Problem 12 - Calculus and its applications

Question

Evaluate. $$ \int_{0}^{1} \int_{-1}^{x}\left(x^{2}+y^{2}\right) d y d x $$

Step-by-Step Solution

Answer

The value of the integral is 1.

Step 1: Identify the Order of Integration

The given double integral is \[ \int_{0}^{1} \int_{-1}^{x}(x^{2}+y^{2}) \ d y \ d x \] This tells us to integrate with respect to \( y \) first, while keeping \( x \) constant, and then integrate the result with respect to \( x \).

Step 2: Integrate with Respect to y

Integrate the inner integral \( \int_{-1}^{x}(x^{2}+y^{2}) \ d y \) which breaks down to:\( \int_{-1}^{x} x^{2} \ d y + \int_{-1}^{x} y^{2} \ d y \)Calculate each component:- \( \int x^{2} \ d y = x^{2}y \) - \( \int y^{2} \ d y = \frac{y^3}{3} \)Evaluate from \( y = -1 \) to \( y = x \):\[ x^{2}x + \frac{x^3}{3} - \left(x^{2}(-1) + \frac{(-1)^3}{3}\right) \] simplifying gives:\[ x^{3} + \frac{x^{3}}{3} + x^{2} + \frac{1}{3} \]

Step 3: Integrate with Respect to x

Now integrate\[ \int_{0}^{1}\left(x^{3} + \frac{x^{3}}{3} + x^{2} + \frac{1}{3}\right) \ d x \]Break it into parts:- \( \int x^{3}\ d x = \frac{x^4}{4} \)- \( \int \frac{x^{3}}{3}\ d x = \frac{x^4}{12} \)- \( \int x^{2}\ d x = \frac{x^3}{3} \)- \( \int \frac{1}{3}\ d x = \frac{x}{3} \)Evaluate from \( x = 0 \) to \( x = 1 \):\[ \left(\frac{1^4}{4} + \frac{1^4}{12} + \frac{1^3}{3} + \frac{1}{3}\right) - \left(\frac{0^4}{4} + \frac{0^4}{12} + \frac{0^3}{3} + \frac{0}{3}\right) \]Simplifying gives\[ \frac{1}{4} + \frac{1}{12} + \frac{1}{3} + \frac{1}{3} \]Combine the fractions to find the final result.

Step 4: Simplify the Result

Add the fractions obtained:- Convert each fraction to have the same denominator:\[ \frac{1}{4} = \frac{3}{12} \]\[ \frac{1}{3} = \frac{4}{12} \]Add the fractions:\[ \frac{3}{12} + \frac{1}{12} + \frac{4}{12} + \frac{4}{12} = \frac{12}{12} \]Therefore, the value of the original integral is 1.

Key Concepts

Order of Integration

When faced with a double integral, it is crucial to understand the order in which you should perform the integrations. The notation of the integral gives us a clue. In our given problem, we have two integrals written as \( \int_{0}^{1} \int_{-1}^{x}(x^{2}+y^{2}) \, d y \, d x \). The innermost integral \( \int_{-1}^{x} (x^2 + y^2) \, dy \) suggests starting with integration with respect to \( y \).

This order of operations - integrating first with respect to \( y \) and subsequently with \( x \) - ensures computations align with the specified limits of integration.

• Integrating with respect to \( y \) means treating \( x \) as a constant while varying \( y \) from \( -1 \) to \( x \).
• Once the \( y \)-integration completes, the resultant expression involves \( x \) and is integrated from \( 0 \) to \( 1 \).

This sequence forms the correct process of evaluating double integrals by respecting their integration order.

Iterated Integrals

Double integrals, such as the one we are working with, can be conceptualized as iterated integrals. Iterated integrals refer to performing integration over repeated applications in a sequence. In this exercise, we ran through nested integrals, where we first integrated with respect to \( y \) and then with \( x \).

Iterated integrals can be understood as layers:

• The "inner layer" is \( \int_{-1}^{x} (x^2 + y^2) \, dy \), where only \( y \) varies. The limits here depend on \( x \), indicating a \( y \)-changing range between \( -1 \) and \( x \).
• The "outer layer" involves \( \int_{0}^{1} \cdots \ dx \), processing the \( y \)-integrated result across the \( x \) bounds from \( 0 \) to \( 1 \).

This approach segregates complex calculations into simpler, single-variable integrals, making it easier to solve composite regions by processing iterated limits.

Integration with Respect to y

In working through double integrals, the first integration step, as prescribed by the order, was with respect to \( y \). This involves treating any other variables, in this case \( x \), as constants. The integral \( \int_{-1}^{x} x^{2} \, dy + \int_{-1}^{x} y^{2} \, dy \) separates our problem into digestible parts.

To integrate with respect to \( y \):

• Integrating \( x^2 \, dy \), which treats \( x^2 \) as a constant, produces \( x^2y \).
• Integrating \( y^2 \, dy \) results in \( \frac{y^3}{3} \), following the power rule for integration.
• The evaluation occurs between \( y = -1 \) and \( y = x \) to incorporate specific limits.

The process is vital for correctly setting up the next steps in the iterative solution, ensuring each piece affects the final function appropriately as input for subsequent \( x \)-integration.