Problem 35 - University Calculus: Early Transcendentals

Question

Guess an antiderivative for the integrand function. Validate your guess by differentiation and then evaluate the given definite integral. (Hint: Keep in mind the Chain Rule in guessing an antiderivative. You will learn how to find such antiderivatives in the next section.) $$\int_{0}^{1} x e^{x^{2}} d x$$

Step-by-Step Solution

Answer

The integral results in \( \frac{1}{2} (e - 1) \).

Step 1: Guess an Antiderivative

For the integrand function \( x e^{x^2} \), try to guess an antiderivative by inspecting it. Notice that the derivative of \( x^2 \) is \( 2x \), and the function \( e^{x^2} \) is related to the Chain Rule. Therefore, a reasonable guess for the antiderivative is \( \frac{1}{2} e^{x^2} \).

Step 2: Differentiate the Antiderivative

Differentiate your guessed antiderivative \( \frac{1}{2} e^{x^2} \) using the chain rule. The derivative is: \( \frac{d}{dx}\left(\frac{1}{2} e^{x^2}\right) = \frac{1}{2} \cdot e^{x^2} \cdot 2x = x e^{x^2} \). This shows that our guess is correct because we get back the original integrand.

Step 3: Evaluate the Definite Integral

Now that we have confirmed our antiderivative, evaluate the definite integral \( \int_{0}^{1} x e^{x^2} \, dx \) by calculating \( \left[ \frac{1}{2} e^{x^2} \right]_{0}^{1} \).Calculate at the upper limit: \( \frac{1}{2} e^{1^2} = \frac{1}{2} e \). Calculate at the lower limit: \( \frac{1}{2} e^{0^2} = \frac{1}{2} \times 1 = \frac{1}{2} \).Subtract: \( \frac{1}{2} e - \frac{1}{2} = \frac{1}{2} (e - 1) \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus used for finding the derivative of composite functions, where one function is nested inside another. When dealing with integrals like \(\int x e^{x^2} \, dx\), the Chain Rule helps us identify connections between the integrand and possible antiderivatives.

For example, notice in the problem how \(e^{x^2}\) is a "nested" function within the problem. Here, the derivative of the inner function \(x^2\) is \(2x\), which hints at using the Chain Rule to guess a suitable antiderivative. This is because \(x\) appears in the original expression, implying that differentiating \(e^{x^2}\) will naturally lead to including \(x\) in the expression.

• Recognizing these nested patterns is critical for applying the Chain Rule efficiently.
• It provides insight into guessing antiderivatives, helping to check if they differentiate back to the original integrand.

Antiderivative

In Integral Calculus, finding an antiderivative means identifying a function whose derivative is the given function. For \(x e^{x^2}\), we guessed \(\frac{1}{2} e^{x^2}\) as an antiderivative. This step often involves:

• Recognizing patterns and connections, such as relationships to common derivative forms.
• Using knowledge of differentiation techniques like the Chain Rule to hypothesize an expression.

Guessing an antiderivative might feel like a trial and error process at first, but with practice, it becomes intuitive. We apply this by recognizing that the antiderivative, once differentiated, should restore the original integrand. Validation through differentiation ensures the guess is correct and not merely coincidental. For example:

After differentiating \(\frac{1}{2} e^{x^2}\), we found \(x e^{x^2}\) as expected, establishing the correctness of our guess.

Definite Integral

Definite integrals represent the accumulation of a quantity between two points, providing a numerical value. In this problem, after confirming the antiderivative, use it to evaluate the definite integral \(\int_{0}^{1} x e^{x^2} \, dx\). This process involves:

• Substituting the upper and lower bounds into the antiderivative to find the accumulation over that interval.
• Calculating the difference between these two values.

For \(\int_{0}^{1} x e^{x^2} \, dx\), it means calculating \(\left[ \frac{1}{2} e^{x^2} \right]_0^1\), resulting in:

Calculate at \(x = 1\): \(\frac{1}{2} e^{1^2} = \frac{1}{2} e\).
Calculate at \(x = 0\): \(\frac{1}{2} e^{0^2} = \frac{1}{2}\). Subtracting gives us the definite integral's value: \(\frac{1}{2} (e - 1)\).

• This value represents the total area under the curve from \(x = 0\) to \(x = 1\) for the given function.