Problem 35 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int x^{3} e^{x^{2}} d x \quad\left[\right.\)and use a substitution to find \(v\) from \(d v\).

Step-by-Step Solution

Answer

The integral is \( \frac{1}{2} x^2 e^{x^2} - \int e^{x^2} dx + C \).

Step 1: Choose substitution for integration by parts

We will use integration by parts, which is based on the formula \( \int u \, dv = uv - \int v \, du\). First, identify \(u\) and \(dv\). We choose \(u = x^2\) and therefore \(du = 2x\, dx\). This choice makes substitution easier in later steps.

Step 2: Substitute and differentiate given expressions

With the chosen \(u\), \(du\) becomes \(2x \, dx\). Also, express \(dv = x \, e^{x^2} \, dx\). We need to find \(v\) by integrating \(dv\).

Step 3: Use substitution to find \(v\)

To find \(v\), perform the substitution \(w = x^2\), which gives \(dw = 2x \, dx\) or \(x \, dx = \frac{1}{2} \, dw\). Hence, \(dv = \frac{1}{2} e^w \, dw\), so integrating gives \(v = \frac{1}{2} \int e^w \, dw = \frac{1}{2} e^w = \frac{1}{2} e^{x^2}\).

Step 4: Apply the integration by parts formula

Replace \(u\), \(v\), \(du\), and \(dv\) into the integration by parts formula. Substitute back into the expression: \( \int x^3 e^{x^2} dx = \left( x^2 \cdot \frac{1}{2} e^{x^2} \right) - \int \left( \frac{1}{2} e^{x^2} \cdot 2x \right) dx \).

Step 5: Simplify and solve the remaining integral

Simplify the equation to get \( \frac{1}{2} x^2 e^{x^2} - \int e^{x^2} dx \). The integral \(\int e^{x^2} dx\) cannot be expressed in terms of elementary functions, so it remains part of the solution.

Step 6: Compile the final solution

The integral of \( \int x^3 e^{x^2} dx \) gives \( \frac{1}{2} x^2 e^{x^2} - \int e^{x^2} dx + C\), where \(C\) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on integrating functions to find areas, volumes, and other quantities under curves. It is often paired with differential calculus, as both explore change and accumulation, respectively.
One of the key objectives when solving integrals is to find an

• antiderivative, which is the function that was differentiated to produce a given derivative.
• For instance, integrating a constant function yields a linearly increasing antiderivative.
• Similarly, integrating a polynomial yields a higher degree polynomial.

Understanding integrals is crucial for solving complex calculus problems and real-world applications, from physics to engineering.

Integration Techniques

Integration techniques are methods used to solve complex integrals when they cannot be solved directly. Several common techniques include

• substitution, which simplifies an integral by changing variables,
• integration by parts, perfect for products of functions,
• partial fraction decomposition, breaking down rational functions,
• Trigonometric substitution for integrals involving square roots.

Integration by parts, for instance, is particularly helpful when dealing with products of algebraic and exponential or trigonometric functions. Knowing which technique to apply can simplify integration significantly and lead to successful problem solving.

Definite Integrals

Definite integrals represent the exact area under a curve between two specified limits. This area is calculated using the limits of integration, denoted as the upper and lower limits. Unlike indefinite integrals, definite integrals do not include a constant of integration, and are represented as \[ \\int_{a}^{b} f(x) \, dx \\]where \(a\) and \(b\) are the limits.
Definite integrals are widely applicable:

• They provide exact physical quantities, such as displacements and total stored energy.
• This makes them valuable in sciences and engineering for precise measurements.
• For example, finding the area under a velocity-time curve gives the object's displacement.

Understanding definite integrals bridges the gap between theoretical calculus and practical implementation.

Substitution Method

The substitution method, or u-substitution, is akin to reverse differentiation. This important technique simplifies integrals by changing variables. It is particularly useful when functions are difficult to integrate directly.
Here's a general guide:

• Identify a part of the integral, \( u \), that simplifies the integral once substituted.
• Calculate \( du \), which is the differential of \( u \).
• Replace the original variable and integral with \( u \) and \( du \).
• Integrate the transformed function, which is often easier.
• Re-substitute the original variable back into the integral.

The substitution method is like a puzzle. Its solution often requires a keen understanding of function relationships to find the simplest form.