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Chapter 4: Problem 42

Determine these indefinite integrals. $$\int(x+4)^{2} d x$$

Short Answer

Expert verified
\[ \int (x+4)^2 \, dx = \frac{x^3}{3} + 4x^2 + 16x + C \]

Step by step solution

01

Expand the Integrand

First, expand the integrand \(x+4\)^2. To do this, use the binomial theorem or simply expand it as \(a+b\)^2 = a^2 + 2ab + b^2. Thus: \[ (x+4)^2 = x^2 + 2 \times x \times 4 + 4^2 = x^2 + 8x + 16 \] So, the integral becomes: \[ \int (x^2 + 8x + 16)\,dx \]
02

Integrate Term by Term

Integrate each term in the expanded integrand separately. Recall the basic power rule for integration: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] Applying this rule to each term: \[- \int x^2 \, dx = \frac{x^{3}}{3} + C_1 \- \int 8x \, dx = 8 \cdot \frac{x^{2}}{2} + C_2 = 4x^2 + C_2 \- \int 16 \, dx = 16x + C_3 \- \]
03

Combine the Results and Simplify

Combine the results of the individual integrals to find the total indefinite integral. Since the constant of integration can be combined into a single constant: \[ \int (x+4)^2 \, dx = \frac{x^3}{3} + 4x^2 + 16x + C \] where \C = C_1 + C_2 + C_3\ is a constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
Integration is the process of finding the integral of a function. There are various techniques used in integration depending on the complexity of the function. For this exercise, we utilized the technique of expanding the integrand first. This simplifies the polynomial expression and makes it easier to compute the indefinite integral term by term. Other common techniques include substitution, integration by parts, and partial fraction decomposition. Always choose the technique that simplifies the integrand the most.
Binomial Theorem
The binomial theorem provides a formula for expanding expressions that are raised to a power, such as \( (a + b)^n \). Specifically, for \( (x + 4)^2 \), we utilize \( (a + b)^2 = a^2 + 2ab + b^2 \). By applying this, we expanded \( (x + 4)^2 \) as \[ x^2 + 2 \times x \times 4 + 4^2 = x^2 + 8x + 16 \]. This step simplifies the original integrand \((x+4)^2\) into a sum that鈥檚 more manageable to integrate term by term.
Power Rule for Integration
The power rule for integration is fundamental when dealing with polynomials. This rule states that the integral of \( x^n \) is given by \[ \int x^n \ \, dx = \ \frac{x^{n+1}}{n+1} + C \]. In our solution, we applied this rule to each term obtained after expanding the binomial. For instance: \[ \- \int x^2 \ \, dx = \ \frac{x^{3}}{3} + C \- \int 8x \ \, dx = 8 \ \, \frac{x^{2}}{2} = 4x^2 + C \- \int 16 \ \, dx = 16x + C \]. When applying the power rule, carefully adjust the exponent and remember to add the constant of integration.
Constant of Integration
In each integral, the constant of integration \(C\) represents an arbitrary constant because indefinite integrals have infinite possible answers differing only by a constant. This constant arises because differentiation of a constant is zero, making it invisible in the reverse process. When combining the individual integrals, \(C_1, C_2, \) and \(C_3 \) merge into a single constant \(C\). Therefore, we end our solution with a general expression reflecting that: \[\ \int (x+4)^2 \, dx = \ \frac{x^3}{3} + 4x^2 + 16x + C \]. Always remember to include \(C\) in your final answer for indefinite integrals.

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Most popular questions from this chapter

Show that, for any function \(f\) defined for all \(x_{1}\) 's, and any constant \(k\), we have $$ \sum_{i=1}^{4} k f\left(x_{i}\right)=k \sum_{i=1}^{4} f\left(x_{i}\right) $$ Then show that, in general, $$ \sum_{i=1}^{n} h f\left(x_{i}\right)=h \sum_{i=1}^{n} f\left(x_{i}\right) $$ for any constant \(k\) and any function \(f\) defined for all \(x_{1}^{\prime}\) s.

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