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Magazine Chapter 7: Problem 11
Zeigen Sie, daB die Funktion \(f(x)=\sqrt{x}\) integrierbar 眉ber \([O, b]\) ist und bestimmen Sie mit Satz (7.17) den Fl盲cheninhalt \(\int_{0}^{b} \sqrt{x} d x\).
Short Answer
Expert verified
The area under the curve \( f(x) = \sqrt{x} \) over \([0, b]\) is \( \frac{2}{3} b^{3/2} \).
Step by step solution
01
Understanding the Problem
We need to show that the function \( f(x) = \sqrt{x} \) is integrable over the interval \([0, b]\), and then calculate the definite integral \( \int_{0}^{b} \sqrt{x} \, dx \). In the context of integration theory, a function is integrable over an interval if its definite integral exists.
02
Ensure Integrability of \( f(x) \)
The function \( f(x) = \sqrt{x} \) is continuous on the interval \([0, b]\) since \( \sqrt{x} \) is defined and continuous for all \( x \geq 0 \). As per the properties of continuous functions on closed intervals, \( f(x) \) is integrable over \([0, b]\).
03
Set up the Integral Calculation
We are required to calculate \( \int_{0}^{b} \sqrt{x} \, dx \). This involves finding the antiderivative of \( \sqrt{x} = x^{1/2} \).
04
Find the Antiderivative
Use the power rule for integration, which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \). For \( x^{1/2} \), we have:\[\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} + C\]
05
Evaluate the Definite Integral
To find \( \int_{0}^{b} \sqrt{x} \, dx \), substitute the bounds into the antiderivative:\[\int_{0}^{b} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{b}\]This evaluates to:\[\frac{2}{3} b^{3/2} - \frac{2}{3} \cdot 0^{3/2} = \frac{2}{3} b^{3/2}\]
06
Conclusion on Area
The area under the curve \( \sqrt{x} \) from \( x = 0 \) to \( x = b \) is \( \frac{2}{3} b^{3/2} \). This confirms that the integral exists and provides the area calculation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrability
To determine whether a function is integrable, we need to check if its definite integral can be calculated on a given interval. A key criteria for integrability is continuity. If a function is continuous over a closed interval \([a, b]\), it is integrable over that interval. This means we can find the area under the curve formed by the function between these points. For our function \(f(x) = \sqrt{x}\), it is continuous for \(x \geq 0\), thus ensuring integrability over \[0, b\].
In simpler terms, if a function does not have any breaks or jumps in its defined range, like \(\sqrt{x}\), you're good to go for integration!
In simpler terms, if a function does not have any breaks or jumps in its defined range, like \(\sqrt{x}\), you're good to go for integration!
Definite Integral
The definite integral is all about finding the total area under a curve between two points on the x-axis, which, in not-too-complicated math terms, is between two bounds like \(0\) and \(b\) for \(\int_0^b \sqrt{x} \, dx\).
This process gives us a number representing this area, minus any constant because it's canceled out in definite integration. For the definite integral \(\int_0^b \sqrt{x} \, dx\), it results in \(\frac{2}{3} b^{3/2}\), deleting the constant that typically shows in indefinite integrals. This is indeed our total area from \(x = 0\) to \(x = b\).
This process gives us a number representing this area, minus any constant because it's canceled out in definite integration. For the definite integral \(\int_0^b \sqrt{x} \, dx\), it results in \(\frac{2}{3} b^{3/2}\), deleting the constant that typically shows in indefinite integrals. This is indeed our total area from \(x = 0\) to \(x = b\).
- # Area is finite and measurable.
- # Upper and lower bounds limit the integration.
Antiderivative
Finding an antiderivative, also known as an indefinite integral, is like reverse-engineering a derivative. For the function \(\sqrt{x} = x^{1/2}\), we're looking for a function whose derivative brings us back to \(\sqrt{x}\). This journey backwards employs the **Power Rule** for integration.
In our case, the antiderivative of \(x^{1/2}\) is \(\frac{2}{3} x^{3/2} + C\), a necessary step leading us to evaluate our definite integral. Think of the antiderivative as preparing the groundwork needed for finding definite integrals.
Remember:
In our case, the antiderivative of \(x^{1/2}\) is \(\frac{2}{3} x^{3/2} + C\), a necessary step leading us to evaluate our definite integral. Think of the antiderivative as preparing the groundwork needed for finding definite integrals.
Remember:
- The antiderivative comes with a constant \(C\), accounting for all possible primitive functions.
- It forms the backbone of evaluating definite integrals by solving it at the boundaries.
Power Rule
The Power Rule is a fantastic tool in calculus for simplifying the process of finding antiderivatives. It states: When integrating a term \(x^n\), where \(n \, eq -1\), the formula \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) applies, making integration a much smoother ride. For \(x^{1/2}\), we plug in the power \(n = 1/2\), yielding \(\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C\).
Simply, you add 1 to the exponent, divide by the new exponent, and attach \(+C\), representing an infinite number of antiderivatives due to the constant.
Simply, you add 1 to the exponent, divide by the new exponent, and attach \(+C\), representing an infinite number of antiderivatives due to the constant.
- The power rule simplifies the integration of polynomial functions.
- It provides a fast track to solving and finding antiderivatives, especially for non-complex functions.
