lose any constant information in the derivative process.Here’s a quick checklist about antiderivatives:
  • Antiderivatives are the foundation for integral calculus.
  • A function can have many antiderivatives differing by a constant.
  • Finding an antiderivative is like solving the reverse problem of differentiation.
In the context of the Fundamental Theorem of Calculus, antiderivatives play a critical role in calculating definite integrals, helping us transition from finding the function back from its rate of change.
Definite Integral
A definite integral represents the signed area under a curve of a function between two points. Unlike indefinite integrals, which focus on finding functions, definite integrals yield a number representing an area.In the example we evaluate \( \int_{-\pi / 2}^{\pi / 2} \sin \theta \, d\theta \), where the interval is from \(-\pi / 2\) to \(\pi / 2\). Here, we:
  • Find the antiderivative \( F(\theta) = -\cos \theta \).
  • Evaluate it at the upper and lower bounds of the interval (\( F(\pi/2) \) and \( F(-\pi/2) \)).
  • Subtract the two evaluations to get the area under the curve within these limits.
This process demonstrates the practical use of the Fundamental Theorem of Calculus, which tells us that the definite integral is calculated by the difference \( F(b) - F(a) \), where \( F \) is an antiderivative of the integrand function.
Trigonometric Integrals
Trigonometric integrals involve integration of trigonometric functions like \(\sin(\theta)\), \(\cos(\theta)\), etc. These integrals often appear in problems related to periodic phenomena, such as waves.To solve these integrals effectively:
  • Use known antiderivatives for trig functions, such as \(\int \sin(\theta) \, d\theta = -\cos(\theta) + C\).
  • Remember identities that might simplify integrals, like \(\sin^2(x) + \cos^2(x) = 1\).
In solving the definite integral \( \int_{-\pi / 2}^{\pi / 2} \sin \theta \, d\theta \), using the antiderivative \(-\cos(\theta)\) is crucial. Often, trigonometric identities can simplify the integration of more complex expressions, though they were not necessary in this straightforward example.

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Most popular questions from this chapter

Let \(y(t)\) denote the number of \(E .\) coli cells in a container of nutrient solution \(t\) minutes after the start of an experiment. Assume that \(y(t)\) is modeled by the initial-value problem $$\frac{d y}{d t}=(\ln 2) 2^{t / 20}, \quad y(0)=20$$ Use this model to estimate the number of \(E .\) coli cells in the container 2 hours after the start of the experiment.

(a) Find the limit $$\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{\sin (k \pi / n)}{n}$$ by evaluating an appropriate definite integral over the interval [0,1] (b) Check your answer to part (a) by evaluating the limit directly with a CAS.

Solve the initial-value problems. $$\frac{d y}{d x}=\sqrt{5 x+1}, y(3)=-2$$

A traffic engineer monitors the rate at which cars enter the main highway during the afternoon rush hour. From her data she estimates that between 4: 30 PM. and 5: 30 P.M. the rate \(R(t)\) at which cars enter the highway is given by the formula \(R(t)=100\left(1-0.0001 t^{2}\right)\) cars per minute, where \(t\) is the time (in minutes) since 4: 30 PM. (a) When does the peak traffic flow into the highway occur? (b) Estimate the number of cars that enter the highway during the rush hour.

Define the function \(f\) by $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{x}, & x \neq 0 \\ 0, & x=0 \end{array}\right. $$ It follows from Theorem \(5.5 .8(b)\) that \(f\) is not integrable on the interval \([0,1] .\) Prove this to be the case by applying Definition \(5.5 .1 .\) [Hint: Argue that no matter how small the mesh size is for a partition of [0,1] , there will always be a choice of \(x_{1}^{*}\) that will make the Riemann sum in Definition \(5.5 .1 \text { as large as we like. }]\)

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