Question

Evaluate the following integrals as they are written. $$\int_{0}^{2} \int_{0}^{4-y^{2}} y d x d y$$

Short answer

Question: Evaluate the double integral $$\int_{0}^{2} \int_{0}^{4-y^{2}} y d x d y$$ Answer: 4

Step-by-step solution

Evaluate the inner integral with respect to x

We have: $$\int_{0}^{2} \bigg(\int_{0}^{4-y^{2}} y d x \bigg) d y$$ First, let's evaluate \(\int y dx\): Since \(y\) is a constant with respect to \(x\), the integral will be: $$y \int_{0}^{4-y^2} dx$$ Evaluating the integral and applying the limits, we get: $$y \left[x\right]_0^{4-y^2} = y((4-y^2)-0) = y(4-y^2)$$

Evaluate the outer integral with respect to y

Now, let's evaluate the outer integral with respect to \(y\): $$\int_{0}^{2} y(4-y^2)d y$$ We can split the integral into two parts: $$\int_{0}^{2} 4y d y - \int_{0}^{2} y^3 d y$$ Now, let's evaluate each integral separately: $$\left[ 2y^2 \right]_0^2 - \left[ \frac{y^4}{4} \right]_0^2$$ Applying the limits of integration: $$ \left( 2(2)^2 - 0 \right) - \left( \frac{(2)^4}{4} - 0 \right) = 8 - 4 = 4$$ Therefore, the value of the double integral is 4.

Key concepts

Definite Integrals

Definite integrals are a core concept in calculus. They represent the accumulation of a quantity, such as area, over a certain interval. Unlike indefinite integrals, which come with a constant of integration, definite integrals yield a numerical value as a result. This is because they are evaluated over specific limits of integration, providing a precise numerical output.

The notation for a definite integral is given by \( \int_{a}^{b} f(x) \, dx \), where \( f(x) \) is the function to be integrated, and \( a \) and \( b \) are the limits of integration. These limits specify the interval along the x-axis over which we need to calculate the area under the curve \( f(x) \).

• The function \( f(x) \): This function is continuous and integrable over the interval [a, b].
• The limits \( a \) and \( b \): These are the boundary points between which the integral is calculated.

Evaluating a definite integral involves finding the antiderivative of \( f(x) \), computing it at the upper limit \( b \), and subtracting its value at the lower limit \( a \). This process gives the accumulated value, like finding the total accumulated change, value, or improvement in a system.

Limits of Integration

The limits of integration are pivotal in determining the interval over which a definite integral is evaluated. For a double integral, such as in this problem, limits are crucial for both variables involved, commonly denoted as \( x \) and \( y \). These limits can be constants or functions themselves, defining the boundary within which the variables can vary.

In the exercise provided, we observe the following limits:

• Inner Integral (with respect to \( x \)): The limits are 0 to \( 4-y^2 \). These change based on the value of \( y \), reflecting the dependency within the region of integration. The expression \( 4-y^2 \) adjusts the boundary of \( x \) dynamically as \( y \) varies.
• Outer Integral (with respect to \( y \)): The limits here are 0 to 2, providing a constant boundary for \( y \) over which the integration occurs.

Understanding these limits is crucial as they define the domain of integration, helping to establish the area or volume being considered for the integral. Adjusting these limits directly impacts the evaluated integral, highlighting the importance of correctly setting these boundaries.

Integration Techniques

There are various integration techniques to efficiently solve difficult or complex integrals. Each technique is chosen based on the function's structure to simplify the calculation and facilitate the integration process.

Let's explore two key techniques used in the given exercise:

• Substitution Method: Not explicitly used here, but understanding substitution helps in recognizing simple forms within complex functions, simplifying them for integration.
• Decomposition/Writing as Sum of Simpler Integrals: When you encounter a polynomial, as with \( y(4-y^2) \), splitting it into two separate integrals, such as \( 4y \) and \( y^3 \), helps break down the problem. This is evident in our exercise when \( \, \large{ \int_{0}^{2} 4y \, dy - \int_{0}^{2} y^3 \, dy } \) is split to simplify evaluation.

Through these techniques, complex integrals become more approachable, transforming them into more manageable parts that can be evaluated using basic integral rules. Choosing the right technique is crucial to swift and correct integration, emphasizing practice and experience in solving integrals effectively.