Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals in Problems 1-14. \(\int_{0}^{2} \int_{-x}^{x} e^{-x^{2}} d y d x\)

Step-by-Step Solution

Answer

\(1 - \frac{1}{e^4}\)

Step 1: Analyze the Inner Integral

First, look at the inner integral: \[ \int_{-x}^{x} e^{-x^{2}} dy \].Since there is no \(y\) in the integrand, \(e^{-x^2}\) is a constant with respect to \(y\). The integral of a constant \(c\) with respect to \(y\) over the interval \([-x, x]\) is simply \(c(y_{upper} - y_{lower})\).

Step 2: Evaluate the Inner Integral

Now we evaluate the inner integral:\[ \int_{-x}^{x} e^{-x^{2}} dy = e^{-x^2} [y]_{-x}^{x} = e^{-x^2} (x - (-x)) = e^{-x^2} (2x).\]

Step 3: Replace the Inner Integral Result into the Outer Integral

With the result from the inner integral, we replace it into the outer integral:\[ \int_{0}^{2} 2x e^{-x^2} dx.\]

Step 4: Apply Substitution for Simplification

Use the substitution method where let \(u = -x^2\), giving us \(du = -2x dx\). Therefore, \(2x dx = -du\).Change the limits of integration. When \(x = 0\), \(u = 0\) and when \(x = 2\), \(u = -4\).The integral becomes:\[ \int_{0}^{-4} -e^{u} du.\] Use the property \( \int_{a}^{b} -f(u) du = \int_{b}^{a} f(u) du \), which allows us to rewrite it as:\[ \int_{-4}^{0} e^{u} du. \]

Step 5: Evaluate the Integral

Now, integrate \( e^u \):\[ \int_{-4}^{0} e^{u} du = [e^{u}]_{-4}^{0} = e^{0} - e^{-4} = 1 - \frac{1}{e^4}.\]

Step 6: Finalize the Solution

The value of the iterated integral is thus \( 1 - \frac{1}{e^4} \).

Key Concepts

Inner Integral Evaluation

When dealing with iterated integrals, it's essential to understand the concept of evaluating an inner integral first. In this context, the inner integral is \( \int_{-x}^{x} e^{-x^2} dy \).
The term \( e^{-x^2} \) doesn't depend on \( y \), so we can treat it as a constant concerning this particular integrand.
This simplifies our problem significantly. When you integrate a constant \( c \) with respect to the variable \( y \) over an interval \([y_{lower}, y_{upper}]\), it simplifies to \( c(y_{upper} - y_{lower}) \).

• In this example, the limits of integration for \( y \) are \( -x \) to \( x \), leading to \( x - (-x) \).
• Hence, the evaluation of the integral becomes \( e^{-x^2} (2x) \).

This technique allows you to focus on the variable of interest and treat all other parts as constants, streamlining the computation process.

Substitution Method

The substitution method is a powerful tool in the integration toolkit, particularly useful when simplifying complex integrals. In our exercise, we're looking to solve \( \int_{0}^{2} 2x e^{-x^2} dx \). To simplify this, we use substitution.

• Let \( u = -x^2 \), which necessitates that \( du = -2x dx \).
• This transforms \( 2x dx \) into \( -du \).

Now, you have a simpler integral: \( \int_{0}^{-4} -e^{u} du \). Don’t forget to change the integration limits.
With \( u = -x^2 \), the original limits of \( x = 0 \) and \( x = 2 \) change \( u \) to \( 0 \) and \( -4 \) respectively.
After applying substitution, you're ready to evaluate the new, and often much simpler, integral.

Integration Limits

Understanding and adjusting the integration limits is crucial when changing variables in an integral.
When using the substitution method, as shown in our exercise with \( u = -x^2 \), it changes the limits from the \( x \) variable to the \( u \) variable.

• Originally, when \( x = 0 \), substitute into \( u = -x^2 \) and find \( u = 0 \).
• When \( x = 2 \), substitute similarly to find \( u = -4 \).

Thus, the limits shift, and your integral becomes \( \int_{0}^{-4} -e^{u} du \).
However, to maintain conventional order (lower to upper), utilize the property: \( \int_{a}^{b} -f(u) du = \int_{b}^{a} f(u) du \), transforming it to \( \int_{-4}^{0} e^{u} du \).
Tracking limits correctly ensures your integral remains valid and is evaluated over the right interval.

Definite Integrals

Lastly, let's dive into definite integrals, which compute the exact area under a curve between two limits.
For the integral \( \int_{-4}^{0} e^{u} du \), perform straightforward integration.
The antiderivative of \( e^{u} \) is itself \( e^{u} \). So integrating from \( -4 \) to \( 0 \), we have:

• The evaluation becomes \( [e^{u}]_{-4}^{0} \).
• This computes as \( e^{0} - e^{-4} = 1 - \frac{1}{e^4} \).

Definite integrals provide a finite result, unlike indefinite integrals, which provide a general form with an added constant \( C \).
They are the backbone of finding areas and understanding the accumulation of quantities across intervals.