/*! 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 54 Evaluate the integral by first r... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integral by first reversing the order of integration. $$ \int_{0}^{2} \int_{y / 2}^{1} \cos \left(x^{2}\right) d x d y $$

Short Answer

Expert verified
The integral evaluates to \( \sin(1) \).

Step by step solution

01

Understand the Region of Integration

The given integral \( \int_{0}^{2} \int_{y/2}^{1} \cos(x^2) \, dx \, dy \) involves integrating first with respect to \(x\)in the limits \(\frac{y}{2}\) to \(1\), and then with respect to \(y\) from \(0\) to \(2\). This describes a region where \(y\) is between \(0\) and \(2\), and for each \(y\), \(x\) varies from \(\frac{y}{2}\) to \(1\).
02

Sketch the Region

To better visualize the region of integration, sketch the lines \(x = \frac{y}{2}\) and \(x = 1\) for \(0 \leq y \leq 2\). The region is bounded by the line \(x = \frac{y}{2}\), the line \(x = 1\), the x-axis, and the line \(y = 2\).
03

Reversing the Order of Integration

The integral needs to be rewritten by reversing the order of integration. In the new order, integrate with respect to \(y\) first and \(x\) second. From the sketch, we see \(y\) ranges from \(0\) to \(2x\) if \(x\) goes from \(0\) to \(1\). Thus, the integral becomes \( \int_{0}^{1} \int_{0}^{2x} \cos(x^2) \, dy \, dx \).
04

Evaluate the Inner Integral

Evaluate \( \int_{0}^{2x} \cos(x^2) \, dy \), which simplifies to \([y \cos(x^2)]_{0}^{2x} = 2x \cos(x^2)\).
05

Evaluate the Outer Integral

Now, evaluate the outer integral \( \int_{0}^{1} 2x \cos(x^2) \, dx \). Use substitution by letting \( u = x^2 \), which gives \( du = 2x \, dx \). Then the integral becomes \( \int u \, du \).
06

Compute the Final Integral

Integrate \( \int \cos(u) \, du \), which is \( \sin(u) + C \). Change back to the variable \(x\), resulting in \( \sin(x^2) \). Evaluate this definite integral from \(0\) to \(1\) to get \( \sin(1) - 0 = \sin(1) \).
07

Final Answer

The final value of the integral is \( \sin(1) \).

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Key Concepts

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

Order of Integration
When dealing with double integrals, like the one in our problem, the order of integration can be reversed to simplify the computation. This means swapping the order of integration between the two variables. In our exercise, the original integral is:\[\int_{0}^{2} \int_{y / 2}^{1} \cos \left(x^{2}\right) d x d y\]This setup implies integrating first in the direction of \(x\) between \(y/2\) and 1, then in the direction of \(y\) from 0 to 2. By drawing the region of integration, we can see it more clearly and help decide how to flip the limits.Steps to Reverse the Order:
  • Identify the limits of each variable to understand the region.
  • Draw the boundaries on a grid (often helpful on paper).
  • Swap the limits: Identify how the limits swap when reversing.
This switch often makes it easier to evaluate the integral, allowing us to tackle it from a more convenient perspective. The reversed order simplifies the problem and sometimes makes direct evaluation possible, which is the case here.
Definite Integrals
A definite integral computes the area under a curve within given limits. In integral calculus, it's important to understand that these bounds define the region over which we integrate. In our exercise, we're dealing with:\[\int_{0}^{2} \int_{y / 2}^{1} \cos \left(x^{2}\right) d x d y\]This tells us to sum up all values from \(y = 0\) to \(y = 2\) and \(x = y/2\) to \(x = 1\). Once reversed, we are integrating:\[\int_{0}^{1} \int_{0}^{2x} \cos(x^2) \, dy \, dx\]With these new limits, the area differs: \(y\) from 0 to \(2x\), and \(x\) from 0 to 1.Importance of Limits:
  • The limits for \(x\) and \(y\) set the specific boundaries to calculate within.
  • Changing limits influences the region of integration, impacting the solution.
  • After reversing, the new limits simplify the solution by reducing complexity.
The concept behind definite integrals remains crucial since it governs how the area is calculated between the selected bounds.
Trigonometric Integrals
Trigonometric integrals involve functions like sine and cosine. They are common in calculus problems, and dealing with them requires understanding both their mathematical behavior and how they interact within the integrals.In our particular example, the expression \(\cos(x^2)\) makes it a trigonometric integral:Using Substitution:
  • Start by identifying a substitution to simplify integration. Here, with \(u = x^2\) yielding \(du = 2x \, dx\), the expression becomes more manageable.
  • This substitution changes our task to integrating \(\int \cos(u) \, du\), a more straightforward function.
  • Integrate \(\cos(u)\), which results in \(\sin(u) + C\). Switch back to \(x\) to express \(\sin(x^2)\).
The process of substitution aids in transforming complex expressions into simpler ones, making evaluation much simpler. Understanding these steps helps in handling trigonometric integrals effectively, leading to the accurate computation of areas as in definite integrals.

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Most popular questions from this chapter

Suppose that the boundary curves of a region \(R\) in the \(x y\) -plane can be described as level curves of various functions. Discuss how this information can be used to choose an appropriate change of variables for a double integral over \(R\). Illustrate your discussion with an example.

The average value or mean value of a continuous function \(f(x, y, z)\) over a solid \(G\) is defined as $$f_{\text {ave }}=\frac{1}{V(G)} \iiint_{G} f(x, y, z) d V$$ where \(V(G)\) is the volume of the solid \(G\) (compare to the definition preceding Exercise 61 of Section \(14.2\) ). Use this definition in these exercises. Find the average value of \(f(x, y, z)=x y z\) over the spherical region \(x^{2}+y^{2}+z^{2} \leq 1\)

The tendency of a lamina to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If a lamina occupies a region \(R\) of the \(x y\) -plane, and if its density function \(\delta(x, y)\) is continuous on \(R\), then the moments of inertia about the \(x\) -axis, the \(y\) -axis, and the \(z\) -axis are denoted by \(I_{x}, I_{y}\), and \(I_{z}\), respectively, and are defined by $$\begin{aligned}&I_{x}=\iint_{R} y^{2} \delta(x, y) d A, \quad I_{y}=\iint_{R} x^{2} \delta(x, y) d A \\\&I_{z}=\iint_{R}\left(x^{2}+y^{2}\right) \delta(x, y) d A \end{aligned}$$ Use these definitions. Consider the circular lamina that occupies the region described by the inequalities \(0 \leq x^{2}+y^{2} \leq a^{2}\). Assuming that the lamina has constant density \(\delta\), show that $$I_{x}=I_{y}=\frac{\delta \pi a^{4}}{4}, \quad I_{z}=\frac{\delta \pi a^{4}}{2}$$

Let \(G\) be the rectangular box defined by the inequalities $$\begin{aligned}&a \leq x \leq b, c \leq y \leq d, k \leq z \leq l . \text { Show that } \\ &\iiint_{G} f(x) g(y) h(z) d V \\ &\quad=\left[\int_{a}^{b} f(x) d x\right]\left[\int_{c}^{d} g(y) d y\right]\left[\int_{k}^{l} h(z) d z\right]\end{aligned}$$

Evaluate the iterated integral. \(\int_{0}^{3} \int_{0}^{\sqrt{9-z^{2}}} \int_{0}^{x} x y d y d x d z\)

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