Chapter 7: Problem 11
\(\int_{0}^{1} \int_{1}^{2} x y d y d x=\ldots \ldots\)
Short Answer
Expert verified
The value of the integral is \( \frac{3}{4} \).
Step by step solution
01
Understand the Exercise
We are asked to evaluate the iterated integral \( \int_{0}^{1} \int_{1}^{2} x y \, dy \, dx \). This involves finding the integral of the function \( x y \) first with respect to \( y \) and then with respect to \( x \).
02
Integrate with Respect to y
First, we fix \( x \) and integrate the inner integral \( \int_{1}^{2} x y \, dy \). This can be solved by integrating \( x \) (considered a constant) multiplied by \( y \). The antiderivative is \( \frac{1}{2}xy^2 \).
03
Evaluate the Inner Integral
We evaluate \( \frac{1}{2}xy^2 \) at \( y = 2 \) and \( y = 1 \). This gives: \( \frac{1}{2}x(2)^2 - \frac{1}{2}x(1)^2 = \frac{1}{2}x(4) - \frac{1}{2}x(1) = 2x - \frac{1}{2}x = \frac{3}{2}x \).
04
Integrate with Respect to x
Now, we integrate the result \( \int_{0}^{1} \frac{3}{2}x \, dx \). The antiderivative of \( \frac{3}{2}x \) is \( \frac{3}{4}x^2 \).
05
Evaluate the Outer Integral
Apply the limits to \( \frac{3}{4}x^2 \) at \( x = 1 \) and \( x = 0 \). This is \( \frac{3}{4}(1)^2 - \frac{3}{4}(0)^2 = \frac{3}{4} \cdot 1 - 0 = \frac{3}{4} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by Parts is a vital technique in calculus, especially when working with complicated integrals. However, in the context of iterated integrals, like in our exercise, we focus instead on breaking down a double integral into more manageable parts.
In a typical double integral problem, the process requires evaluating one integral first (usually the innermost), followed by the second. While Integration by Parts is not directly applied to this process, understanding its mechanics can be beneficial. This technique is based on the product rule for derivatives and expressed by the formula:
\[\int u \, dv = uv - \int v \, du\]Although the problem at hand does not directly use integration by parts, it's important to recognize its utility when facing functions where direct integration isn't straightforward.
If you encounter an integral with a product of functions, Integration by Parts might be necessary, thus broadening your problem-solving toolkit. For now, knowing when and how to swap back to simpler methods, like breaking down iterated integrals, is critical.
In a typical double integral problem, the process requires evaluating one integral first (usually the innermost), followed by the second. While Integration by Parts is not directly applied to this process, understanding its mechanics can be beneficial. This technique is based on the product rule for derivatives and expressed by the formula:
\[\int u \, dv = uv - \int v \, du\]Although the problem at hand does not directly use integration by parts, it's important to recognize its utility when facing functions where direct integration isn't straightforward.
If you encounter an integral with a product of functions, Integration by Parts might be necessary, thus broadening your problem-solving toolkit. For now, knowing when and how to swap back to simpler methods, like breaking down iterated integrals, is critical.
Antiderivative
The antiderivative, also known as an indefinite integral, is essentially the reverse of differentiation. When solving iterated integrals, finding the antiderivative is an essential step.
During Step 2 of our exercise, we compute the antiderivative of the function with respect to its variable. For example, when integrating \(x \cdot y \) with respect to \(y\), it's crucial to treat \(x\) as a constant. This transforms the expression to \(\int xy \, dy\), which necessitates finding the antiderivative of \(xy\) in terms of \(y\).
To do this, the antiderivative turns into:
\[\frac{1}{2}xy^2\]This expression helps us progress through the iterated integral method as we apply further limits or additional variable integration.
Moreover, understanding the antiderivative offers the groundwork for breaking complex functions down into simpler parts, facilitating the evaluation of both definite and indefinite integrals. Always remember to carefully consider constant and variable behavior during integration.
During Step 2 of our exercise, we compute the antiderivative of the function with respect to its variable. For example, when integrating \(x \cdot y \) with respect to \(y\), it's crucial to treat \(x\) as a constant. This transforms the expression to \(\int xy \, dy\), which necessitates finding the antiderivative of \(xy\) in terms of \(y\).
To do this, the antiderivative turns into:
\[\frac{1}{2}xy^2\]This expression helps us progress through the iterated integral method as we apply further limits or additional variable integration.
Moreover, understanding the antiderivative offers the groundwork for breaking complex functions down into simpler parts, facilitating the evaluation of both definite and indefinite integrals. Always remember to carefully consider constant and variable behavior during integration.
Limits of Integration
Limits of Integration define the range over which you're evaluating the function, and are crucial for turning an antiderivative into a definite integral. In our problem, we have separate bounds for both inner and outer integrals, representing different scopes.
For the inner integral \(\int_{1}^{2}\), these bounds indicate evaluating the function from \(y=1\) to \(y=2\). Similarly, for the outer integral \(\int_{0}^{1}\), the range extends from \(x=0\) to \(x=1\).
Evaluating limits involves substituting these boundaries into the antiderivative found, as demonstrated in our problem solved. This step ensures we obtain a specific value rather than an expression, finalizing our calculation:
1. Inner integral is computed from the antiderivative \(\frac{1}{2}xy^2\).2. Outer integral involves \(\frac{3}{4}x^2\) as its antiderivative.These limits are applied directly through evaluation, reducing the function to a single numerical value corresponding to the definite integral on its prescribed interval. Practicing clear boundaries understanding helps accurately capture the essence of the problem and avoid common errors.
For the inner integral \(\int_{1}^{2}\), these bounds indicate evaluating the function from \(y=1\) to \(y=2\). Similarly, for the outer integral \(\int_{0}^{1}\), the range extends from \(x=0\) to \(x=1\).
Evaluating limits involves substituting these boundaries into the antiderivative found, as demonstrated in our problem solved. This step ensures we obtain a specific value rather than an expression, finalizing our calculation:
1. Inner integral is computed from the antiderivative \(\frac{1}{2}xy^2\).2. Outer integral involves \(\frac{3}{4}x^2\) as its antiderivative.These limits are applied directly through evaluation, reducing the function to a single numerical value corresponding to the definite integral on its prescribed interval. Practicing clear boundaries understanding helps accurately capture the essence of the problem and avoid common errors.
