Question

Suppose \(f(x, y)=g(x) h(y),\) where \(g\) and \(h\) are continuous functions for all real values. a. Show that \(\int_{c}^{d} \int_{a}^{b} f(x, y) d x d y=\left(\int_{a}^{b} g(x) d x\right)\left(\int_{c}^{d} h(y) d y\right)\) Interpret this result geometrically. b. Write \(\left(\int_{a}^{b} g(x) d x\right)^{2}\) as an iterated integral. c. Use the result of part (a) to evaluate \(\int_{0}^{2 \pi} \int_{10}^{30} \cos x e^{-4 y^{2}} d y d x\)

Short answer

Question: Show that the double integral of a function that is a product of two other functions can be represented as a product of two integrals, and use this result to evaluate a given double integral. Solution: a. Show that the double integral is a product of two integrals: Step 1: Write down the double integral of f(x, y): The double integral we want to evaluate is \(\int_{c}^{d} \int_{a}^{b} f(x, y) dx dy.\) Step 2: Replace f(x, y) with g(x)h(y): Now, we substitute the function f(x, y) with the product of g(x) and h(y): \(\int_{c}^{d} \int_{a}^{b} g(x)h(y) dx dy.\) Step 3: Separate the x and y integrals: In this step, we'll separate the integrals, moving the factors related to x in the first integral and the factors related to y in the second one: \(\left(\int_{c}^{d} h(y) dy\right) \left(\int_{a}^{b} g(x) dx\right).\) Geometrical interpretation: This result implies that integrating a product of two functions over a rectangular region can be separated into the product of the integral of each individual function over the corresponding intervals. This simplifies the process of calculating higher-dimensional integrals in cases where the function can be represented as a product of separate functions for each coordinate. b. Write the square of the integral as an iterated integral: Step 1: Square the integral of g(x): We have to write \(\left(\int_{a}^{b} g(x) dx\right)^2\) as an iterated integral. Step 2: Rewrite as an iterated integral: To do this, we can use a change of variable: let \(x' = x.\) Then, we can write the square of the integral as an iterated integral with respect to \(x\) and \(x'\): \(\int_{a}^{b} \int_{a}^{b} g(x)g(x') dx dx'.\) c. Evaluate the integral: Step 1: Write down the integral: We want to evaluate \(\int_{0}^{2 \pi} \int_{10}^{30} \cos x e^{-4y^2} dy dx.\) Step 2: Identify the functions g(x) and h(y): We can see that \(g(x) = \cos x\) and \(h(y) = e^{-4y^2}.\) Step 3: Use the result of part (a): Using the result from part (a), we can rewrite the integral as the product of two separate integrals: \(\left(\int_{0}^{2 \pi} \cos x dx\right)\left(\int_{10}^{30} e^{-4 y^2} dy\right).\) Step 4: Evaluate the integrals: Now, we evaluate the separate integrals. First, \(\int_{0}^{2 \pi} \cos x dx = [\sin x]_{0}^{2 \pi} = (\sin 2\pi -\sin 0) = 0.\) Since one of the integral equals to 0, their product will also be 0: Answer: \(\int_{0}^{2 \pi} \int_{10}^{30} \cos x e^{-4 y^2} dy dx = 0\)

Step-by-step solution

Write down the double integral of f(x, y)

The double integral we want to evaluate is \(\int_{c}^{d} \int_{a}^{b} f(x, y) dx dy.\)

Replace f(x, y) with g(x)h(y)

Now, we substitute the function f(x, y) with the product of g(x) and h(y): \(\int_{c}^{d} \int_{a}^{b} g(x)h(y) dx dy.\)

Separate the x and y integrals

In this step, we'll separate the integrals, moving the factors related to x in the first integral and the factors related to y in the second one: \(\left(\int_{c}^{d} h(y) dy\right) \left(\int_{a}^{b} g(x) dx\right).\)

Interpretation (a) geometrically:

The result implies that integrating a product of two functions over a rectangular region can be separated into the product of the integral of each individual function over the corresponding intervals. This simplifies the process of calculating higher-dimensional integrals in cases where the function can be represented as a product of separate functions for each coordinate. #b. Write the square of the integral as an iterated integral#

Square the integral of g(x)

We have to write \(\left(\int_{a}^{b} g(x) dx\right)^2\) as an iterated integral.

