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Chapter 5: Problem 75

Show that the value of \(\int_{0}^{1} \sin \left(x^{2}\right) d x\) cannot possibly be 2 .

Short Answer

Expert verified
The integral is less than or equal to 1, so it cannot be 2.

Step by step solution

01

Understand the Problem

We need to evaluate the integral \( \int_{0}^{1} \sin(x^2) \, dx \) and show that its value cannot be 2. This means finding an estimate of the integral's value directly or indirectly that can demonstrate this claim.
02

Estimate the Integral Through an Inequality

Notice that for all \(x\) in \([0, 1]\), \(0 \leq \sin(x^2) \leq 1 \). Therefore, the integral \( \int_{0}^{1} \sin(x^2) \, dx \) is also bounded as follows:\[0 \leq \int_{0}^{1} \sin(x^2) \, dx \leq \int_{0}^{1} 1 \, dx.\]Evaluating the rightmost integral, we get:\[\int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 - 0 = 1.\]Thus, \( \int_{0}^{1} \sin(x^2) \, dx \leq 1 \).
03

Conclude That the Integral Cannot Be 2

From Step 2, we showed that \( \int_{0}^{1} \sin(x^2) \, dx \leq 1 \). Since the upper bound of the integral is 1, it is impossible for the integral to evaluate to 2 as 2 exceeds this bound.

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Key Concepts

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

Definite Integrals
Definite integrals give us a way to calculate the area under a curve within a specific interval. For the integral \( \int_{a}^{b} f(x) \, dx \), we are looking at the function \( f(x) \) between \( x = a \) and \( x = b \). This means we are considering the summation of infinite tiny slivers that lie underneath the curve from \( a \) to \( b \), and adding them up to get the total area.
  • The limits \( a \) and \( b \) are known as the lower and upper bounds, respectively.
  • The value of a definite integral can give insights into the trends of the function and whether it is mostly positive, negative, or zero over that interval.
When solved, definite integrals not only represent areas but can also provide other physical quantities such as displacement, total growth, and accumulated change at certain points.
Upper and Lower Bounds
In the context of integrals, upper and lower bounds are essential in estimating the value of an integral. By understanding and exploring these bounds, we can establish that certain values are or are not possible. In the given problem, we're using these bounds to disprove the possibility that the integral is equal to 2.
  • The lower bound \( 0 \) was deduced since \( \sin(x^2) \geq 0 \) for \( x \) within the interval \([0, 1]\).
  • The upper bound of \( \int_{0}^{1} 1 \, dx \) is calculated to be 1, as \( \sin(x^2) \leq 1 \) for all \( x \) in this range.
Therefore, by leveraging the inequality \( 0 \leq \int_{0}^{1} \sin(x^2) \, dx \leq 1 \), it's clear that the integral cannot be 2, because 2 exceeds this calculated upper limit of 1.
Trigonometric Functions
Trigonometric functions, like \( \sin(x) \), are mathematical functions that relate the angles of a triangle to the lengths of its sides in a right triangle. The function \( \sin(x^2) \) used in the integral predicts a wave-like function as it operates over the interval \([0,1]\).
  • The sine function oscillates between -1 and 1. However, in the interval \([0, 1]\), when considering \( x^2 \) and its effect on \( \sin \), the values will stay between 0 and 1.
  • These wave-like properties often make trigonometric function integrals more challenging to evaluate precisely.
Understanding the behavior of trigonometric functions on specific intervals helps not only in evaluating them but also in ensuring accurate estimations of their integrals.

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Most popular questions from this chapter

If your CAS can draw rectangles associated with Riemann sums, use it to draw rectangles associated with Riemann sums that converge to the integrals in Exercises \(89-94 .\) Use \(n=4,10,20,\) and 50 subintervals of equal length in each case. $$ \int_{-\pi}^{\pi} \cos x d x=0 $$

In Exercises \(95-98,\) use a CAS to perform the following steps: a. Plot the functions over the given interval. b. Partition the interval into \(n=100,200,\) and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation \(f(x)=(\) average value) for \(x\) using the average value calculated in part (c) for the \(n=1000\) partitioning. $$ f(x)=\sin ^{2} x \quad \text { on } \quad[0, \pi] $$

In Exercises \(95-98,\) use a CAS to perform the following steps: a. Plot the functions over the given interval. b. Partition the interval into \(n=100,200,\) and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation \(f(x)=(\) average value) for \(x\) using the average value calculated in part (c) for the \(n=1000\) partitioning. $$ f(x)=x \sin \frac{1}{x} \quad \text { on } \quad\left[\frac{\pi}{4}, \pi\right] $$

Evaluate the integrals in Exercises \(17-50\) $$ \int 3 x^{5} \sqrt{x^{3}+1} d x $$

Find the areas of the regions enclosed by the lines and curves. $$ y=\sec ^{2}(\pi x / 3) \quad \text { and } \quad y=x^{1 / 3}, \quad-1 \leq x \leq 1 $$
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