Question

In Exercises \(43-46\) , evaluate the integral by using a substitution prior to integration by parts. $$\int x^{7} e^{x^{2}} d x$$

Short answer

The solution to the integral \(\int x^{7} e^{x^{2}} dx\) is \( \frac{1}{2} ( x^6 e^{x^2} - 3x^4 e^{x^2} + 6x^2 e^{x^2} - 6e^{x^2} ) + C \)

Step-by-step solution

Substitute

Firstly, employ the substitution strategy. Let's substitute \(u = x^{2}\). Differentiate both sides with respect to \(x\) to obtain \(du = 2x dx\). This means that \(x dx = \frac{du}{2}\). Substitute into the given integral, replacing \(x^{6} dx\) with \(\frac{u^{3} du}{2}\). Now the integral becomes \(\frac{1}{2} \int u^{3} e^{u} du\).

Integration by Parts

Next, apply the integration by parts formula, which states, \(\int u v dx = u v - \int v du\). In this case, consider \(v = e^{u}\) and \(u = u^{3}\), then \(dv = e^{u} du\) and \(du = 3u^{2} du\). Applying the formula we get: \(\frac{1}{2} ( u^{3} e^{u} - \int 3u^{2} e^{u} du )\).

Repeat Substitution and Integration by Parts

Notice that the result from Step 2 still results in a complex integral that can't be computed directly. Therefore, apply integration by parts again, taking \(v = e^{u}\) and \(u = 3u^{2}\), then \(dv = e^{u} du\) and \(du = 6u du\). Now the integral becomes, \(\frac{1}{2} ( u^3 e^u - 3u^2 e^u + \int 6u e^u du )\).

Simplify and Final Substitution

Further simplification and one last round of integration by parts, this time choosing \(v = e^u\) and \(u = 6u\) yields the simplified integral: \( \frac{1}{2} ( u^3 e^u - 3u^2 e^u + 6u e^u - 6e^u ) \). Now, substitute \(u = x^2\) back into the equation. The result is: \( \frac{1}{2} ( x^6 e^{x^2} - 3x^4 e^{x^2} + 6x^2 e^{x^2} - 6e^{x^2} ) \).

Key concepts

U-Substitution in Integrals

U-substitution is a powerful technique for evaluating integrals that might initially seem intractable. Imagine you're trying to fit a square peg into a round hole — that's what you face when an integral doesn't match any standard form you know. U-substitution is like finding the right key for a lock.

Here's how you can unlock integrals with u-substitution: first, identify a portion of the integrand that can be simplified or transformed into a ‘u’. This usually involves an inside function that’s part of a composition, like the inner 'x' in \( e^{x^2} \) from our exercise. You then differentiate this inner function with respect to 'x' to find 'du'. The goal is to rewrite the entire integral in terms of 'u' and 'du'.

This change of variables simplifies the problem and often reveals a more recognizable form. As seen in the solution provided, by substituting \( u = x^2 \) and \( du = 2x dx \) the intractable integrand becomes a friendlier \( u^3 e^u \) which is easier to integrate.

Integration by Parts

Integration by parts is a go-to method when dealing with products of functions. Think of it as the integration counterpart to the product rule of differentiation. The mnemonic 'UV minus integral V DU' stems from the formula \( \int u v dx = u v - \int v du \).

To deploy this strategy, you split the integrand into two parts: \( u \) (which you will differentiate) and \( dv \) (which you will integrate). It often requires some finesse to choose 'u' and 'dv' wisely. In our example, we opted for \( u = u^3 \) and \( dv = e^u du \) leading to integrating \( v = e^u \) and differentiating \( u = u^3 \) which provides \( du = 3u^2 du \).

Through iterative application, as seen in the example solution, integration by parts can reduce a complex integral to simpler terms or to another integral that is easier to solve.

Indefinite Integral Calculation

An indefinite integral, represented by the integral sign without limits, is essentially the antiderivative of a function. It represents a family of functions rather than a single value and is used to find the general form of the original function before differentiation.

In the context of our exercise, after substitutions and applying integration by parts, we're left with the indefinite integral calculation to find the general antiderivative of \( e^{x^2} \) times a polynomial in \( x \). The calculations reveal the antiderivative up to an arbitrary constant, as is the nature of indefinite integrals. The resulting antiderivative is a blend of an exponential function and a polynomial — a result that would be much harder to achieve without the initial substitution and integration by parts techniques.

Exponential Functions Integration

Exponential functions, such as \( e^{x} \), are unique because they are their own derivative. Integrating exponential functions, particularly those with a twist like \( e^{x^2} \), can pose a significant challenge. In the given exercise, the integration of \( e^{x^2} \) cannot be done directly.

However, the substitution we used simplifies the exponential term into \( e^u \) and from there, it integrates to itself. When dealing with exponential functions in integration, be on the lookout for substitutions or alternative methods like integration by parts that can transform the integral into a friendlier form. The final expression after integrating an exponential function often retains the exponential form, showcasing this special property.