Rewrite as an iterated integral

To do this, we can use a change of variable: let \(x' = x.\) Then, we can write the square of the integral as an iterated integral with respect to \(x\) and \(x'\): \(\int_{a}^{b} \int_{a}^{b} g(x)g(x') dx dx'.\) #c. Evaluate the integral#

Write down the integral

We want to evaluate \(\int_{0}^{2 \pi} \int_{10}^{30} \cos x e^{-4y^2} dy dx.\)

Identify the functions g(x) and h(y)

We can see that \(g(x) = \cos x\) and \(h(y) = e^{-4y^2}.\)

Use the result of part (a)

Using the result from part (a), we can rewrite the integral as the product of two separate integrals: \(\left(\int_{0}^{2 \pi} \cos x dx\right)\left(\int_{10}^{30} e^{-4 y^2} dy\right).\)

Evaluate the integrals

Now, we evaluate the separate integrals. First, \(\int_{0}^{2 \pi} \cos x dx = [\sin x]_{0}^{2 \pi} = (\sin 2\pi -\sin 0) = 0.\) Since one of the integral equals to 0, their product will also be 0: Answer: \(\int_{0}^{2 \pi} \int_{10}^{30} \cos x e^{-4 y^2} dy dx = 0\)

Key concepts

Continuous Functions

Any function that can be drawn without lifting a pen from the paper is considered a continuous function. In mathematical terms, a function is continuous if, at every point within its domain, the limit of the function as you approach that point equals the function's actual value at that point.

Continuous functions are particularly important in calculus and analysis because they guarantee that we can apply the Fundamental Theorem of Calculus as well as various integration techniques. You can think of continuous functions as being smooth. There are no sudden jumps, breaks, or holes in the function. This characteristic makes them predictable and manageable in calculus.

In our exercise, we deal with continuous functions specifically because integrals involving these functions are well-defined, ensuring precise calculations without discrepancies.

Iterated Integrals

Iterated integrals are a method of calculating double or multiple integrals by successively integrating one variable at a time. The double integral \(\int_{c}^{d} \int_{a}^{b} f(x, y) \, dx \, dy\), for example, represents a technique where you first integrate with respect to \(x\) over the interval \([a, b]\) and then integrate the result with respect to \(y\) over \([c, d]\).

Iterated integrals are straightforward once you recognize that they're just repeated applications of single-variable integration rules.

This approach simplifies the complexity of multivariate integrals by breaking them down into step-by-step procedures that can be individually solved and combined. In our exercise, transitioning from a double integral to iterated single-variable integrals aids in not only understanding but also organizing our computations for easier handling.

Geometric Interpretation

Geometrically interpreting integrals, especially multiple integrals, can provide a clearer understanding of what the calculation represents. When we compute \(\int_{c}^{d} \int_{a}^{b} f(x, y) \, dx \, dy\), we're essentially looking to find the volume under the surface described by \(f(x, y)\) over the rectangle bounded by \([a, b]\) in the \(x\)-direction and \([c, d]\) in the \(y\)-direction.

This result is particularly useful when \(f(x, y)\) can be broken into the product of two separate continuous functions, \(g(x)\) and \(h(y)\), as seen in our exercise. Such a breakdown allows us to visualize the integration process as calculating the area under curves individually along \(x\) and \(y\), then combining these in a multiplicative manner to find the sought volume.

The realization that a complex area or volume can be computed through simpler, independent calculations is a powerful tool in mathematics and helps streamline what might otherwise be an unwieldy computation.

Integration Techniques

Integration is a fundamental element of calculus that's used to find areas under curves. Understanding various techniques of integration can significantly ease the process of evaluating integrals.

Some key integration techniques include:

• Substitution: This involves changing variables to simplify the integral.
• Integration by Parts: Useful when dealing with products of functions, applying the formula \(\int u \, dv = uv - \int v \, du\) can help simplify the process.
• Separation of Variables: When faced with a function of multiple variables that can be expressed as a product of separate functions, it can be advantageous to split the integral into separate parts. This was the technique employed in our exercise to solve the multiple integral more easily.

Mastering these techniques allows for efficient handling of integration problems and is particularly helpful in dealing with more complex integrals encountered in higher dimensions.