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Chapter 10: Problem 2

Describe the region of integration and evaluate. (Show the details.) $$\int_{0}^{1} \int_{x}^{2 x}(x+y)^{2} d y \ d x$$

Short Answer

Expert verified
The region is bounded by \(y = x\), \(y = 2x\), \(x = 0\), \(x = 1\). The integral evaluates to \(\frac{19}{12}\).

Step by step solution

01

Identify Limits of Integration

First, examine the limits of integration given in the iterated integral. The outer integral is with respect to \(x\) and ranges from \(0\) to \(1\). The inner integral is with respect to \(y\) and ranges from \(x\) to \(2x\). This means for each \(x\), \(y\) varies between \(x\) and \(2x\).
02

Describe the Region of Integration

The region of integration can be described in the \(xy\)-plane. Here, \(x\) varies from \(0\) to \(1\), and for a fixed \(x\), \(y\) varies from \(x\) to \(2x\). This forms a region that is bounded by the lines \(y=x\), \(y=2x\), \(x=0\), and \(x=1\).
03

Evaluate Inner Integral

The inner integral is \(\int_{x}^{2x} (x+y)^{2} \, dy\). To solve, we find the antiderivative of \((x+y)^2\) with respect to \(y\). The antiderivative is \(\frac{1}{3}(x+y)^3\). Evaluating this from \(y=x\) to \(y=2x\), we get: \[ \left. \frac{1}{3}(x+y)^3 \right|_{x}^{2x} = \frac{1}{3}((x+2x)^3 - (x+x)^3) = \frac{1}{3}((3x)^3 - (2x)^3) = \frac{1}{3}(27x^3 - 8x^3) = \frac{19}{3}x^3 \].
04

Evaluate Outer Integral

With the result from the inner integral, we evaluate the outer integral: \(\int_{0}^{1} \frac{19}{3}x^3 \, dx\). Find the antiderivative of \(\frac{19}{3}x^3\), which is \(\frac{19}{12}x^4\). Evaluate this from \(0\) to \(1\): \[ \left. \frac{19}{12}x^4 \right|_{0}^{1} = \frac{19}{12}(1)^4 - \frac{19}{12}(0)^4 = \frac{19}{12} \].

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Key Concepts

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

Region of Integration
Understanding the region of integration is a fundamental concept in solving double integrals. In the given exercise, our region is determined by the limits of integration. We identify these limits from the original integral:
  • The outer integral ranges from \(x = 0\) to \(x = 1\).
  • The inner integral ranges from \(y = x\) to \(y = 2x\).
This essentially describes a region on the \(xy\)-plane. Visualizing, \(x\) varies between \(0\) and \(1\), and for each fixed \(x\), \(y\) stretches between \(x\) and \(2x\). When graphed, this looks like a triangular region bounded by:
  • The line \(y = x\)
  • The line \(y = 2x\)
  • The vertical line \(x = 0\)
  • The vertical line \(x = 1\)
Understanding this bounded area is vital as it provides the defined space where we're integrating our function \((x + y)^2\).
Iterated Integral
An iterated integral is a way to evaluate a double integral by breaking it down into two single integrals. The process involves integrating one variable at a time, hence the term "iterated." In our exercise, the integral is split into:
  • The inner integral, \(\int_{x}^{2x} (x+y)^{2} \, dy\), which evaluates the expression with respect to \(y\) while considering \(x\) as a constant.
  • The outer integral, \(\int_{0}^{1} \), which operates on the result of the first integral with respect to \(x\).
This step-by-step calculation helps simplify the problem by focusing on one dimension at a time. The key here is to solve the inner integral, substitute the limits, and then use this result in the outer integral to get the final solution.
Antiderivative
In calculus, finding an antiderivative is critical in solving integrals. It involves determining a function whose derivative matches the given expression. In our exercise:
  • For the inner integral \( \int_{x}^{2x} (x+y)^{2} \, dy \), we needed to find the antiderivative of \((x+y)^2\) with respect to \(y\). This was determined as \(\frac{1}{3}(x+y)^3\).
  • For the outer integral, \(\int_{0}^{1} \frac{19}{3}x^3 \, dx\), we identified \(\frac{19}{12}x^4\) as the antiderivative.
Evaluating these antiderivatives over their respective limits gives the solution to the original problem. Mastering finding antiderivatives is essential for solving integrals quickly and correctly.

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