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Chapter 5: Problem 34

Find each indefinite integral. \(\int(x+5)(x-3) d x\)

Short Answer

Expert verified
The indefinite integral is \(\frac{x^3}{3} + x^2 - 15x + C\).

Step by step solution

01

Expand the Integrand

First, expand the expression \((x+5)(x-3)\) by distributing the terms:\(x \cdot x + x \cdot (-3) + 5 \cdot x + 5 \cdot (-3)\).This simplifies to \(x^2 - 3x + 5x - 15\).Combine like terms to get the expanded form: \(x^2 + 2x - 15\).
02

Integrate Each Term

Now, find the indefinite integral of the expanded polynomial, term by term:The integral of \(x^2\) is \(\frac{x^3}{3}\).The integral of \(2x\) is \(x^2\).The integral of \(-15\) is \(-15x\).Don't forget the constant of integration, \(C\).
03

Combine the Integrated Terms

Combine the results of the term-by-term integration:\(\frac{x^3}{3} + x^2 - 15x + C\).This is the indefinite integral of the given function.

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Key Concepts

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

Polynomial Expansion
Polynomial expansion is the process of breaking down a product of expressions into a sum of simpler terms. In the exercise, the expression \((x+5)(x-3)\) is expanded using distribution. This involves multiplying each term in the first polynomial by every term in the second polynomial:
  • Multiply the \(x\) from \((x+5)\) by \(x\) from \((x-3)\), which equals \(x^2\).
  • Multiply \(x\) by \(-3\), resulting in \(-3x\).
  • Next, multiply \(5\) by \(x\), which gives \(5x\).
  • Finally, multiply \(5\) by \(-3\), resulting in \(-15\).
Combine these terms to obtain the expanded polynomial: \(x^2 + 2x - 15\). This step simplifies the integrand, making it easier to integrate term by term.
Integration Techniques
Integration is the process of finding a function's antiderivative. For polynomials, this involves reversing the process of differentiation, term by term. Let's look at the integration of the expanded polynomial \(x^2 + 2x - 15\). Each term is integrated individually:
  • The integral of \(x^2\) is found using the power rule: increase the exponent by one and divide by the new exponent. Thus, \(\int x^2 dx = \frac{x^3}{3}\).
  • For the linear term \(2x\), apply the power rule to get \(\int 2x \, dx = x^2\), because \(\int x \, dx = \frac{x^2}{2}\), and multiplying by \(2\) gives \(x^2\).
  • The constant term \(-15\) integrates to \(-15x\) because the integral of a constant \(c\) is \(cx\).
By integrating each term separately and then combining them, we arrive at the solution: \(\frac{x^3}{3} + x^2 - 15x + C\).
Constant of Integration
After finding the indefinite integral, it's essential to include the constant of integration, represented by \(C\). This constant accounts for the fact that antiderivatives are not unique. Each has infinite versions differing only by a constant.
  • If you differentiate a function, any constant term disappears (since the derivative of a constant is zero).
  • When finding the antiderivative, we must account for this "lost" constant by adding \(C\).
  • Thus, the general solution to each indefinite integral includes this constant to represent all possible antiderivatives.
In the exercise solution, including \(C\) is crucial. It ensures you have the most general form of the antiderivative: \(\frac{x^3}{3} + x^2 - 15x + C\). This reflects all functions differentiating to the original integrand.

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Most popular questions from this chapter

For each definite integral: a. Evaluate it "by hand." b. Check your answer by using a graphing calculator. $$ \int_{0}^{4} \sqrt{x^{2}+9} x d x $$

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int x(x-2)^{6} d x $$

If the values of a function on an interval are always positive, can the average value of the function over that interval be negative?

Which one of these formulas is correct? a. \(\int \ln x d x=\frac{1}{|x|}+C\) b. \(\int \ln |x| d x=\frac{1}{x}+C\) c. \(\int \frac{1}{x} d x=\ln |x|+C\) d. \(\int \frac{1}{\ln x} d x=|x|+C\)

An average child of age \(x\) years grows at the rate of \(6 x^{-1 / 2}\) inches per year (for \(2 \leq x \leq 16\) ). Find the total height gain from age 4 to age \(9 .\)
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