91Ó°ÊÓ

The antiderivative, also known as the indefinite integral, is the reverse process of differentiation. It helps us to find a function whose derivative is the given function. When dealing with iterated integrals, particularly the inner integral, finding the antiderivative is crucial.

For example, if you have the function \(x + y\), to integrate it with respect to \(y\), you treat \(x\) as a constant. The antiderivative of \(x + y\) with respect to \(y\) is \(xy + \frac{y^2}{2}\).

Notice that each component of \(x + y\) is integrated separately.

• The constant term \(x\) becomes \(xy\), since the integral of a constant \(k\) with respect to \(y\) is \(ky\).
• The term \(y\) becomes \(\frac{y^2}{2}\), using the power rule \(\int y^n dy = \frac{y^{n+1}}{n+1}\), where \(n=1\).

Understanding antiderivatives is essential for calculating integrals, as it's the main tool we use to reverse the derivative operation and solve integrals.

Integration bounds determine the limits over which the integral is calculated and are represented by the numbers next to the integral sign. They define the interval for the variable you are integrating.

In the context of iterated integrals, this process occurs twice.

• For the inner integral, for instance, \(\int_{0}^{2}(x+y)\ dy\), the bounds \(0\) and \(2\) refer to the values of \(y\) from \(0\) to \(2\).
• The outer integral \(\int_{1}^{3}(2x + 2)\\ dx\) uses bounds \(1\) to \(3\) to evaluate \(x\).

Each boundary helps in capturing the full area or volume beneath a curve or surface over the specified interval.

Without proper bounds, the integration would not provide the precise value within the intended range.

The inner integral is the first integration to be evaluated in an iterated integral problem, focusing on one variable while treating others as constants.

For the provided problem, consider the inner integral first: \(\int_{0}^{2}(x+y)\ dy\).

• It involves integrating the function \(x+y\) with respect to \(y\), and treating \(x\) as a constant throughout.
• The antiderivative is calculated as \(xy + \frac{y^2}{2}\) before applying the bounds.
• After calculating the antiderivative, evaluate it from \(y=0\) to \(y=2\) giving you \(2x + 2\).

This resulting expression \(2x + 2\) then becomes the integrand for the outer integral.

The outer integral is the second and final step in evaluating an iterated integral. It uses the result from the inner integral as its integrand, and integrates over the second set of bounds.

In our problem, after solving the inner integral \(\int_{0}^{2}(x+y)\\ dy\) to be \(2x + 2\), the outer integral becomes: \(\int_{1}^{3}(2x + 2)\ dx\).

• This involves integrating with respect to \(x\).
• The antiderivative of \(2x + 2\) with respect to \(x\) is \(x^2 + 2x\).
• After calculating the antiderivative, apply the bounds \(x=1\) to \(x=3\), resulting in a final value of \(12\).

Understanding the role of the outer integral is important as it completes the process of finding the area or volume defined by the iterated integral.