Chapter 4: Problem 24
Evaluate the iterated integrals. $$ \int_{1}^{2} \int_{-u^{2}-1}^{-u}(8 u v) d v d u $$
Short Answer
Expert verified
The value of the iterated integral is \(-62.3333\) or \(-\frac{187}{3}\).
Step by step solution
01
Understand the Problem
We are given a double integral \( \int_{1}^{2} \int_{-u^{2}-1}^{-u}(8 u v) \, dv \, du \). We need to evaluate this integral by performing the inner integration first, followed by the outer integration.
02
Perform the Inner Integral
First, integrate the inner integral with respect to \( v \):\[\int_{-u^{2}-1}^{-u}(8 u v) \, dv \]The antiderivative of \(8uv\) with respect to \(v\) is \(4uv^2\). Evaluating this from \(-u^{2}-1\) to \(-u\) yields:\[4u(-u)^2 - 4u(-u^2-1)^2 \]Simplifying gives:\[4u(u^2) - 4u((u^2 + 1)^2)\]which simplifies to:\[4u^3 - 4u(u^4 + 2u^2 + 1)\]
03
Simplify the Inner Result
Simplify the expression:\[4u^3 - 4u(u^4 + 2u^2 + 1) = 4u^3 - 4u^5 - 8u^3 - 4u\]which simplifies further to:\[-4u^5 - 4u^3 - 4u\]
04
Perform the Outer Integral
Now, integrate the result \(-4u^5 - 4u^3 - 4u\) with respect to \(u\):\[\int_{1}^{2} (-4u^5 - 4u^3 - 4u) \, du\]The antiderivative is:\[-\frac{4}{6}u^6 - \frac{4}{4}u^4 - \frac{4}{2}u^2 = -\frac{2}{3}u^6 - u^4 - 2u^2\]Evaluate this from 1 to 2:\[\left(-\frac{2}{3}(2)^6 - (2)^4 - 2(2)^2\right) - \left(-\frac{2}{3}(1)^6 - (1)^4 - 2(1)^2\right)\]
05
Calculate and Simplify
Substitute and simplify the expression:\[\left(-\frac{2}{3} imes 64 - 16 - 8\right) - \left(-\frac{2}{3} - 1 - 2\right)\]This simplifies to:\[\left(-\frac{128}{3} - 24\right) - \left(-\frac{2}{3} - 3\right)\]Combine like terms:\[-\frac{128}{3} - \frac{72}{3} + \frac{2}{3} + 3\]This results in:\[-\frac{198}{3} + \frac{2}{3} + 3 = -66 + 0.6667 + 3\]Which finally gives:\(-62.3333\) or \(-\frac{187}{3}\).
06
Present the Final Answer
The final answer to the double integral is -62.3333, or in fractional form, \(-\frac{187}{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Iterated Integrals
When dealing with double integrals, particularly iterated integrals, the process involves integrating a function over a two-dimensional region. The key is to perform the integration one variable at a time.
- Start with the inner integral, integrating with respect to one variable while treating the other as a constant.
- After integrating the inner variable over its respective bounds, move on to the outer integral, addressing the remaining variable.
Antiderivative
Finding an antiderivative is a fundamental step in calculus, particularly within integral calculations. An antiderivative is a function that, when differentiated, yields the original function you started with.
- For the inner integration step \( \int_{-u^{2}-1}^{-u}(8 u v) \ dv \), we find the antiderivative of \( 8uv \) with respect to \( v \).
- This antiderivative results in \( 4uv^2 \).
Integration Limits
Integration limits define the range over which the integrals are evaluated. They are crucial in accurately capturing the area or volume under study within the function.
- The inner integral limits, \(-u^{2}-1\) to \(-u\), govern the variable \(v\). These boundaries may vary with respect to the other variable, \(u\), indicating that the integration range changes as you move along the second variable.
- For the outer integral, the fixed bounds from 1 to 2 describe where \( u \) should be evaluated.