Chapter 6: Problem 20
Using the Fundamental Theorem, evaluate the definite integrals in Problems \(1-20\) exactly. $$\int_{4}^{9} \sqrt{x} d x$$
Short Answer
Expert verified
The integral evaluates to \( \frac{38}{3} \).
Step by step solution
01
Identify the Problem
We need to evaluate the definite integral \( \int_{4}^{9} \sqrt{x} \, dx \). This involves finding the antiderivative (or indefinite integral) of \( \sqrt{x} \) and then applying the limits of integration.
02
Express the Integrand in a Simpler Form
The function \( \sqrt{x} \) can be rewritten as \( x^{1/2} \). This change will make it easier to apply the power rule for integration.
03
Find the Antiderivative
Use the power rule for integration, which states that the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. Thus, the antiderivative of \( x^{1/2} \) is \( \frac{x^{3/2}}{3/2} \). Simplifying, this becomes \( \frac{2}{3}x^{3/2} \).
04
Apply the Fundamental Theorem of Calculus
According to the Fundamental Theorem of Calculus, the definite integral \( \int_{a}^{b} f(x) \, dx \) is equal to \( F(b) - F(a) \), where \( F \) is an antiderivative of \( f \). We apply this to \( \frac{2}{3}x^{3/2} \).
05
Evaluate at the Upper Limit
Substitute the upper limit of integration into the antiderivative: \( F(9) = \frac{2}{3}(9)^{3/2} \). Calculating this gives \( 18 \).
06
Evaluate at the Lower Limit
Substitute the lower limit of integration into the antiderivative: \( F(4) = \frac{2}{3}(4)^{3/2} \). Calculating this gives \( \frac{16}{3} \).
07
Compute the Definite Integral
Subtract the value at the lower limit from the value at the upper limit: \( 18 - \frac{16}{3} = \frac{54}{3} - \frac{16}{3} = \frac{38}{3} \).
08
Provide the Final Answer
The exact value of the definite integral \( \int_{4}^{9} \sqrt{x} \, dx \) is \( \frac{38}{3} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a crucial link in calculus that connects differentiation and integration, two core concepts. It asserts that the definite integral of a function over a particular interval can be found using any of its antiderivatives. This makes solving definite integrals significantly more straightforward. Particularly, the theorem consists of two parts:
- First, it states that if a function is continuous over a segment, then its indefinite integral is differentiable, and its derivative is the original function.
- Second, it establishes that for any antiderivative \( F \) of a continuous function \( f \), the integral from \( a \) to \( b \) of \( f \) is given by \( F(b) - F(a) \).
Antiderivative
An antiderivative is essentially a reverse operation to differentiation. It is a function whose derivative returns the original function. Understanding antiderivatives is vital for solving integral problems, particularly indefinite integrals, which do not have limits of integration.
- Finding an antiderivative can sometimes be straightforward, especially with functions like polynomials or those with well-known integration rules.
- In our problem, the antiderivative of \( \sqrt{x} \) was determined using known formulas, specifically the power rule for integration.
Power Rule for Integration
The Power Rule for Integration is a fundamental technique in calculus used to find antiderivatives of polynomial functions. The rule simplifies the integration of power functions, and it states that for any real number \( n eq -1 \), \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \\)Here, \( C \) is the constant of integration, which accounts for any vertical shifts in the original function.
- The power rule is particularly useful because it provides a direct method to handle polynomials and functions expressed as powers.
- In the given exercise, rewriting \( \sqrt{x} \) as \( x^{1/2} \) allowed us to directly apply the power rule. This resulted in \( \frac{x^{3/2}}{3/2} \) or equivalently \( \frac{2}{3}x^{3/2} \).