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Magazine Chapter 15: Problem 12
Compute the integrals. HINT [See Example 1.] $$ \int_{1}^{4} \int_{-\sqrt{x}}^{\sqrt{x}} \frac{1}{x} d y d x $$
Short Answer
Expert verified
The value of the double integral \(\int_{1}^{4} \int_{-\sqrt{x}}^{\sqrt{x}} \frac{1}{x} dy dx\) is calculated as \(8-4 = \boxed{4}\).
Step by step solution
01
Integrate with respect to y
We will first integrate the function \(\frac{1}{x}\) with respect to \(y\):
\[
\int_{-\sqrt{x}}^{\sqrt{x}} \frac{1}{x} dy
\]
Since the function does not contain \(y\), it will act as a constant during integration. The integral becomes:
\[
\frac{1}{x} \int_{-\sqrt{x}}^{\sqrt{x}} dy
\]
Now, we can integrate with respect to \(y\):
\[
\frac{1}{x} \left[ y \right]_{-\sqrt{x}}^{\sqrt{x}}
\]
02
Apply limits and integrate with respect to x
Next, we apply the limits of integration for \(y\):
\[
\frac{1}{x} (\sqrt{x} - (-\sqrt{x})) = \frac{1}{x}(2\sqrt{x})
\]
Now, we integrate the resulting expression with respect to \(x\):
\[
\int_{1}^{4} \frac{1}{x}(2\sqrt{x}) dx
\]
Simplify the expression inside the integral:
\[
\int_{1}^{4} 2\sqrt{x} \cdot \frac{1}{x} dx = \int_{1}^{4} 2\frac{\sqrt{x}}{x} dx
\]
We can further simplify the expression by realizing that \(\frac{\sqrt{x}}{x}\) can be written as \(\frac{x^{\frac{1}{2}}}{x^{1}} = x^{-\frac{1}{2}}\). Now, the integral becomes:
\[
\int_{1}^{4} 2x^{-\frac{1}{2}} dx
\]
To solve this integral, use the power rule of integration:
\[
\left[ 2 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right]_{1}^{4}
\]
We can simplify the expression as:
\[
\left[ 4x^{\frac{1}{2}} \right]_{1}^{4}
\]
Now, apply the limits of integration:
\[
4(4^{\frac{1}{2}}) - 4(1^{\frac{1}{2}}) = 4(2) - 4(1) = 8 - 4
\]
03
Final Answer
So, the value of the double integral is:
\[
\int_{1}^{4} \int_{-\sqrt{x}}^{\sqrt{x}} \frac{1}{x} dy dx = 8-4 = \boxed{4}
\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration with Respect to y
When tackling double integrals, it's often beneficial to integrate one variable at a time. Let's start by focusing on integration with respect to \( y \). Imagine you have a function where \( y \) is just an additional dimension or direction, but not directly involved in the function itself. In our example, the integrand \( \frac{1}{x} \) doesn't actually include \( y \). This means \( \frac{1}{x} \) is considered a constant during integration with respect to \( y \).
Here鈥檚 a helpful tip: when integrating a constant over \( y \), simply multiply the constant by the length of the interval defined by the limits of \( y \).
Here鈥檚 a helpful tip: when integrating a constant over \( y \), simply multiply the constant by the length of the interval defined by the limits of \( y \).
- The integral: \( \int_{-\sqrt{x}}^{\sqrt{x}} dy \) gives us \( [y]_{-\sqrt{x}}^{\sqrt{x}} = \sqrt{x} - (-\sqrt{x}) \), which simplifies to \( 2\sqrt{x} \).
Integration with Respect to x
Next, we move on to integrating with respect to \( x \). At this stage, you will consider the results you obtained from the first integration. Integrating with respect to \( x \) essentially means calculating the accumulated area or volume under the curve, taking \( x \) as the variable.
After substituting the results of the \( y \)-integral from our previous step, we get:
This step ultimately brings us closer to finding the exact area under the surface defined by the original double integral setup.
After substituting the results of the \( y \)-integral from our previous step, we get:
- \( \int_{1}^{4} 2 \frac{\sqrt{x}}{x} \, dx \)
This step ultimately brings us closer to finding the exact area under the surface defined by the original double integral setup.
Power Rule of Integration
The power rule is a fundamental tool in integration, especially useful when dealing with polynomial expressions or their derivatives. When you have an integral of the form \( \int x^n \, dx \), you apply the power rule. According to this rule, the integral becomes \( \frac{x^{n+1}}{n+1} \).
Using the power rule for our simplification:
Using the power rule for our simplification:
- We had \( 2 \int x^{- rac{1}{2}} \, dx \)
- Applying the rule, it becomes: \( 2 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 4x^{\frac{1}{2}} \)
Limits of Integration
Understanding and applying limits of integration is crucial in defining the boundaries over which you calculate the integral. The limits essentially slice a portion of the space you are working in鈥攄efining start and endpoint boundaries.
In our example, the limits of \( y \) are from \( -\sqrt{x} \) to \( \sqrt{x} \), which means we are looking at a symmetrical segment centered around zero, for each \( x \). The limits of \( x \) are from 1 to 4. Each limit determines the specific slice of \( x \) values under consideration for the final integration step after resolving the first dimension.
Once you've integrated within these limits, you substitute them in the final expression as follows:
In our example, the limits of \( y \) are from \( -\sqrt{x} \) to \( \sqrt{x} \), which means we are looking at a symmetrical segment centered around zero, for each \( x \). The limits of \( x \) are from 1 to 4. Each limit determines the specific slice of \( x \) values under consideration for the final integration step after resolving the first dimension.
Once you've integrated within these limits, you substitute them in the final expression as follows:
- \( \left[ 4x^{\frac{1}{2}} \right]_{1}^{4} = 4(4^{\frac{1}{2}}) - 4(1^{\frac{1}{2}}) = 8 - 4 \)
