Problem 20 Let \(F(x)=\int_{0}^{x} \sin \le... [FREE SOLUTION]

91影视

Calculus

Deborah Hughes-Hallett, Andrew M. Gleason, William G. McCallum

$Math Studyset 91影视 Explanations$ Math

7 Edition

Chapter 6: Problem 20

Let $F(x)=\int_{0}^{x} \sin \left(t^{2}\right) d t$. Using a graph of $F^{\prime}(x),$ decide where $F(x)$ is increasing and where $F(x)$ is decreasing for $0 \leq x \leq 2.5$

Short Answer

Expert verified

Increasing on $[0, 1.77)$, decreasing on $(1.77, 2.5]$.

Step by step solution

Understand the Relationship

We need to find where the function $F(x)$ is increasing or decreasing. From calculus, we know that a function is increasing where its derivative is positive, and decreasing where its derivative is negative. So, we first need to compute $F'(x)$.

Derive the Function

Given $F(x)=\int_{0}^{x} \sin(t^2) \, dt$, by the Fundamental Theorem of Calculus, the derivative of $F(x)$ is $F'(x)=\sin(x^2)$.

Analyze $F'(x)$

Now, we need to determine where $F'(x)=\sin(x^2)$ is positive and where it is negative in the interval $0 \leq x \leq 2.5$. Since $F'(x)$ is a sine function applied to $x^2$, it will oscillate between -1 and 1.

Determine Sign of $F'(x)$

To decide on the intervals over $0 \leq x \leq 2.5$, identify where $\sin(x^2) = 0$, where it is positive, and where it is negative. $\sin(x^2)$ is zero at $x^2 = n\pi$ (for integer $n$), positive between $(n-1)\pi < x^2 < n\pi$ for odd $n$, and negative for even $n$. This gives us intervals based on approximations of where $x^2=\pi, 2\pi$ etc.

Approximate Key Points in $0 \leq x \leq 2.5$

Calculate approximate values: $x^2=\pi$ gives $x\approx1.77$, and $x^2=2\pi$ gives $x\approx2.51$. So, $x^2<\pi$ corresponds roughly to $x<1.77$, where $\sin(x^2)$ is positive and $F(x)$ is increasing; $1.77<x<2.51$ is where it is negative, so $F(x)$ is decreasing.

Conclusion from Analysis

Therefore, $F(x)$ is increasing on $[0, 1.77)$ and decreasing on $(1.77, 2.5]$.

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Key Concepts

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

Derivative of an Integral

Understanding the concept of the derivative of an integral requires a dive into the fundamental principles of calculus. The Fundamental Theorem of Calculus connects differentiation and integration, two core operations in calculus. When we have a function, like in our exercise, defined by an integral from a constant to a variable, the theorem states that differentiating this integral brings us back to the original integrand.

Let's see this in action with our function: $ F(x)=\int_{0}^{x} \sin(t^2) \, dt $. Applying the theorem gives us $ F'(x) = \sin(x^2) $. This tells us the rate at which $ F(x) $ changes with $ x $.

To recap,

The derivative of an integral like $ F(x) $ is simply the integrand evaluated at the upper limit.
For $ F(x) = \int_{a}^{x} f(t) \, dt $, we get $ F'(x) = f(x) $.

Increasing and Decreasing Functions

The behavior of a function鈥攚hether it's going upwards or downwards as you move across the x-axis鈥攊s determined by its derivative. To decide where $ F(x) $ is increasing or decreasing, we examine the sign of $ F'(x) $. If $ F'(x) > 0 $, the function is increasing; if $ F'(x) < 0 $, the function is decreasing.

For our specific function, $ F'(x) = \sin(x^2) $, this means examining the sign of the sine of $ x^2 $. Since sine values oscillate between -1 and 1:

$ \, F(x) $ is increasing whenever $ \sin(x^2) > 0 $.
$ \, F(x) $ is decreasing whenever $ \sin(x^2) < 0 $.

This analysis is crucial in various applications, like determining growth or shrinkage of a business modeled by the function.

Sine Function Behavior

The sine function, $ \sin(x) $, is periodic and oscillates between -1 and 1. Understanding this behavior helps make sense of where $ F'(x) = \sin(x^2) $ becomes positive or negative. The sine function's zero-crossing points are at integer multiples of $ \pi $: here, at $ n\pi $ for integer $ n $. These define boundaries for the intervals of increase or decrease in our function.

When $ \sin $ applied to $ x^2 $ (like $ \sin(x^2) $) returns zero for $ x^2 = n\pi $, $ x $ is approximately where it transitions between positive and negative:

Between values where $ (n-1)\pi < x^2 < n\pi $ and $ n $ is odd, $ \sin(x^2) $ is positive.
In regions $ (n-1)\pi < x^2 < n\pi $ with even $ n $, it is negative.

These characteristics are key in analyzing periodic behaviors in various contexts, including physics and engineering.

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91影视

Let \(F(x)=\int_{0}^{x} \sin \left(t^{2}\right) d t\). Using a graph of \(F^{\prime}(x),\) decide where \(F(x)\) is increasing and where \(F(x)\) is decreasing for \(0 \leq x \leq 2.5\)

Short Answer

Step by step solution

Understand the Relationship

Derive the Function

Analyze \(F'(x)\)

Determine Sign of \(F'(x)\)

Approximate Key Points in \(0 \leq x \leq 2.5\)

Conclusion from Analysis

Key Concepts

Derivative of an Integral

Increasing and Decreasing Functions

Sine Function Behavior

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