Problem 33
Question
Find each indefinite integral. \(\int(x-2)(x+4) d x\)
Step-by-Step Solution
Verified Answer
The indefinite integral is \( \frac{x^3}{3} + x^2 - 8x + C \).
1Step 1: Expand the integrand
We begin by expanding the expression inside the integral. The expression is \( (x-2)(x+4) \). We apply the distributive property: \( (x-2)(x+4) = x(x+4) - 2(x+4) = x^2 + 4x - 2x - 8 = x^2 + 2x - 8 \).
2Step 2: Set up the integral with the expanded form
Now we rewrite the integral using the expanded form: \(\int (x^2 + 2x - 8) \, dx\).
3Step 3: Integrate term by term
Integrate each term separately. For \(x^2\), the integral is \(\frac{x^3}{3}\). For \(2x\), the integral is \(x^2\), and for the constant \(-8\), the integral is \(-8x\). Thus, the integral is \( \frac{x^3}{3} + x^2 - 8x + C \), where \(C\) is the constant of integration.
Key Concepts
Distributive propertyIntegral calculusConstant of integration
Distributive property
The distributive property is a fundamental aspect of algebra that allows us to simplify expressions and make calculations easier. When dealing with multiplication, especially involving sums or differences, the distributive property comes into play. It states that for any three numbers or expressions, say a, b, and c, the following is always true: \\[a(b + c) = ab + ac\]
This means you can "distribute" the multiplication of a over the addition of b and c.
In the context of the given exercise, we used the distributive property to expand the expression \\((x-2)(x+4)\). By distributing each term and simplifying, we arrived at \\(x^2 + 2x - 8\). Applying the distributive property is essential because it transforms the integrand into a simpler form that is easier to integrate.
This means you can "distribute" the multiplication of a over the addition of b and c.
In the context of the given exercise, we used the distributive property to expand the expression \\((x-2)(x+4)\). By distributing each term and simplifying, we arrived at \\(x^2 + 2x - 8\). Applying the distributive property is essential because it transforms the integrand into a simpler form that is easier to integrate.
Integral calculus
Integral calculus is a branch of calculus focusing on the concept of an integral. The fundamental idea behind integral calculus is to find the area under a curve. It also helps solve problems related to accumulation of quantities, determining areas, volumes, and other aspects of mathematical analysis.
In the exercise, we are asked to find the indefinite integral of the expanded expression, which is \\(\int (x^2 + 2x - 8) \, dx\). An indefinite integral, often referred to as an antiderivative, is a function that reverses differentiation.
In the exercise, we are asked to find the indefinite integral of the expanded expression, which is \\(\int (x^2 + 2x - 8) \, dx\). An indefinite integral, often referred to as an antiderivative, is a function that reverses differentiation.
- To integrate, we apply the power rule: The integral of \(x^n\) is \(\frac{x^{n+1}}{n+1}\).
- Each term is integrated separately: \(\int x^2 \, dx = \frac{x^3}{3}\), \(\int 2x \, dx = x^2\), \(\int -8 \, dx = -8x\).
Constant of integration
When performing indefinite integration, a critical component to remember is the constant of integration, denoted by \(C\). This constant arises because the process of differentiation (which indefinite integration reverses) would eliminate any constant term. Therefore, when finding an antiderivative, we must account for any constant that could have been present before differentiation.
In the solution to the exercise, after integrating each term, we arrive at the expression \(\frac{x^3}{3} + x^2 - 8x + C\). The \(C\) ensures that the result is complete, reflecting all possible antiderivatives of the integrand.
In the solution to the exercise, after integrating each term, we arrive at the expression \(\frac{x^3}{3} + x^2 - 8x + C\). The \(C\) ensures that the result is complete, reflecting all possible antiderivatives of the integrand.
- The constant \(C\) is indispensable in indefinite integrals.
- It symbolizes an infinite set of possible solutions, each differing by a constant.
Other exercises in this chapter
Problem 33
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