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For Activities 21 through 24 a. write the general antiderivative, b. evaluate the expression. \(\int_{0}^{4} 2 e^{2 x} d x\)

Short Answer

Expert verified
The definite integral evaluates to \(e^8 - 1\).

Step by step solution

01

Identify the Function to Integrate

We need to find the antiderivative of the function \(2e^{2x}\).
02

Determine the General Antiderivative

To integrate \(2e^{2x}\), we use the rule: The antiderivative of \(e^{ax}\) is \(\frac{1}{a}e^{ax}\). Here, \(a = 2\), so the antiderivative is \(\frac{1}{2}e^{2x}\). Therefore, the antiderivative of \(2e^{2x}\) is \(e^{2x}\), plus a constant \(C\).
03

Set Up the Definite Integral with Limits

The definite integral requires evaluating from 0 to 4. The setup is \[ \int_{0}^{4} 2e^{2x} \, dx = \left[ e^{2x} \right]_{0}^{4}. \]
04

Evaluate the Definite Integral

Substitute the upper limit (4) and lower limit (0) into the antiderivative: \[ e^{2(4)} - e^{2(0)} = e^{8} - e^{0}. \] Since \(e^{0} = 1\), the result is \(e^{8} - 1\).
05

Conclusion

The value of the definite integral \( \int_{0}^{4} 2e^{2x} \, dx = e^8 - 1 \).

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Key Concepts

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

Antiderivative
The antiderivative, also known as the indefinite integral, is a fundamental concept in calculus. It is the reverse process of differentiation. While differentiation involves finding the rate at which a function changes, finding the antiderivative means determining the original function given its rate of change. In simpler terms, if you have a function's derivative, the antiderivative is the function you started with, except it's usually expressed with an added constant, since the derivative of that constant is zero when reversed.

When given a function like \(2e^{2x}\), solving for the antiderivative involves applying basic integration rules. One common rule is that the antiderivative of \(e^{ax}\) is \(\frac{1}{a}e^{ax}\). This means for \(2e^{2x}\), you divide by the coefficient of \(x\) inside \(e^{ax}\) which is 2, giving us \(\frac{1}{2}e^{2x}\). Therefore, the general antiderivative of \(2e^{2x}\) is \(e^{2x} + C\), where \(C\) is a constant of integration. This constant represents any value that, when differentiated, becomes zero but changes the vertical position of the function on a graph.
Definite Integral
A definite integral, unlike an antiderivative, is an evaluation process that provides a specific numerical value. It calculates the area under the curve of a function, between two specified limits. This area is essential in various fields, including physics and engineering, as it helps in quantifying things like displacement and total quantity changes.

To solve a definite integral like \(\int_{0}^{4} 2e^{2x} \, dx\), you first find its antiderivative. We've determined that the antiderivative of \(2e^{2x}\) is \(e^{2x}\). Once you have the antiderivative, you evaluate it at the upper limit and the lower limit specified in the integral, which are 4 and 0 in this case.

After substituting, you compute: \(e^{2(4)} - e^{2(0)}\). Since \(e^{0} = 1\), this simplifies to \(e^{8} - 1\). Thus, the definite integral provides the exact area between the function \(2e^{2x}\) and the x-axis from 0 to 4.
Exponential Function
Exponential functions are a key concept in calculus, characterized by the constant ratio of change, where the variable exponent is the argument of an exponential base, usually \(e\), which is approximately equal to 2.71828. Exponential functions grow or decay at a rapid rate, making them useful in modeling natural growth processes like populations or radioactive decay.

The function \(2e^{2x}\) is an example of an exponential function, driven by the base \(e\). In this case, "2" acts as a coefficient, determining the initial height or how steep the exponential curve starts. The term "2x" means the growth rate itself is multiplied, causing the function to increase more rapidly. This manipulation of exponential functions is crucial in both theoretical research and practical applications, allowing dynamic scaling, modification, and understanding of various natural phenomena.

Understanding the growth pattern of exponential functions through calculus techniques, like finding antiderivatives and evaluating definite integrals, provides insights into how rapidly quantities can change over time and space.

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Most popular questions from this chapter

\(\begin{array}{llll}\text { Newspaper } & \text { Circulation } & \text { The circulation } & \text { (as } & \text { of }\end{array}\) September 20 of each year) of daily English-language newspapers in the United States between 1986 and 2000 can be modeled as $$ n(x)=0.00792 x^{3}-0.32 x^{2}+3.457 x+51.588 $$ where \(x\) is the number of years since 1980 . (Source: Based on data from Statistical Abstract, 1995 and 2001\()\) a. What was the average newspaper circulation from 1986 through \(2000 ?\) b. In what year was the newspaper circulation closest to the average circulation from 1986 through \(2000 ?\)

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Write the general antiderivative. \(\int 2 x \ln \left(x^{2}+1\right) d x\)

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