Chapter 6: Problem 74
The area under \(1 / \sqrt{x}\) on the interval \(1 \leq x \leq b\) is equal to \(6 .\) Find the value of \(b\) using the Fundamental Theorem.
Short Answer
Expert verified
The value of \( b \) is 16.
Step by step solution
01
Set Up the Integral
To find the area under the curve \( \frac{1}{\sqrt{x}} \) from \( x = 1 \) to \( x = b \), we set up the improper integral as follows: \[ \int_{1}^{b} \frac{1}{\sqrt{x}} \, dx = 6. \]
02
Integrate the Function
First, notice that the integral of \( \frac{1}{\sqrt{x}} \) can be rewritten as \( x^{-1/2} \). This can be integrated using the power rule of integrals. The integral is: \[ \int x^{-1/2} \, dx = 2x^{1/2} + C. \]
03
Apply the Fundamental Theorem of Calculus
According to the Fundamental Theorem of Calculus, we evaluate the antiderivative from step 2 at the bounds 1 and \( b \):\[ \left[ 2 x^{1/2} \right]_{1}^{b} = 2b^{1/2} - 2 \cdot 1^{1/2} = 6. \]
04
Simplify and Solve for \(b\)
We are given that the result of the definite integral equals 6. Substitute and solve for \( b \):\[ 2b^{1/2} - 2 = 6. \]Add 2 to both sides:\[ 2b^{1/2} = 8. \]Divide by 2:\[ b^{1/2} = 4. \]Square both sides to solve for \( b \):\[ b = 16. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
A definite integral helps us find the area under a curve within a given interval on the x-axis. It essentially sums up all the infinitesimal slices under the curve from one point to another, giving us a total area. In our exercise, we are finding the area under the curve \( \frac{1}{\sqrt{x}} \) from \( x = 1 \) to \( x = b \). This is set up by the integral \( \int_{1}^{b} \frac{1}{\sqrt{x}} \, dx = 6 \).
- The number "6" represents the total area under the curve between the points 1 and \( b \).
- The process involves evaluating the integral with specific limits.
Area Under a Curve
The concept of the area under a curve is fundamental in understanding real-world phenomena, such as physics, engineering, or economics. In essence, it quantifies the space that lies beneath a given curve and above the x-axis between two points.
- In our example, we look at the curve \( \frac{1}{\sqrt{x}} \).
- We calculate the area from \( x = 1 \) to some unknown \( x = b \).
Power Rule of Integration
The power rule of integration is a straightforward method to integrate functions of the form \( x^n \). When you have a function \( x^n \), where \( n \) is any real number except -1, the integral is given by:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C.\]In our exercise, we needed to integrate \( \frac{1}{\sqrt{x}} \), which can be rewritten as \( x^{-1/2} \).
- Using the power rule, the integral becomes \( 2x^{1/2} + C \).
- This rule greatly simplifies finding antiderivatives for polynomial functions.