Problem 1 - Calculus and its applications

Question

Evaluate. $$ \int_{0}^{4} \int_{0}^{3} 3 x d x d y $$

Step-by-Step Solution

Answer

The double integral evaluates to 54.

Step 1: Understand the Integral

We are asked to evaluate a double integral. The double integral is of the form \( \int_{a}^{b} \int_{c}^{d} f(x,y) \, dx \, dy \), where in our case \( f(x,y) = 3x \), the inner integral bounds for \( x \) are from 0 to 3, and the outer integral bounds for \( y \) are from 0 to 4.

Step 2: Evaluate the Inner Integral

First, evaluate the inner integral \( \int_{0}^{3} 3x \, dx \). The antiderivative of \( 3x \) with respect to \( x \) is \( \frac{3}{2}x^2 \). Evaluate this from 0 to 3:\[ \left. \frac{3}{2}x^2 \right|_{0}^{3} = \frac{3}{2}(3)^2 - \frac{3}{2}(0)^2 = \frac{3}{2} \times 9 = 13.5. \]

Step 3: Evaluate the Outer Integral

Now, substitute the result of the inner integral into the outer integral: \( \int_{0}^{4} 13.5 \, dy \). Since 13.5 is a constant with respect to \( y \), the antiderivative is \( 13.5y \). Evaluate this from 0 to 4:\[ \left. 13.5y \right|_{0}^{4} = 13.5 \times 4 - 13.5 \times 0 = 54. \]

Step 4: Interpretation of the Result

The value 54 represents the volume underneath the surface defined by the function \( 3x \) over the rectangle defined by the limits of \( x \) and \( y \).

Key Concepts

Multiple Integration

Multiple integration extends the concept of integration to functions of multiple variables. Unlike single integrals, which find the area under a curve, multiple integrals help us find quantities like volume or mass within a region.
In double integration, two integrals are used in succession to integrate functions of two variables over a specific region. Here, we are working with a double integral of the form \[ \int_{a}^{b} \int_{c}^{d} f(x,y)\, dx\, dy \], where

• \( f(x,y) = 3x \),
• the inner integral bounds \( x \) between 0 and 3,
• and the outer integral bounds \( y \) between 0 and 4.

Evaluating a double integral typically involves calculating the inner integral first, followed by the evaluation of the outer integral. This systematic approach helps in managing the complexity, as seen in our exercise.

Antiderivative

The antiderivative, or indefinite integral, is the reverse process of differentiation. It helps us evaluate integrals by finding a function whose derivative yields the original function. This is particularly useful in solving definite integrals, where bounds are applied.
For our exercise, the function \( 3x \) is integrated with respect to \( x \).

• The antiderivative of \( 3x \) is \( \frac{3}{2}x^2 \).
• This function represents the area under the curve \( 3x \) in terms of \( x \).
• Once the antiderivative is determined, we evaluate it over the specified limits, yielding the result 13.5.

Recognizing and calculating antiderivatives is a fundamental skill in integration, as it allows us to solve not just for areas, but for other geometrical properties as well, such as volumes.

Volume Interpretation

Double integrals provide a powerful way to interpret geometric shapes in terms of volume. For a given surface, like \( 3x \) in our exercise, the double integral calculates the volume beneath this surface over a defined region.
The function \( 3x \) forms a surface that stretches over the rectangle defined by the limits of \( x \) and \( y \). By integrating over this region:

• The inner integral \( \int_{0}^{3} 3x \, dx \) finds the area under the curve for each slice along \( y \).
• This process effectively stacks these slices, building up a volume as defined by the outer integral \( \int_{0}^{4} \, dy \).
• The result, 54, is the total volume under the surface, showing how integration extends our understanding from simple 2D shapes to complex 3D volumes.

Thus, the double integral ties together calculus and geometry, offering a visual interpretation that simplifies and enriches our comprehension of mathematical concepts.