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Chapter 7: Problem 7

Evaluate. \(\int_{0}^{1} \int_{x^{2}}^{x}(x+y) d y d x\)

Short Answer

Expert verified
The value of the integral is \(\frac{3}{20}\).

Step by step solution

01

Understand the limits of integration

The given integral is \(\int_{0}^{1} \int_{x^{2}}^{x}(x+y) dy dx\). The outer integral runs from 0 to 1, and for each value of x, the inner integral runs from \(y = x^{2}\) to \(y = x\).
02

Integrate with respect to y

First, integrate the function \( x + y \) with respect to y. \(\int_{x^{2}}^{x} (x + y) dy\). The integral of \(x + y\) with respect to y is \[xy + \frac{y^{2}}{2}\].
03

Evaluate the inner integral

Now evaluate \[xy + \frac{y^{2}}{2}\] from \(y = x^{2}\) to \(y = x\): \[\left. \left( xy + \frac{y^{2}}{2} \right) \right|_{y = x^{2}}^{y = x} = \left( x \cdot x + \frac{x^{2}}{2} \right) - \left( x \cdot x^{2} + \frac{(x^{2})^{2}}{2} \right)\]. After simplification, it becomes: \(\left( x^{2} + \frac{x^{2}}{2} \right) - \left( x^{3} + \frac{x^{4}}{2} \right) = \frac{3x^{2}}{2} - x^{3} - \frac{x^{4}}{2}\).
04

Integrate with respect to x

Now integrate \[\frac{3x^{2}}{2} - x^{3} - \frac{x^{4}}{2}\] with respect to x from 0 to 1: \(\int_{0}^{1} \left(\frac{3x^{2}}{2} - x^{3} - \frac{x^{4}}{2}\right) dx\).
05

Find the antiderivative

The antiderivative of \(\frac{3x^{2}}{2}\) is \[\frac{3x^{3}}{6} = \frac{x^{3}}{2}\].The antiderivative of \(-x^{3}\) is \[\frac{-x^{4}}{4}\].The antiderivative of \(-\frac{x^{4}}{2}\) is \[\frac{-x^{5}}{10}\].
06

Evaluate the antiderivatives

Evaluate the antiderivative from 0 to 1: \(\left[ \frac{x^{3}}{2} - \frac{x^{4}}{4} - \frac{x^{5}}{10} \right]_{0}^{1} = \left[ \frac{1}{2} - \frac{1}{4} - \frac{1}{10} \right] - \left[ 0 \right] = \frac{1}{2} - \frac{1}{4} - \frac{1}{10} = \frac{5}{10} - \frac{2.5}{10} - \frac{1}{10} = \frac{1.5}{10} = \frac{3}{20}\).

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Key Concepts

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

Calculus
Calculus is the branch of mathematics that studies change. It deals with derivatives and integrals. This helps us understand concepts like motion, area under curves, and growth rates. In our example, we use a double integral to find the volume under a surface. Double integrals are a way to add up values over a two-dimensional area. This can be a bit complex but breaks down into manageable steps.
Integration
Integration is the process of finding the area, volume, or total value by summing small pieces. With \(\text{double integrals}\), we're adding up smaller parts over a region. For the given example, we first integrate the function \(x + y\) with respect to \(\text{y}\), and then with respect to \(\text{x}\). This breaks down an initially complex problem into simpler parts. This process often involves understanding integration limits, which tell us the boundaries for the variables.
Antiderivative
The antiderivative is the reverse process of finding a derivative. When we integrate, we often find the antiderivative of a function. For example, the antiderivative of \(x^{2}\) is \(\frac{x^{3}}{3}\). In our exercise, we find the antiderivative of \( \frac{3x^{2}}{2} \) as \(\frac{x^{3}}{2} \). Similarly, we handle other terms. This step is crucial to solving integrals as it helps to reverse the differentiation process.
Evaluation of Integrals
Evaluation of integrals involves computing the exact value of an integral. Once we find the antiderivative, we substitute the upper and lower limits of integration. For instance, substituting \(\text{x} = 1\) and \(\text{x} = 0\) into \(\frac{x^{3}}{2} - \frac{x^{4}}{4} - \frac{x^{5}}{10}\) gives us \(\frac{1}{2} - \frac{1}{4} - \frac{1}{10}\). Simplifying this results in \(\frac{3}{20}\). This final step gives us the numerical value for the area under the curve or the volume we are interested in.

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Most popular questions from this chapter

The following [ormula is used by psychologists and educators to predict the reading ease E of a passage of words: $$ E=206.835-0.846 w-1.015 s $$ where \(w\) is the number of syllables in a 100 -word section and \(s\) is the average number of words per sentence. Find the reading ease in each case. $$ w=146 \text { and } s=5 $$

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Because wind speed enhances the loss of heat from the skin, we feel colder when there is wind than when there is not. The wind chill temperature is what the temperature would have to be with no wind in order to give the same chilling effect. The wind chill temperature \(W\) is given by \(W(v, T)=\) $$ 91.4-\frac{(10.45+6.68 \sqrt{v}-0.447 v)(457-5 T)}{110}, $$ where \(T\) is the actual temperature as given by a thermometer, in degrees Fahrenheit, and \(v\) is the speed of the wind, in miles per hour. Find the wind chill temperature in each case. Round to the nearest one degree. $$ T=-10^{\circ} \mathrm{F}, \mathrm{v}=30 \mathrm{mph} $$
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