Problem 61

Question

Choose the correct answer. \(\int x^{e} d x=?\) a. \(\frac{1}{e+1} x^{e+1}+C\) b. \(x^{e}+C\) c. \(x^{e} x+C\)

Step-by-Step Solution

Verified
Answer
The correct answer is a: \( \frac{1}{e+1} x^{e+1} + C \).
1Step 1: Understand the Problem
The integral given is \( \int x^{e} \, dx \). We need to integrate an exponential function where the exponent is \( e \), which is a constant.
2Step 2: Apply the Power Rule of Integration
The power rule for integration states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for any constant \( n eq -1 \). Here, \( n = e \), so the integral becomes \( \int x^{e} \, dx = \frac{x^{e+1}}{e+1} + C \).
3Step 3: Verify the Solution
Check each option: - a. \( \frac{1}{e+1} x^{e+1} + C \) matches the result from applying the power rule. - b. \( x^{e} + C \) doesn't follow the power rule.- c. \( x^e x + C \) is incorrect as it doesn't factor properly from the integration, also doesn't follow the rule.
4Step 4: Conclusion
After comparing each option with the derived integral, it is confirmed that the correct answer aligns with option a: \( \frac{1}{e+1} x^{e+1} + C \).

Key Concepts

Power Rule of IntegrationExponentiationCalculus
Power Rule of Integration
The power rule of integration is a fundamental concept in calculus that makes integrating polynomial expressions straightforward. It states that when you have an expression of the form \(x^n\), where \(n\) is any constant except \(-1\), its integral is
  • \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
Here, \(C\) represents the constant of integration, which you always add to the result of an indefinite integral.
Why should the power rule be true? When you differentiate \( \frac{x^{n+1}}{n+1} \), you bring down the exponent \(n+1\) as a coefficient and reduce the power by one, resulting in \(x^n\). That's the reverse of what integration does, which is why this rule works beautifully without any additional fuss.
For this exercise, where \(n = e\) and \(e\) is a constant, applying this rule is straightforward. You simply add 1 to the exponent \(e\) to get \(e+1\) and then divide by this new exponent to get \(\frac{x^{e+1}}{e+1} + C\). That's why option (a) is correct.
Exponentiation
Exponentiation is a crucial operation in mathematics, mainly dealing with powers, which can range from whole numbers to constants like \(e\). In the context of calculus, exponentiation plays a vital role when dealing with integrals and derivatives.
When you see \(x^e\), this means that \(x\) is raised to the power of \(e\). If \(e\) were a natural number, you'd multiply \(x\) by itself \(e\) times. However, since \(e\) is a mathematical constant, often representing the base of natural logarithms, it behaves consistently under the rules of exponents.
Handling exponentiation within integrals involves applying rules like the power rule. The key is understanding that even though \(e\) is a number, it doesn't affect the form of the power rule itself, which still treats \(e\) as any other non-negative constant that isn't \(-1\). That knowledge gives you a consistent way to integrate, even with different exponents like \(e\).
Calculus
Calculus is the branch of mathematics focusing on changes, a concept crucial when studying integrals and derivatives. The integration is the process you use to find the area under curves, count summations of infinitely small points, or reverse differentiation.
In this problem, calculus employs its integration techniques to solve for \(\int x^e \, dx\). Integrals bring a rich understanding of how functions accumulate values over intervals, which is fundamentally linked to the changing aspect calculus deals with.
  • It bridges simple geometric areas to complex problem-solving situations.
  • Derivatives focus on rates of change, while integrals focus on total accumulation.
In a nutshell, calculus helps us not just solve this problem but provide a framework to understand natural phenomena, model changes, and predict outcomes. Using the power rule as an integration tool is an example of how calculus is wielded practically.