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Chapter 4: Problem 6

$$ \text { Calculate } \int_{0}^{1} \int_{0}^{1}|x-y| d x d y . \text { Can you confirm your result by a geometric argument? } $$

Short Answer

Expert verified
\( \frac{1}{3} \)

Step by step solution

01

- Understand the Absolute Value

Analyze the absolute value function \( |x-y| \). The absolute value function makes the integrand equal to either \( x-y \) when \( x \geq y \) or \( y-x \) when \( y > x \).
02

- Split the Integral

Split the double integral into two regions based on the condition \( x \geq y \) and \( y > x \). \[\int_{0}^{1} \int_{0}^{1} |x-y| \, dx \, dy = \int_{0}^{1} \int_{0}^{y} (y-x) \, dx \, dy + \int_{0}^{1} \int_{y}^{1} (x-y) \, dx \, dy \]
03

- Evaluate the First Integral

Evaluate \( \int_{0}^{1} \int_{0}^{y} (y-x) \, dx \, dy \). \[ \int_{0}^{1} \left[ yx - \frac{x^2}{2} \right]_{0}^{y} dy = \int_{0}^{1} \left( y^2 - \frac{y^2}{2} \right) dy = \int_{0}^{1} \frac{y^2}{2} dy \] Calculate the remaining integral: \[ \int_{0}^{1} \frac{y^2}{2} \, dy = \frac{1}{2} \int_{0}^{1} y^2 \, dy = \frac{1}{2} \cdot \frac{y^3}{3} \Bigg|_{0}^{1} = \frac{1}{6} \]
04

- Evaluate the Second Integral

Evaluate \( \int_{0}^{1} \int_{y}^{1} (x-y) \, dx \, dy \). \[ \int_{0}^{1} \left[ \frac{x^2}{2} - yx \right]_{y}^{1} dy = \int_{0}^{1} \left( \frac{1}{2} - y - \frac{y^2}{2} + y^2 \right) dy = \int_{0}^{1} \left( \frac{1}{2} - y + \frac{y^2}{2} \right) dy \] Combine and simplify: \[ \int_{0}^{1} \left( \frac{1}{2} - y + \frac{y^2}{2} \right) dy = \frac{1}{2} y - \frac{y^2}{2} + \frac{y^3}{6} \Bigg|_{0}^{1} = \frac{1}{2} - \frac{1}{2} + \frac{1}{6} = \frac{1}{6} \]
05

- Summarize the Results

Sum the results of the two regions: \[ \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \] So, the value of the integral \( \int_{0}^{1} \int_{0}^{1} |x-y| \, dx \, dy \) is \ \frac{1}{3} \.
06

- Geometric Interpretation

Consider the function \( |x-y| \) which represents the area between the lines \( x = y \) over the region \( [0,1] \times [0,1] \). This can be visualized as two right triangles with combined area \( \int_{0}^{1} \int_{0}^{y} |x-y| \, dx \, dy + \int_{0}^{1} \int_{y}^{1} |x-y| \, dx \, dy \). The combined area confirms the integral calculation.

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Key Concepts

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

Integral Calculus
Integral calculus is a critical branch of mathematics that deals with integrals and their properties. Essentially, integrals help in finding quantities when given rates of change or the accumulation of a quantity. There are two main types of integrals: definite and indefinite integrals.

Definite integrals calculate the net area under a curve within specified limits, like from point a to point b. Indefinite integrals, on the other hand, represent the family of functions whose derivative matches the given function, often including a constant of integration.

In this exercise, we deal with a definite double integral, which extends the notion of integrals to functions of multiple variables. By computing the double integral \(\text{Calculate} \int_{0}^{1} \int_{0}^{1}|x-y| dx dy\), we're aggregating a two-dimensional area under the function \(|x-y|\) within the square region [0,1] 脳 [0,1]. Double integrals are particularly useful in calculating volumes beneath surfaces and areas in multivariable contexts.
Absolute Value Function
The absolute value function is essential in various mathematical contexts. It measures the distance of a number from zero on a number line, disregarding the direction. Formally, the absolute value of a real number x is given by \(|x| = \begin{cases} x, & \text{if} \ x \geq 0 \ x/y\ -x, & \text{if} \ x < 0 \end{cases}\).

