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Chapter 5: Problem 59

Calculate the integrals. $$ \int(x+2) \sqrt{x-5} d x $$

Short Answer

Expert verified
The integral is \( \frac{2}{5}(x-5)^{5/2} + \frac{14}{3}(x-5)^{3/2} + C \).

Step by step solution

01

Substitution

Let's start by using substitution. Set \( u = x - 5 \), so \( du = dx \). This means we can rewrite \( x \) in terms of \( u \): \( x = u + 5 \). Thus the integral becomes \( \int (u + 7) \sqrt{u} \ du \).
02

Distribute and Simplify

Expand the expression \( (u + 7) \sqrt{u} \). This equals \( u \sqrt{u} + 7\sqrt{u} \). Using the property of exponents, rewrite it as \( u^{3/2} + 7u^{1/2} \).
03

Integrate Each Term

Now integrate each term separately. The integral of \( u^{3/2} \) is \( \frac{2}{5}u^{5/2} \), and the integral of \( 7u^{1/2} \) is \( 7 \times \frac{2}{3} u^{3/2} = \frac{14}{3}u^{3/2} \).
04

Substitute Back

Substitute back \( u = x - 5 \) into the antiderivative. So we have \( \frac{2}{5}(x-5)^{5/2} + \frac{14}{3}(x-5)^{3/2} + C \), where \( C \) is the integration constant.

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Key Concepts

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

Understanding Integral Calculus
Integral calculus is a branch of calculus that deals with the accumulation of quantities and the areas under and between curves. Unlike differential calculus, which focuses on rates of change, integral calculus emphasizes the summing up of infinitely small data points to find a total.
A key concept in integral calculus is the integral, which can be thought of as an antiderivation. There are two main types of integrals:
  • Indefinite Integrals: These represent a family of functions and include a constant of integration, usually denoted as \( C \).
  • Definite Integrals: These calculate the net area between the graph of a function and the x-axis, over a specified interval \([a, b]\).
Integrals are often used to calculate areas, volumes, central points, and many useful things in physics and engineering. The power of integral calculus lies in its ability to solve complex problems in these fields.
Mastering U-Substitution
U-substitution, also known as substitution method, is a technique used to simplify the process of finding antiderivatives and evaluating integrals. It involves changing the variable of integration to make an integral easier to solve. This technique is particularly useful when dealing with composite functions.
The general idea is to choose a substitution that simplifies the integral into a basic form. Here are common steps:
  • Pick a substitution \( u = g(x) \) such that it simplifies the function.
  • Calculate \( du \), which is the derivative of \( u \) with respect to \( x \).
  • Replace \( dx \) in the integral with \( du \) and re-express the other variables in terms of \( u \).
After finding the integral in terms of \( u \), the last step involves substituting back to express the answer in terms of the original variable. This approach was used in the exercise when we set \( u = x - 5 \) to simplify the given integral.
Discovering Antiderivatives
Antiderivatives are the reverse process of differentiation. While differentiation splits a function into its rate of change, an antiderivative compiles those changes back together into a smooth curve or function. An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \). Finding antiderivatives is central to solving indefinite integrals.
A few key points about antiderivatives:
  • The constant of integration \( C \) represents an infinite family of solutions because adding a constant does not affect differentiation.
  • Using substitution can make finding antiderivatives simpler for more complex functions.
In the solved exercise, terms were broken down using properties of exponents to find the antiderivative of each part separately, simplifying the process of integrating the original function.
Exploring Definite Integrals
Definite integrals give us a way to calculate the area under a curve between two limits \([a, b]\). Unlike indefinite integrals, definite integrals are evaluated over a specific interval, which results in a real number representing the accumulated quantity.
The process often involves:
  • Finding the antiderivative of the given function.
  • Evaluating this antiderivative at the upper and lower limits of the integral.
  • Subtracting the value of the function at the lower limit from its value at the upper limit.
Though the exercise provided dealt with an indefinite integral, understanding definite integrals is critical as they have practical applications in calculating physical quantities like distances, area, and even probabilities.

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Most popular questions from this chapter

Determine the area between the two curves over the range of \(x\). $$ f(x)=\sin (2 x) \quad g(x)=\cos (x), 0 \leq x \leq \pi / 2 $$

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