/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Evaluate the integrals. $$ \... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integrals. $$ \int x^{7} d x $$

Short Answer

Expert verified
The integral of the function \(x^7\) is evaluated by applying the power rule for integration: \(\int x^7 dx = \frac{x^8}{8} + C\), where C is the constant of integration.

Step by step solution

01

Apply the power rule for integration

To find the integral of the function \(x^7\), we will apply the power rule for integration. To do this, we add 1 to the exponent and then divide by the new exponent.
02

Evaluate the integral

Applying the power rule for integration, we get: \[ \int x^7dx = \frac{x^{7 + 1}}{7 + 1} + C \] where C is the constant of integration. You can now simplify the expression: \[ \int x^7 dx = \frac{x^8}{8} + C \] So, the integral of the function \(x^7\) is: \[ \int x^7 dx = \frac{x^8}{8} + C. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule for Integration
The power rule for integration is one of the core principles used to find the integral of a function with a variable raised to a certain power. It is the counterpart to the power rule for differentiation and is essential for solving many basic integrals in calculus. Here's the rule in its general form:
When dealing with a function of the form \( x^n \), where \( n \) is any real number except -1, you can integrate it by applying the following formula:
\[ \int x^n \, dx = \frac{x^{n + 1}}{n + 1} + C \]
In this formula, \( n + 1 \) in the numerator indicates that you increase the original power by one. Then, you divide the result by \( n + 1 \), which is now the new power. Lastly, you add the constant of integration, \( C \), to account for any constant which was differentiated to produce the original function. It’s crucial to remember that this rule does not apply when \( n = -1 \), which would lead to the natural logarithm function rather than a power function.
Integral Calculation
The process of integral calculation or integration is central to calculus. It’s the mathematical way of finding the accumulation of quantities, and it can be thought of as the reverse process of derivation. When you integrate a function, you're essentially looking for the antiderivative or the original function whose derivative gives you the function you started with.
In the context of the given exercise, the integral calculation involves the power rule, which simplifies the process considerably. To perform an integral calculation, you typically follow these steps:
  • Identify the type of function you are integrating.
  • Determine the appropriate integration rule, such as the power rule for integration here.
  • Apply the rule to calculate the antiderivative of the function.
  • Don’t forget to add the constant of integration, \( C \), as any number of functions could be the original before differentiation.
Integral calculations are foundational for solving problems in areas such as physics, economics, and engineering where the accumulation of quantities like distance, area, and others, over an interval, is key.
Constant of Integration
The constant of integration, represented as \( C \), is a critical component of antiderivatives. When you differentiate a function that has a constant, the constant disappears. Therefore, when you are finding the antiderivative, you must consider that there could have been any constant value there initially.
Here's the conceptual takeaway: Since derivatives of constants are zero, undetermined constants are a natural part of indefinite integrals. They remind us of the family of functions that have the same derivative. The constant of integration ensures that all possibilities are covered.
For instance, if we have \( F(x) \) as an antiderivative of \( f(x) \), then \( F(x) + C \) represents the most general antiderivative of \( f(x) \), where \( C \) can be any real number. In the context of the exercise, once you've applied the power rule to find the antiderivative of \( x^7 \), incorporating \( C \) finalizes the solution by acknowledging that the original function could include any constant value.

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Most popular questions from this chapter

According to the special theory of relativity, the apparent mass of an object depends on its velocity according to the formula $$ m=\frac{m_{0}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{1 / 2}} $$ where \(v\) is its velocity, \(m_{0}\) is the "rest mass" of the object (that is, its mass when \(v=0\) ), and \(c\) is the velocity of light: approximately \(3 \times 10^{8}\) meters per second. a. Show that, if \(p=m v\) is the momentum, $$ \frac{d p}{d v}=\frac{m_{0}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{3 / 2}} $$ b. Use the integral formula for \(W\) in the preceding exercise, together with the result in part (a) to show that the work required to accelerate an object from a velocity of \(v_{0}\) to \(v_{1}\) is given by $$ W=\frac{m_{0} c^{2}}{\sqrt{1-\frac{v_{1}^{2}}{c^{2}}}}-\frac{m_{0} c^{2}}{\sqrt{1-\frac{v_{0}^{2}}{c^{2}}}} . $$ We call the quantity \(\frac{m_{0} c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\) the total relativistic energy of an object moving at velocity \(v\). Thus, the work to accelerate an object from one velocity to another is given by the change in its total relativistic energy. c. Deduce (as Albert Einstein did) that the total relativistic energy \(E\) of a body at rest with rest mass \(m\) is given by the famous equation $$ E=m c^{2} $$.

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