In this particular problem, we need to evaluate the absolute value of the difference between two variables, i.e., \(|x-y|\). It's crucial to understand that \(|x-y|\) simplifies to \(x-y\) when \(x \geq y \) and to \(y-x\) when \(y > x\).

This split in the function, based on the relationship between x and y, allows us to decompose the double integral into more manageable pieces. By understanding where \(x \geq y\) and where \(y > x\) within the given limits of integration over [0,1] x [0,1], we can correctly interpret and compute each segment of the integral.
Geometric Interpretation
Geometric interpretation adds a visual perspective to our integral calculus problem. It helps us understand abstract mathematical concepts through shapes and areas.

For the given problem, we are asked to interpret the integral of \(|x-y|\) over the square region [0,1] 脳 [0,1] geometrically.

Visualizing this, consider the function \(|x-y|\) which represents the vertical distance between the points \((x, y)\) and the line \(y=x\). This line divides our region into two triangles; one where \(x \geq y\) and the other where \(y > x\).

Evaluating the integral over these regions, we calculate the area of two right triangles. Each of these triangles contributes equally to the overall area, confirming the integral value, which is \( \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \). Hence, the geometric interpretation aligns with our step-by-step calculation, serving as a neat verification of our result.

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Most popular questions from this chapter

(From Johansen (1972).) For fixed positive values of \(q_{1}\) and \(q_{2}\), consider the set \(G\left(q_{1}, q_{2}\right)\) in the \(\xi_{1} \xi_{2}\)-plane given by (draw a sketch!) $$ G\left(q_{1}, q_{2}\right)=\left\\{\left(\xi_{1}, \xi_{2}\right): q_{1} \xi_{1}+q_{2} \xi_{2} \leq 1, \xi_{1} \geq 0, \xi_{2} \geq 0\right\\} $$ Let \(f\left(\xi_{1}, \xi_{2}\right)\) be a continuous function defined over \(G\left(q_{1}, q_{2}\right)\). (a) Write down the double integral of \(f\) over \(G\left(q_{1}, q_{2}\right)\) when integrating first w.r.t. \(\xi_{2}\). (b) Write down the corresponding expression, integrating first w.r.t. \(\xi_{1}\). (c) The value of the double integral in (a) and in (b) will depend on \(q_{1}\) and \(q_{2}\), denote it by \(g\left(q_{1}, q_{2}\right)\). Compute \(\partial g / \partial q_{1}\).

(a) Sketch the domain of integration and compute the integral \(\int_{0}^{1}\left(\int_{x^{2}}^{x}\left(x^{2}+x y\right) d y\right) d x\). (b) Change the order of integration and verify that you obtain the same result as in (a).

Let \(f(x, y)=e^{x^{2}}\) be defined over the triangle \(A=\\{(x, y): x \in[0,1], 0 \leq y \leq x\\}\) Find the volume \(V\) under the graph of \(f\) over \(A\). (Hint: Integrate first w.r.t. \(y .\) If you try to integrate w.r.t. \(x\) first, there is no expression for the relevant integral in terms of elementary functions.)

(a) \(\int_{0}^{10}\left(10 t^{2}-t^{3}\right) d t\) (b) \(\int_{0}^{10} 4 t e^{-2 t} d t\) (c) \(\int_{0}^{10} \frac{10 t^{2}-t^{3}}{t+1} d t\)

$$ \text { Compute the double integral } I=\int_{0}^{2}\left(\int_{-2}^{1}\left(x^{2} y^{3}-(y+1)^{2}\right) d y\right) d x \text {. } $$
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