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Chapter 5: Problem 50

Sketch the curve and find the total area between the curve and the given interval on the \(\bar{x}\) -axis. \(y=\sin x ;[0,3 \pi / 2]\)

Short Answer

Expert verified
The total area between the curve and \( x \)-axis is 3 units squared.

Step by step solution

01

Understand the Function and Interval

We are given the function \( y = \sin x \) and the interval \( [0, \frac{3\pi}{2}] \). The goal is to sketch the curve of \( y = \sin x \) over this interval and calculate the total area between the curve and the \( x \)-axis.
02

Sketch the Sine Curve

Sketch the curve \( y = \sin x \) starting from \( x = 0 \) to \( x = \frac{3\pi}{2} \). The sine curve starts at 0, reaches 1 at \( \frac{\pi}{2} \), returns to 0 at \( \pi \), and reaches -1 at \( \frac{3\pi}{2} \).
03

Identify the Partitions and Areas

Identify different partitions of the interval based on the zeros and peaks of the sine curve. Between \( 0 \) and \( \pi \), the curve is positive. Between \( \pi \) and \( \frac{3\pi}{2} \), the curve is negative.
04

Calculate the Area Under Each Partition

Calculate the area under the curve from \( 0 \) to \( \pi \). This is the integral \( \int_0^\pi \sin x \, dx \), which gives 2 units squared. Then calculate the area from \( \pi \) to \( \frac{3\pi}{2} \), \( \int_\pi^{\frac{3\pi}{2}} \sin x \, dx \), which gives \( -1 \) unit squared.
05

Find the Total Area

Sum the absolute values of the calculated areas: \( |2| + |-1| = 2 + 1 = 3\). The total area between the curve and the \( x \)-axis over the interval \( [0, \frac{3\pi}{2}] \) is 3 units squared.

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Key Concepts

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

Sine Function
The sine function, denoted as \( y = \sin x \), is a fundamental trigonometric function that describes how an angle in a triangle relates to its opposite side and hypotenuse. It originates from the unit circle, where for any angle \( x \), the sine value corresponds to the y-coordinate of a point on the circle's circumference.
  • The graph of \( y = \sin x \) is periodic with a period of \( 2\pi \), which means it repeats every \( 2\pi \) radians.
  • The range of the sine function is \([-1, 1]\), indicating it oscillates between these two values.
  • Key points on the sine curve include:
    • Starts at 0 when \( x = 0 \)
    • Peaks at 1 when \( x = \frac{\pi}{2} \)
    • Returns to 0 at \( x = \pi \)
    • Has a minimum at -1 at \( x = \frac{3\pi}{2} \)
The smooth, wave-like appearance of the sine curve makes it intuitive for understanding oscillatory phenomena, such as sound waves or light waves.
Definite Integrals
In calculus, definite integrals are a method to calculate the area under a curve over a specific closed interval. The concept is essential for finding the actual accumulated quantity, like total area or volume.
  • The notation \( \int_a^b f(x) \, dx \) is used, where \( a \) and \( b \) define the interval's endpoints.
  • Geometrically, this integral provides the signed area between the curve \( y = f(x) \) and the x-axis from \( x = a \) to \( x = b \).
  • When a curve dips below the x-axis, the integral becomes negative, reflecting the area lying below the axis.
In solving the exercise, definite integrals help calculate the area under the sine curve from \( 0 \) to \( \pi \) and then from \( \pi \) to \( \frac{3\pi}{2} \). By treating these areas separately and considering their signs, understanding the complete area is achieved.
Area Under Curve
Finding the area under a curve is a common problem in calculus that applies integrals to calculate the region enclosed by the curve and the x-axis.
  • For positive areas, the curve lies above the axis, contributing positively to the overall sum.
  • For negative areas, the curve dips below, leading to negative contributions.
To find the total area from an interval that includes both positive and negative segments, sum the absolute values of each partitioned area's integral. In the exercise, the area from \( 0 \) to \( \pi \) is entirely above the x-axis and positive, while from \( \pi \) to \( \frac{3\pi}{2} \) it is below and negative. This requires determining the absolute values to combine them effectively and find the total area, giving \( 3 \) units squared.

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Most popular questions from this chapter

$$ \int \frac{d x}{\sqrt{x} e^{(2 \sqrt{x})}} $$

Consider the sum \(\sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=\left[\begin{array}{rl} & {\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right]} \\ & +\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right]\end{array}\right.\) \(=5^{3}-1^{3}=124\) For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$ \sum_{k=1}^{100}\left(2^{k+1}-2^{k}\right) $$

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