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Find the most general antiderivative of the function.(Check your answer by differentiation.) \(v(s)=4 s+3 e^{s}\)

Short Answer

Expert verified
The most general antiderivative is \( V(s) = 2s^2 + 3e^s + C \).

Step by step solution

01

Identify the Function Components

The function given is \( v(s) = 4s + 3e^s \). It consists of two parts: the term \( 4s \) and the term \( 3e^s \). We will find the antiderivative of each part separately.
02

Antiderivative of Linear Term

The term \( 4s \) is linear and has an antiderivative: \( \int 4s \, ds = 4 \cdot \frac{s^2}{2} = 2s^2 \).
03

Antiderivative of Exponential Term

The exponential term \( 3e^s \) has a straightforward antiderivative, since the antiderivative of \( e^s \) is itself. Thus, \( \int 3e^s \, ds = 3e^s \).
04

Combine the Antiderivatives

Combine the antiderivatives of both parts to form the most general antiderivative: \( V(s) = 2s^2 + 3e^s + C \), where \( C \) is the constant of integration.
05

Check by Differentiation

Differentiate the antiderivative \( V(s) = 2s^2 + 3e^s + C \) to verify it produces the original function: \( \frac{d}{ds}[2s^2 + 3e^s + C] = 4s + 3e^s \), which matches the original function \( v(s) \). The check confirms the solution is correct.

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Key Concepts

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

Differentiation
Differentiation is the process of finding the derivative of a function. A derivative is simply a way to measure how a function changes as its input changes. For instance, when you have a function like \( v(s) = 4s + 3e^s \), finding its derivative involves calculating how the output changes with respect to changes in \( s \).

In differentiation, we use certain rules and formulae that make the process straightforward for specific types of functions:
  • For linear functions like \( 4s \), we apply the power rule where the derivative of \( s^n \) is \( ns^{n-1} \). So, \( 4s \) becomes 4 as its derivative.
  • For exponential functions such as \( e^s \), the derivative is simply itself, meaning \( \frac{d}{ds}[e^s] = e^s \).
Differentiation is crucially important because it enables us to confirm the correctness of an antiderivative by checking if differentiating the antiderivative gives us back the original function.
Linear Functions
Linear functions are the simplest type of functions. They have a constant rate of change and can be written in the form of \( ax + b \), where \( a \) is the slope, and \( b \) is the y-intercept.

In the context of antiderivatives, a linear function like \( 4s \) can be integrated using the power rule. During integration, the process involves increasing the power by one and then dividing by this new power. Consequently, the antiderivative of \( 4s \) becomes \( 2s^2 \).

Understanding linear functions is essential since they are foundational elements in calculus, often appearing in more complex functions as components to be integrated separately.
Exponential Functions
Exponential functions are functions that contain the term \( e^x \), which is the mathematical constant approximately equal to 2.71828. They are unique because their rate of growth is proportional to their current value.

In calculus, exponential functions are special due to their antiderivative property. The antiderivative of \( e^s \) is itself, \( e^s \), making them straightforward for integration. This property is vital in solving differential equations and appears frequently in modeling real-life situations such as population growth and radioactive decay.

Antiderivatives involving exponential functions often include a constant multiple, such as in \( 3e^s \), where the antiderivative simply scales with the coefficient, resulting in \( 3e^s \). This simplicity is part of what makes exponential functions an exciting and powerful tool in mathematics.

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Most popular questions from this chapter

Find the most general antiderivative of the function.(Check your answer by differentiation.) \(f(x)=\frac{1+x-x^{2}}{x}\)

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