Problem 8 - APEX Calculus

Question

Evaluate the integral and subsequent iterated integral. (a) \(\int_{y}^{y^{2}}(x-y) d x\) (b) \(\int_{-1}^{1} \int_{y}^{y^{2}}(x-y) d x d y\)

Step-by-Step Solution

Answer

(a) \( \frac{y^4}{2} - y^3 + \frac{y^2}{2} \); (b) \( \frac{8}{15} \).

Step 1: Evaluate the Inner Integral

For part (a), we want to evaluate the integral \( \int_{y}^{y^2} (x-y) \, dx \). We perform the integration with respect to \(x\). The integral of \(x-y\) with respect to \(x\) is \(\frac{x^2}{2} - yx\). Apply the definite limits:\[ \int_{y}^{y^2} (x-y) \, dx = \left[ \frac{x^2}{2} - yx \right]_y^{y^2} = \left( \frac{y^4}{2} - y \cdot y^2 \right) - \left( \frac{y^2}{2} - y^2 \right) \].

Step 2: Simplify the Expression

Now, substitute the limits into the expression:\[ \left( \frac{y^4}{2} - y^3 \right) - \left( \frac{y^2}{2} - y^2 \right) = \left( \frac{y^4}{2} - y^3 \right) - \frac{y^2}{2} + y^2 \].Simplifying further, we get:\[ \frac{y^4}{2} - y^3 - \frac{y^2}{2} + y^2 = \frac{y^4}{2} - y^3 + y^2 - \frac{y^2}{2} \].Combine like terms:\[ \frac{y^4}{2} - y^3 + \frac{y^2}{2} \].

Step 3: Evaluate the Iterated Integral

For part (b), we need to evaluate the iterated integral \( \int_{-1}^{1} \int_{y}^{y^2} (x-y) \, dx \, dy \). We already found the inner integral in Step 1 as \( \frac{y^4}{2} - y^3 + \frac{y^2}{2} \). Now integrate this expression with respect to \(y\):\[ \int_{-1}^{1} \left( \frac{y^4}{2} - y^3 + \frac{y^2}{2} \right) \, dy \].The antiderivative is:\[ \frac{1}{2} \cdot \frac{y^5}{5} - \frac{y^4}{4} + \frac{1}{2} \cdot \frac{y^3}{3} \].Evaluate from \(-1\) to \(1\):\[ \left( \frac{y^5}{10} - \frac{y^4}{4} + \frac{y^3}{6} \right) \bigg|_{-1}^{1} \].

Step 4: Calculate the Definite Iterated Integral

Substitute the limits into the expression:At \(y=1\):\[ \frac{1^5}{10} - \frac{1^4}{4} + \frac{1^3}{6} = \frac{1}{10} - \frac{1}{4} + \frac{1}{6} \].Find a common denominator (60):\[ \frac{6}{60} - \frac{15}{60} + \frac{10}{60} = \frac{1}{60} \].At \(y=-1\):\[ \frac{(-1)^5}{10} - \frac{(-1)^4}{4} + \frac{(-1)^3}{6} = \frac{-1}{10} - \frac{1}{4} - \frac{1}{6} \].Find a common denominator (60):\[ \frac{-6}{60} - \frac{15}{60} - \frac{10}{60} = \frac{-31}{60} \].Now compute the difference:\[ \frac{1}{60} - \left(\frac{-31}{60}\right) = \frac{1}{60} + \frac{31}{60} = \frac{32}{60} = \frac{8}{15} \].

Key Concepts

definite integrals

A definite integral is a crucial concept in calculus that calculates the area under a curve within a certain interval. To evaluate a definite integral, you find the antiderivative (or the integral) of the function, then substitute the upper and lower limits of the interval into this antiderivative, and calculate the difference.

In the original problem, the expression involved evaluating the definite integral \[ \int_{y}^{y^2} (x-y) \, dx \]First, we determined the antiderivative of the given function with respect to \(x\) and then evaluated it at the bounds. This process applies the Fundamental Theorem of Calculus, which links the concept of integration with differentiation.

• You first integrate the function.
• Next, apply the upper and lower limits of integration.
• Find the difference.

Understanding definite integrals helps you handle problems that involve finding total quantities such as areas, volumes, and other accumulations.

antiderivatives

An antiderivative is essentially the reverse of differentiation. It is a function whose derivative is the original function we started with. Calculating an antiderivative is a foundational step in solving definite integrals.

To find the antiderivative of \(x-y\) in the expression \[ \int (x-y) \, dx \]we perform integration term-by-term.

• The antiderivative of \(x\) with respect to \(x\) is \(\frac{x^2}{2}\).
• The antiderivative of \(-y\) with respect to \(x\) is \(-yx\), as \(-y\) is treated as a constant in this context.

Combining these results gives us: \[ \frac{x^2}{2} - yx \].
Antiderivatives are vital because they serve as a basis for calculating definite integrals by allowing you to find exact values over an interval.

integration by substitution

Integration by substitution is a technique used to simplify integrals by changing variables, which makes it easier to integrate certain functions. It's a useful method when a direct integration approach seems complicated or impossible.

Though substitution was not explicitly used in the solution of the original problem, it is worth noting that substitution could simplify similar integrals. The technique involves:

• Selecting a part of the integrand to replace with a new variable.
• Changing the limits of integration accordingly if they are definite integrals.
• Transforming the differential accordingly.

For example, if you have an integral like \(\int (x - y) dx\), you might set \(u = x - y\) and then integrate in terms of \(u\). This approach makes the integration process manageable.
Understanding integration by substitution not only broadens your integration tool belt but also deepens your comprehension of how integrals can transform under variable changes.

limits of integration

Limits of integration determine the interval over which you evaluate a definite integral. They are pivotal in computing definite integrals, as they define the start and end points of the area or accumulation you are interested in.

In the original problem, the inner integral \[ \int_{y}^{y^2} (x-y) \, dx \] involves limits from \(y\) to \(y^2\). These limits might depend on another variable which is often seen in nested or iterated integrals. The limits dictate how you evaluate the antiderivative, producing specific values when applied.

• Upper and lower limits indicate the exact portion of the curve considered.
• These limits must reflect any changes due to substitution.

In the context of iterated integrals, like part (b) of the problem, each integration step has its own limits. You process the inner integral limits first, then proceed to the next integration layer. Calculating the difference of the evaluated antiderivative at these limits ultimately provides the definite integral's value.