Question

Find the following: (a) \(\int 16 x^{-3} d x \quad(x \neq 0)\) (b) \(\int 9 x^{8} d x\) (c) \(\int\left(x^{5}-3 x\right) d x\) \((d) \int 2 e^{-2 x} d x\) (e) \(\int \frac{4 x}{x^{2}+1} d x\) (f) \(\int(2 a x+b)\left(a x^{2}+b x\right)^{7} d x\)

Short answer

(a) \(-8x^{-2} + C\); (b) \(x^9 + C\); (c) \(\frac{x^6}{6} - \frac{3x^2}{2} + C\); (d) \(-e^{-2x} + C\); (e) \(2\ln|x^2 + 1| + C\); (f) \(\frac{(ax^2+bx)^8}{8} + C\).

Step-by-step solution

Solve (a) Integral Using Power Rule

The integral to solve is \( \int 16x^{-3} \, dx \). Apply the power rule for integration: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). Here, \( n = -3 \), so the integral becomes \( 16 \cdot \frac{x^{-2}}{-2} + C = -8x^{-2} + C \).

Solve (b) Integral Using Power Rule

The integral is \( \int 9x^8 \, dx \). Again, apply the power rule for integration: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). Here, \( n = 8 \), so the integral becomes \( 9 \cdot \frac{x^{9}}{9} + C = x^9 + C \).

Solve (c) Integral with Linearity of Integration

The integral is \( \int(x^5 - 3x) \, dx \). Use the linearity property of integrals: \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \). Solve each separately: \( \int x^5 \, dx = \frac{x^6}{6} + C_1 \) and \( \int (-3x) \, dx = -\frac{3x^2}{2} + C_2 \). The result is: \( \frac{x^6}{6} - \frac{3x^2}{2} + C \).

Solve (d) Exponential Integral

The integral is \( \int 2e^{-2x} \, dx \). Use the integral formula \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \). Here, \( a = -2 \), so the integral becomes \( 2 \cdot \left( \frac{1}{-2} e^{-2x} \right) + C = -e^{-2x} + C \).

Solve (e) with Substitution Method

The integral is \( \int \frac{4x}{x^2 + 1} \, dx \). Use substitution: let \( u = x^2 + 1 \), then \( du = 2x \, dx \). Thus, \( \int \frac{4x}{u} \cdot \frac{1}{2x} \, du = \int \frac{2}{u} \, du \). Solving gives \( 2 \ln|u| + C = 2 \ln|x^2 + 1| + C \).

Solve (f) Using Substitution for Power Function

The integral is \( \int(2ax + b)(ax^2 + bx)^7 \, dx \). Use substitution: let \( u = ax^2 + bx \), and \( du = (2ax + b) \, dx \). This simplifies to \( \int u^7 \, du = \frac{u^8}{8} + C \). Substitute back for \( u \): \( \frac{(ax^2+bx)^8}{8} + C \).

Key concepts

Power Rule

The power rule is a fundamental technique in calculus that simplifies the integration of functions that are power functions. To integrate a function using the power rule, apply the formula:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

Here, \( n \) represents the exponent of the term being integrated, and \( C \) is the constant of integration. Let's consider an example:
In part (a) of the exercise, we have \( \int 16x^{-3} \, dx \). Applying the power rule helps you find that integral as follows:

• Start with the exponent \( n = -3 \).
• Using the power rule gives \( \int x^{-3} \, dx = \frac{x^{-2}}{-2} + C \).
• Multiplied by 16, this becomes \( -8x^{-2} + C \).

This method is really efficient for polynomials and ensures accurate results for each power term by incrementing the exponent by one and dividing by the new exponent.

Linearity of Integration

The linearity of integration is a property that makes tackling complex expressions easier by breaking them down into simpler parts. It is based on the principle that the integration of a sum of functions is the sum of their integrals:

• \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \)

In part (c) of our example, the integral \( \int(x^5 - 3x) \, dx \) uses this property:

• Separate into two integrals: \( \int x^5 \, dx \) and \( \int (-3x) \, dx \).
• Integrate each separately: \( \int x^5 \, dx = \frac{x^6}{6} + C_1 \), and \( \int (-3x) \, dx = -\frac{3x^2}{2} + C_2 \).
• Combine the results: \( \frac{x^6}{6} - \frac{3x^2}{2} + C \).

Using linearity can immensely simplify work on polynomial functions by allowing integration in parts and then combining the results seamlessly.

Exponential Integration

Integrating exponential functions is straightforward when you know the formula for the exponential integral:

• \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \)

The distinct feature of exponential functions is that their integral remains an exponential function. As shown in part (d) with \( \int 2e^{-2x} \, dx \), here's what happens:

• Identify \( a = -2 \) from \( e^{-2x} \).
• Apply the formula: \( \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} + C \).
• Multiply by 2 to complete: \(-e^{-2x} + C \).

Integrating exponentials is heavily used in natural growth and decay models, as these functions replicate real-world exponential changes.

Substitution Method

The substitution method is crucial for integrating more complex functions by simplifying them through a clever substitution. This technique involves substituting a part of the integral with a new variable:

• Identify a substitution that simplifies the integral.
• Differentiate and replace \( dx \) in terms of the new variable.
• Solve the resulting integral before swapping back the original variable.

Take part (e) as our example, with \( \int \frac{4x}{x^2 + 1} \, dx \):

• Choose \( u = x^2 + 1 \), thus \( du = 2x \, dx \).
• Rephrase the integral: \( \int \frac{4x}{u} \cdot \frac{1}{2x} \, du = \int \frac{2}{u} \, du \).
• Solve: \( 2 \ln|u| + C = 2 \ln|x^2 + 1| + C \).

This method is versatile, making it powerful for any integral where direct integration seems complicated, allowing flexibility in dealing with differentials and composite functions.