Problem 47 - AP Calculus with 8 Practice Tests 15th

Question

\(\int x^{2} e^{x} d x=\) (A) \(e^{x}\left(x^{2}+2 x\right)+C\) (B) \(e^{x}\left(x^{2}-2 x-2\right)+C\) (C) \(e^{x}\left(x^{2}-2 x+2\right)+C\) (D) \(e^{x}(x-1)^{2}+C\)

Step-by-Step Solution

Answer

The correct option is (C): \( e^{x}(x^{2}-2x+2)+C \).

Step 1: Identify the Method

When you encounter an integral of the form \( \int x^n e^x \, dx \), it's typically solved using integration by parts. This method is suitable because we have a product of two functions: a polynomial and an exponential function.

Step 2: Set Up Integration by Parts

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). We choose \( u = x^2 \) and \( dv = e^x \, dx \). Thus, \( du = 2x \, dx \) and \( v = e^x \).

Step 3: Apply Integration by Parts

Substitute into the integration by parts formula: \[ \int x^2 e^x \, dx = x^2 e^x - \int e^x \cdot 2x \, dx. \]Now, we need to calculate \( \int 2x e^x \, dx \) using integration by parts again.

Step 4: Integration by Parts for \( \int 2x e^x \, dx \)

Use integration by parts again, choosing \( u = 2x \) and \( dv = e^x \, dx \). So, \( du = 2 \, dx \) and \( v = e^x \). The integral becomes: \[ \int 2x e^x \, dx = 2x e^x - \int 2e^x \, dx. \] This simplifies to: \[ 2x e^x - 2e^x + C. \]

Step 5: Substitute Back and Simplify

Substitute back into our first application of integration by parts: \[ \int x^2 e^x \, dx = x^2 e^x - (2x e^x - 2e^x) = e^x (x^2 - 2x + 2) + C. \]

Step 6: Choose the Correct Option

From our simplified result, we have: \[ e^x (x^2 - 2x + 2) + C. \] Comparing with the options, the correct choice is (C): \( e^{x}(x^{2}-2x+2) + C \).

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of calculus focused on finding the antiderivative or integral of functions. The goal is to determine the area under a curve on a graph, which is especially useful in physics and engineering.
For integration, we usually look for functions whose derivative gives us the integrand. However, when dealing with more complex expressions, we often require specialized techniques, such as integration by parts.

• Integration by parts is a key technique derived from the product rule for differentiation. It leverages the relationship: \[ \int u \, dv = uv - \int v \, du \]
• This method decomposes a difficult integral into simpler parts, often allowing us to evaluate integrals involving the product of different types of functions.
• In our example, integration by parts was used twice, first to handle the initial polynomial-exponential product, and again to simplify further.

Integral calculus is a powerful tool that allows us to solve problems involving areas, volumes, and many other concepts.

Exponential Functions

Exponential functions, such as \( e^x \), are mathematical expressions where a constant base is raised to a variable exponent. These functions have unique properties that make them particularly interesting in calculus.
When integrating or differentiating exponential functions, they exhibit consistent and simple behavior:

• The derivative of \( e^x \) is itself, \( e^x \).
• Similarly, the integral of \( e^x \) is \( e^x + C \), where \( C \) is the constant of integration.

In our integral, \( x^2 e^x \), the presence of \( e^x \) allows us to use integration by parts effectively. The simplicity of exponential functions makes complex problems more manageable. They frequently appear in growth and decay models, making them vital across scientific disciplines.

Polynomial Integration

Polynomial functions are algebraic expressions consisting of variables raised to whole number exponents. Integrating polynomial functions involves finding a function whose derivative is the polynomial in question.
In our exercise with \( x^2 e^x \), we face a polynomial \( x^2 \) that multiplies an exponential function. The polynomial component requires us to apply special techniques like integration by parts because:

• Direct integration of polynomials simply involves increasing the exponent by one and dividing by the new exponent. For example, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
• When combined with other functions, like exponents, this simplicity is disrupted, necessitating methods that handle the interaction between polynomial and non-polynomial terms.

In polynomial integration, mastering techniques like integration by parts allows you to handle these interactions effectively, making complex integrals more approachable and solvable.