Chapter 82: Problem 6
Für \(p \in \mathbf{N}\) ist D- \(\int_{0}^{a} x^{p} \mathrm{~d} x=\frac{a^{p+1}}{p+1} .\) Hinweis: Aufgabe \(1 \mathrm{~d}, \mathrm{~A} 27.3\).
Short Answer
Expert verified
The definite integral is \( \frac{a^{p+1}}{p+1} \).
Step by step solution
01
Understand the Exercise
The problem provides a definite integral \( \int_{0}^{a} x^{p} \, \mathrm{d}x \) and asks to compute it with the hint that it equals \( \frac{a^{p+1}}{p+1} \). Our task is to verify this formula for the integral.
02
Recall Power Rule for Integrals
The power rule for integration states that \( \int x^n \, \mathrm{d}x = \frac{x^{n+1}}{n+1} + C \) for any real number \( n eq -1 \). This is the fundamental theorem we will use to evaluate the integral.
03
Evaluate the Indefinite Integral
Applying the power rule to \( x^{p} \), we get the indefinite integral: \( \int x^{p} \, \mathrm{d}x = \frac{x^{p+1}}{p+1} + C \), where \( C \) is the constant of integration.
04
Apply the Definite Integral Limits
To find the definite integral \( \int_{0}^{a} x^{p} \, \mathrm{d}x \), we apply the limits to the indefinite integral found in Step 3: \( \left[ \frac{x^{p+1}}{p+1} \right]_{0}^{a} \).
05
Compute the Definite Integral Value
Substitute the upper limit \( a \) and the lower limit \( 0 \) into the integral expression: \( \frac{a^{p+1}}{p+1} - \frac{0^{p+1}}{p+1} \). The latter term is 0, simplifying the expression to \( \frac{a^{p+1}}{p+1} \) as the result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Rule for Integrals
The Power Rule for Integrals is a fundamental concept in calculus which helps us integrate expressions of the form \( x^n \), where \( n \) is any real number except -1. When integrating an expression like \( x^n \), you essentially increase the power of \( x \) by one and then divide by the new power. This can be expressed as:
- \( \int x^n \, \mathrm{d}x = \frac{x^{n+1}}{n+1} + C \)
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration. It asserts that the process of taking a derivative is essentially the reverse of integration. The theorem is divided into two parts:
- The First Part of the theorem connects the definite integral of a function to its antiderivative.
- The Second Part guarantees that every continuous function has an antiderivative, and the integral can be evaluated using this antiderivative.
Indefinite Integral
An indefinite integral, of a function, represents the collection of all antiderivatives of that function. When you perform an indefinite integral, you will get an expression with a constant of integration, \( C \), symbolizing any constant that could have been differentiated away.
For instance, integrating \( x^p \) gives us:
For instance, integrating \( x^p \) gives us:
- \( \int x^p \, \mathrm{d}x = \frac{x^{p+1}}{p+1} + C \)
Limits of Integration
Limits of integration specify the range over which a definite integral is calculated, and they are indicated by a lower and an upper limit. When you compute a definite integral, such as \( \int_{0}^{a} x^p \, \mathrm{d}x \), the result represents the net area under the curve \( x^p \) from \( x=0 \) to \( x=a \).
- The upper limit, \( a \), is the point to which the integral is computed.
- The lower limit, \( 0 \), is where the integral calculation begins.