Problem 17
Question
Evaluate the double integral. $$ \int_{1}^{2} \int_{0}^{4}\left(3 x^{2}-2 y^{2}+1\right) d x d y $$
Step-by-Step Solution
Verified Answer
Result of the given double integral is \[128 - 64/3 + 8 - 64 + 8/3 - 4 = \frac{76}{3}\]
1Step 1: Setup of the first integral
Integrating \(3x^2 - 2y^2 + 1\) with respect to \(x\) from 0 to 4 gives us the first single integral to deal with: \[\int_{0}^{4}(3x^2 - 2y^2 + 1) dx\]
2Step 2: Evaluate the first integral
Applying the power rule for integration, we get: \[3*\frac{1}{3}*x^3 - 2y^2*x + x\] Evaluating this from 0 to 4 results in: \[64 - 8*y^2 + 4 - 0 = 64 - 8*y^2 + 4\]
3Step 3: Setup the second integral
Now we need to integrate the result from Step 2 with respect to \(y\) from 1 to 2: \[\int_{1}^{2}(64 - 8*y^2 + 4) dy\]
4Step 4: Evaluate the second integral
Integral of a number is number*y and integral of y^2 is (1/3)*y^3. So, \[64*y - 8*\frac{1}{3}*y^3 + 4*y\] evaluates from 1 to 2 results in: \[128 -64/3 + 8 - (64 - 8/3 + 4) = 128 - 64/3 + 8 - 64 + 8/3 - 4\]
Key Concepts
Definite Integral CalculusIntegration TechniquesPower Rule for Integration
Definite Integral Calculus
In the realm of calculus, a definite integral represents the signed area under a curve. More precisely, it calculates the accumulation of quantities, where the rate of accumulation varies. To calculate the exact area, we fix the interval, hence the term 'definite'. For example, the double integral exercise we are looking at:
\[\begin{equation}\int_{1}^{2} \int_{0}^{4}\left(3 x^{2}-2 y^{2}+1\right) dx dy\end{equation}\]
Here, we are finding the volume under the surface described by the function \(3 x^{2}-2 y^{2}+1\) for x ranging from 0 to 4, and y ranging from 1 to 2. To make definite integrals more approachable, break down the problem into steps, like the provided solution does—first integrating with respect to \(x\), then with respect to \(y\). This not only simplifies the process but also helps you understand the behavior of the function over different dimensions.
\[\begin{equation}\int_{1}^{2} \int_{0}^{4}\left(3 x^{2}-2 y^{2}+1\right) dx dy\end{equation}\]
Here, we are finding the volume under the surface described by the function \(3 x^{2}-2 y^{2}+1\) for x ranging from 0 to 4, and y ranging from 1 to 2. To make definite integrals more approachable, break down the problem into steps, like the provided solution does—first integrating with respect to \(x\), then with respect to \(y\). This not only simplifies the process but also helps you understand the behavior of the function over different dimensions.
Integration Techniques
When dealing with integrals, various techniques can be employed to simplify the process. These include substitution, partial fractions, integration by parts, and recognizing patterns that correspond to known integrals.
Identifying the Technique
For our double integral exercise, the technique used is straightforward. Since we are integrating polynomials, we primarily rely on the power rule. In situations where the integrand is a product of functions or a more complex function, techniques such as integration by parts or substitution may be more appropriate. The choice of technique is crucial for simplifying the integral to an easily manageable form.Application of Techniques
This step-by-step process prevents unnecessary complexity and promotes a clear understanding, which is essential when learning how to tackle integrals. Additionally, with integrals often representing physical quantities such as areas, volumes, or other accumulations, mastering these techniques is fundamental for the application of calculus in real-world scenarios.Power Rule for Integration
One of the most basic and frequently used rules in integration is the power rule. It is crucial for evaluating the integral of polynomial functions, as seen in our exercise. The power rule simply states:
\[\begin{equation}\int x^n dx = \frac{1}{n+1}x^{n+1} + C\end{equation}\]
where \(n\) is any real number except -1, and \(C\) represents the constant of integration. In the context of definite integrals, the constant \(C\) gets canceled out when we evaluate the integral at the upper and lower bounds.
\[\begin{equation}\int x^n dx = \frac{1}{n+1}x^{n+1} + C\end{equation}\]
where \(n\) is any real number except -1, and \(C\) represents the constant of integration. In the context of definite integrals, the constant \(C\) gets canceled out when we evaluate the integral at the upper and lower bounds.
Correct Application
In the solution provided, the power rule lets us quickly find the antiderivative of \(x^2\) to be \(\frac{1}{3}x^3\), and similarly, the antiderivative of \(y^2\) becomes \(\frac{1}{3}y^3\) when integrating with respect to \(y\). Errors in applying the power rule are common, and double-checking work is always encouraged; for instance, ensure the new exponent is indeed one higher than the original, and that the coefficient is adjusted appropriately. Understanding and correctly applying the power rule can significantly expedite the process of integration and avoid common pitfalls.Other exercises in this chapter
Problem 16
Find the distance between the two points. $$ (8,-2,2),(8,-2,4) $$
View solution Problem 17
Examine the function for relative extrema and saddle points. $$ f(x, y)=(x+y) e^{1-x^{2}-y^{2}} $$
View solution Problem 17
Use the regression capabilities of a graphing utility or a spreadsheet to find the least squares regression line for the given points. $$ (0,6),(4,3),(5,0),(8,-
View solution Problem 17
Let \(f(x, y)=3 x^{2} y e^{x-y}\) and \(g(x, y)=\) \(3 x y^{2} e^{y-x} .\) Find each of the following. $$ g_{x}(x, y) $$
View solution