Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. $$ \int_{1}^{2} \int_{0}^{3}\left(x y+y^{2}\right) d x d y $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{55}{4} \).

Step 1: Understand the Problem

We have an iterated integral \( \int_{1}^{2} \int_{0}^{3} (xy + y^2) \,dx \,dy \). This means we will integrate with respect to \( x \) first, then with respect to \( y \).

Step 2: Integrate with respect to x

Treat \( y \) as a constant and integrate \( xy + y^2 \) with respect to \( x \) over the interval [0, 3].\[ \int_{0}^{3} (xy + y^2) \, dx = \int_{0}^{3} xy \, dx + \int_{0}^{3} y^2 \, dx \]- \( \int_{0}^{3} xy \, dx = \left[ \frac{1}{2} x^2 y \right]_{0}^{3} = \frac{1}{2} (3^2 y) - \frac{1}{2} (0^2 y) = \frac{9}{2} y \)- \( \int_{0}^{3} y^2 \, dx = [x y^2]_{0}^{3} = 3y^2 \)Combining both results, we get \( \frac{9}{2} y + 3y^2 \).

Step 3: Integrate with respect to y

Now integrate \( \frac{9}{2} y + 3y^2 \) with respect to \( y \) over the interval [1, 2].\[ \int_{1}^{2} \left( \frac{9}{2} y + 3y^2 \right) \, dy \]- \( \int_{1}^{2} \frac{9}{2} y \, dy = \left[ \frac{9}{4} y^2 \right]_{1}^{2} = \frac{9}{4} (4) - \frac{9}{4} (1) = \frac{27}{4} \)- \( \int_{1}^{2} 3y^2 \, dy = \left[ y^3 \right]_{1}^{2} = (8) - (1) = 7 \)Combining both results, we get \( \frac{27}{4} + 7 = \frac{27}{4} + \frac{28}{4} = \frac{55}{4} \).

Step 4: Final Result

The value of the iterated integral \( \int_{1}^{2} \int_{0}^{3} (xy + y^2) \, dx \, dy \) is \( \frac{55}{4} \).

Key Concepts

Multivariable Calculus

Multivariable Calculus is a branch of calculus that deals with functions of more than one variable. In the context of our exercise, we're working with the function \(f(x, y) = xy + y^2 \), which depends on both \(x\) and \(y\). This introduces the concept of an iterated integral.
The iterated integral \( \int_{1}^{2} \int_{0}^{3} (xy + y^2) \, dx \, dy \) means we first integrate with respect to \(x\) while treating \(y\) as a constant, and then integrate the result with respect to \(y\). This step-by-step process reflects the nature of working with multiple variables to evaluate a definite integral.
Multivariable calculus is essential for understanding how changes in one variable affect another, especially in physical and engineering contexts. By mastering iterated integrals, you can solve various real-world problems involving surface areas, volumes, and more complex interactions between multiple variables.

Definite Integration

Definite integration helps calculate the accumulated value, or the "net" area under a curve, within predetermined bounds. In our exercise, it means finding the value of the integral across two specified ranges: \(x\) from 0 to 3, and \(y\) from 1 to 2.
In simpler terms, definite integration sums up all the tiny rectangles that fit under the curve \(xy + y^2\), giving us a total value over the chosen interval.

• For \( \int_{0}^{3} \), we find how the function behaves as \(x\) changes, holding \(y\) constant.
• Afterward, \( \int_{1}^{2} \) lets us integrate the result as \(y\) varies, making sure we've accumulated everything the function covers from one end of \(y\) to the other.

This process not only gives a concrete number, but also provides a deeper understanding of how the integration limits define the region over which we're summing changes.

Integration by Parts

While integration by parts isn't directly used in our specific example, it's a valuable technique in calculus for tackling integrals involving products of functions. It is based on the product rule of differentiation.
In general terms, integration by parts states: \[\int u \, dv = uv - \int v \, du\]where \(u\) and \(dv\) are parts of the original integral. This approach can simplify complex integrals significantly by reducing them to simpler forms or even iteratively breaking them down into parts.
Although our iterated integral \(\int_{1}^{2} \int_{0}^{3} (xy + y^2) \) doesn't require integration by parts due to its straightforward nature, being familiar with this technique is crucial. It offers an additional method to handle more complicated functions where basic integration methods aren't sufficient.