91Ó°ÊÓ

When tackling an iterated integral such as \( \int_{-2}^{2} \int_{0}^{2} x e^{-y} \, dx \, dy \), the first step involves handling the integration with respect to \(x\). In this phase, everything except \(x\) is considered a constant. For the given integral, this means treating \(e^{-y}\) as constant. To find the integral of \(x e^{-y}\) with respect to \(x\), focus on the integration of \(x\).

• The antiderivative of \(x\) is \( \frac{x^2}{2} \).
• Apply the limits from 0 to 2, converting \( \int_{0}^{2} x e^{-y} \, dx \) into:

Replacing \(x\) with its upper and lower limits results in: \[\int_{0}^{2} x e^{-y} \, dx = e^{-y} \cdot \left[ \frac{x^2}{2} \right]_0^2 = e^{-y} \cdot \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = 2 e^{-y}.\] Here, the integral evaluates to \(2 e^{-y}\), simplifying the expression significantly for the next integration step.

Upon successfully integrating with respect to \(x\), the task now is to integrate the resulting expression \(2 e^{-y}\) with respect to \(y\). This involves calculating the antiderivative of \(2 e^{-y}\). Remember, the exponential function \(e^{-y}\) simplifies to a neat expression when integrated.

• The antiderivative of \(2 e^{-y}\) is \(-2 e^{-y}\).
• Apply the limits of integration from \(-2\) to \(2\).

Computing this, we have:\[\int_{-2}^{2} 2 e^{-y} \, dy = \left[ -2 e^{-y} \right]_{-2}^2 = -2 e^{-2} - (-2 e^{2}) = -2 e^{-2} + 2 e^{2}.\]The output of this integration step is \(-2 e^{-2} + 2 e^{2}\), which captures the variation across the defined range of \(y\). This final result is elegant and encapsulates the nature of exponential decay and growth.

Understanding the antiderivative is fundamental when dealing with iterated integrals. An antiderivative is a reverse engineering of differentiation, leading to the original function before differentiation. For each function you are integrating, discover its antiderivative to proceed with integration.
To break this down:

• The antiderivative of \(x\) is \( \frac{x^2}{2} \), useful in integration with respect to \(x\).
• The antiderivative of \(e^{-y}\) as seen in \(2 e^{-y}\) is \(-2 e^{-y}\).

Remember that integrating involves reversing the rules used in differentiation. Identifying the antiderivatives lets you evaluate definite integrals accurately. They define how the function changes over an interval, leading to the final solution of the integral. In our example, using antiderivatives led us to a simplified answer: \(2 (e^{2} - e^{-2})\). Integrating successfully relies on these calculations, guiding you through each step efficiently.