Problem 21 - Applied Calculus

Question

For each of the following integrals, indicate whether integration by substitution or integration by parts is more appropriate. Do not evaluate the integrals. (a) \(\int \frac{x^{2}}{1+x^{3}} d x\) (b) \(\int x e^{x^{2}} d x\) (c) \(\int x^{2} \ln \left(x^{3}+1\right) d x\) (d) \(\int \frac{1}{\sqrt{3 x+1}} d x\) (e) \(\int x^{2} \ln x d x\) (f) \(\int \ln x d x\)

Step-by-Step Solution

Answer

(a) Substitution (b) Substitution (c) Integration by Parts (d) Substitution (e) Integration by Parts (f) Integration by Parts

Step 1: Recognize the Pattern for Substitution

For substitution, look for integrands where the derivative of a function is a multiplicative component of the integrand itself. The integrals where substitution is suitable often have the form \( f(g(x)) g'(x) \) allowing for \( u=g(x) \).

Step 2: Apply to Integral (a)

Integral \( \int \frac{x^{2}}{1+x^{3}} \, dx \) suggests substitution. Choose \( u = 1 + x^3 \), hence \( du = 3x^2 \, dx \), which aligns well with the existing \( x^2 \) term.

Step 3: Apply to Integral (b)

Integral \( \int x e^{x^{2}} \, dx \) is suitable for substitution. Let \( u = x^2 \) so \( du = 2x \, dx \); the integrand \( x \) fits as part of \( du \).

Step 4: Recognize the Pattern for Integration by Parts

Integration by parts is most suitable for integrals of the form \( \int u \, dv \) where parts can be chosen easily as \( u \) and \( dv \). The formula is \( \int u \, dv = uv - \int v \, du \).

Step 5: Apply to Integral (c)

Integral \( \int x^{2} \ln(x^{3}+1) \, dx \) suggests parts due to the presence of \( \ln(x^{3}+1) \). Choose \( u = \ln(x^{3}+1) \) and \( dv = x^2 \, dx \).

Step 6: Apply to Integral (d)

Integral \( \int \frac{1}{\sqrt{3x+1}} \, dx \) suggests substitution. Choose \( u = 3x+1 \), hence \( du = 3 \, dx \); the integrand simplifies with this substitution.

Step 7: Apply to Integral (e)

Integral \( \int x^2 \ln x \, dx \) is appropriate for parts. Set \( u = \ln x \) and \( dv = x^2 \, dx \) to apply the formula effectively.

Step 8: Apply to Integral (f)

Integral \( \int \ln x \, dx \) requires integration by parts since \( \ln x \) does not straightforwardly fit substitution. Choose \( u = \ln x \), thus \( dv = dx \).

Key Concepts

Integration by Substitution

Integration by substitution is a clever technique used to simplify integrals, especially when the integral contains a composition of functions. Essentially, it involves changing the variable of integration to transform the integral into a simpler form. Here's how it works:

• Look for a part of the integrand that resembles a derivative of another part.
• Introduce a new variable, often called "u," to substitute for a function within the integral.
• Calculate the derivative of this substitution as "du."
• Rewrite the integral in terms of "u" and "du," then solve.

For example, in the integral \( \int \frac{x^{2}}{1+x^{3}} \, dx \), substitution by setting \( u = 1 + x^3 \) works because the derivative \( du = 3x^2 \, dx \) closely matches a component in the integrand, precisely the \( x^2 \, dx \) term, allowing the integral to simplify neatly.

Integration by Parts

Integration by parts is used when substitution doesn't easily simplify the integral. The method is based on the product rule of differentiation and is particularly useful for integrands that are products of functions. The basic formula is:
\[\int u \, dv = uv - \int v \, du\]
Here's how you apply integration by parts:

• Identify "u" as a part of the integrand that becomes simpler when differentiated.
• Let "dv" be the remaining part of the integrand, simplifying when integrated.
• Differentiate "u" to find "du," and integrate "dv" to find "v."
• Substitute into the integration by parts formula.

Consider \( \int x^{2} \ln x \, dx \). Choose \( u = \ln x \) since its derivative \( \frac{1}{x} \) simplifies, and \( dv = x^2 \, dx \) easily integrates to \( \frac{x^3}{3} \). This setup allows the integral to be solved using the integration by parts formula.

Definite Integrals

Definite integrals calculate the net area under a curve from a specific extit{a} to extit{b}. Unlike indefinite integrals, they yield a numerical value representing this area.

• The limits of integration, extit{a} and extit{b}, are crucial in determining the region under the curve.
• Often, you first find the indefinite integral, then apply the limits through the Fundamental Theorem of Calculus.
• This theorem states that if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_a^b f(x) \, dx = F(b) - F(a) \).

For instance, to find \( \int_0^1 x^2 \, dx \), first calculate the antiderivative, \( F(x) = \frac{x^3}{3} \), and then apply the limits: \( F(1) - F(0) = \frac{1^3}{3} - 0 = \frac{1}{3} \). This process provides the total area beneath the curve from 0 to 1.

Indefinite Integrals

Indefinite integrals, often called antiderivatives, are more abstract compared to definite integrals. They represent a family of functions whose derivatives yield the original function being integrated.

• Indefinite integrals do not have limits of integration and include a constant of integration, \( C \).
• Finding an indefinite integral is synonymous with finding an antiderivative.
• They are crucial for solving differential equations and for forming the basis of evaluating definite integrals.

For example, to solve \( \int x^2 \, dx \), you find its antiderivative: \( \frac{x^3}{3} + C \). The "+ C" is essential as it accounts for all possible vertical shifts of the antiderivative function, due to the differentiation process losing information about constant terms.