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Evaluate the iterated integrals. \(\int_{0}^{1} \int_{2 x}^{3 x}\left(x+y^{2}\right) d y d x\)

Short Answer

Expert verified
The value of the iterated integral is \(\frac{23}{12}\).

Step by step solution

01

Understand the Order of Integration

The problem is asking for evaluation of an iterated integral. The integral we have is \(\int_{0}^{1} \int_{2x}^{3x} (x+y^2) \, dy \, dx\). The outer integral is in terms of \(x\), from 0 to 1, and the inner integral is in terms of \(y\), from \(2x\) to \(3x\). We need to first integrate with respect to \(y\), and then \(x\).
02

Perform the Inner Integration

Start with the inner integral \(\int_{2x}^{3x} (x+y^2) \, dy\). Treat \(x\) as a constant and integrate \((x+y^2)\) with respect to \(y\). The antiderivative of \(x\) with respect to \(y\) is \(xy\), and the antiderivative of \(y^2\) with respect to \(y\) is \(\frac{y^3}{3}\). Evaluate this from \(2x\) to \(3x\):\[ \left[ xy + \frac{y^3}{3} \right]_{y=2x}^{y=3x} \]
03

Substitute and Evaluate the Inner Integral

Substitute the bounds into the antiderivative:\[ \left[ x(3x) + \frac{(3x)^3}{3} \right] - \left[ x(2x) + \frac{(2x)^3}{3} \right] \]This simplifies to:\[ 3x^2 + \frac{27x^3}{3} - (2x^2 + \frac{8x^3}{3}) \]Which simplifies to:\[ 3x^2 + 9x^3 - 2x^2 - \frac{8x^3}{3} = x^2 + \frac{27x^3}{3} - \frac{8x^3}{3} \]\[ = x^2 + \frac{19x^3}{3} \]
04

Perform the Outer Integration

Now, integrate \(x^2 + \frac{19x^3}{3}\) with respect to \(x\) from 0 to 1:\[ \int_{0}^{1} \left( x^2 + \frac{19x^3}{3} \right) \, dx \]The antiderivative is:\[ \frac{x^3}{3} + \frac{19x^4}{12} \]Evaluate this from 0 to 1:\[ \left[ \frac{1^3}{3} + \frac{19 \cdot 1^4}{12} \right] - \left[ \frac{0^3}{3} + \frac{19 \cdot 0^4}{12} \right] \]This simplifies to:\[ \frac{1}{3} + \frac{19}{12} \]
05

Simplify the Result

Simplify the sum \(\frac{1}{3} + \frac{19}{12}\). First, convert \(\frac{1}{3}\) to have a common denominator with \(\frac{19}{12}\):\[ \frac{1}{3} = \frac{4}{12} \]So, adding them gives:\[ \frac{4}{12} + \frac{19}{12} = \frac{23}{12} \]
06

Final Step: State the Final Result

Hence, the value of the iterated integral is \(\frac{23}{12}\).

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Key Concepts

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

Understanding Integration in Iterated Integrals
When tackling iterated integrals, integration is a crucial tool that allows us to handle multi-variable functions. Iterated integrals involve integrating a function over a two-dimensional region or more. This means we perform integration multiple times, each time concerning a different variable. Typically, for double integrals, this involves integrating first with respect to one variable while treating the other variable(s) as constants, followed by integrating with respect to the next variable.

In our given problem, \[\int_{0}^{1} \int_{2x}^{3x} (x+y^2) \, dy \, dx\], the process of integration starts with the inner integral \[\int_{2x}^{3x} (x+y^2) \, dy\]. Here, integration is performed with respect to the variable \( y \) first, while treating \( x \) as a constant. This is followed by integrating the result with respect to \( x \) over the given limits from 0 to 1.
Integration here is about finding the sum of an infinite number of infinitesimally small areas, which contributes to the final accumulated value over the specified range. Essentially, integration allows us to determine the total accumulation, which, in the case of iterated integrals, provides the area under the surface defined by the function over the specified domain.
The Role of Antiderivatives in Integration
To successfully integrate a function, finding the antiderivative is a key step. An antiderivative of a function \( f(x) \) is another function \( F(x) \) such that the derivative of \( F(x) \) is \( f(x) \). In integrals, especially definite integrals, we use antiderivatives to calculate the area under a curve over a specified range.

For instance, the inner integral, \[\int_{2x}^{3x} (x+y^2) \, dy\] requires finding the antiderivatives of both \( x \) and \( y^2 \) with respect to \( y \). Here, the antiderivative of \( x \) with respect to \( y \) is \( xy \), as \( x \) is treated as a constant. On the other hand, the antiderivative of \( y^2 \) with respect to \( y \) is \( \frac{y^3}{3} \). Finding these antiderivatives allows us to evaluate the function at the given limits of integration and proceed with the problem.
Using antiderivatives simplifies the process of integration by converting a differentiation problem into one involving evaluation and subtraction between bounds. This transition is vital because it translates complex integrals into manageable computations.
Exploring Limits of Integration
Limits of integration define the boundaries over which the integration is performed. In iterated integrals, it's critical to correctly understand these limits for each variable, as they determine the region of integration.

For the given problem, the iterated integral\[\int_{0}^{1} \int_{2x}^{3x} (x+y^2) \, dy \, dx\]involves limits associated with both variables, \( x \) and \( y \). The inner integral, which involves \( y \), has limits from \( 2x \) to \( 3x \). These limits indicate that for each value of \( x \), \( y \) starts at \( 2x \) and ends at \( 3x \).
The outer integral, concerning \( x \), has limits from 0 to 1, suggesting that \( x \) varies between these values across the entire domain specified by the problem. To solve such integrals, it's essential to substitute these boundary values correctly into the antiderivatives to obtain the definite integral's result.
Incorrectly managing the limits during integration can lead to erroneous results, as they precisely determine the real geometric region over which integration occurs. Understanding and applying these limits correctly ensures the accurate computation of the iterated integral.

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Most popular questions from this chapter

a. Find the volume of the solid \(S_{1}\) bounded by the cylinder \(x^{2}+y^{2}=1\) and the planes \(z=0\) and \(z=1\). b. Find the volume of the solid \(S_{2}\) outside the double cone \(z^{2}=x^{2}+y^{2}\), inside the cylinder \(x^{2}+y^{2}=1\), and above the plane \(z=0\) c. Find the volume of the solid inside the cone \(z^{2}=x^{2}+y^{2}\) and below the plane \(z=1\) by subtracting the volumes of the solids \(S_{1}\) and \(S_{2}\).

Show that the volume of a regular right hexagonal prism of edge length \(\boldsymbol{a}\) is \(\frac{3 a^{3} \sqrt{3}}{2}\) by using triple integrals.

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For the following two exercises, consider a spherical ring, which is a sphere with a cylindrical hole cut so that the axis of the cylinder passes through the center of the sphere (see the following figure). Let \(f\) be a continuous function that can be expressed in polar coordinates as a function of \(\theta\) only; that is, \(f(x, y)=h(\theta)\), where \((x, y) \in D=\left\\{(r, \theta) \mid R_{1} \leq r \leq R_{2}, \theta_{1} \leq \theta \leq \theta_{2}\right\\}\), with \(0 \leq R_{1}

For the following two exercises, consider a spherical ring, which is a sphere with a cylindrical hole cut so that the axis of the cylinder passes through the center of the sphere (see the following figure). If the sphere has radius 4 and the cylinder has radius 2, find the volume of the spherical ring.

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