Question

Find \(F\) as a function of \(x\) and evaluate it at \(x=2, x=5,\) and \(x=8\). $$ F(x)=\int_{1}^{x} \frac{10}{v^{2}} d v $$

Short answer

The integral for the function \( F(x) = \int_{1}^{x} \frac{10}{v^{2}} d v \) is \( F(x) = 10\ln|x|\). When evaluating at \(x = 2\), \(x = 5\) and \(x = 8\), the answers are \( F(2) = 10\ln|2|\), \(F(5) = 10\ln|5|\) and \(F(8) = 10\ln|8|\), respectively.

Step-by-step solution

Finding the Antiderivative

The given function is \( \frac{10}{v^{2}} \) or \( \frac{10}{1/v} \) which can be rewritten as \( 10v^{-1} \). The antiderivative or the primitive of \( v^{-1} \) is \( \ln|v| \), so the antiderivative of the function \(10v^{-1}\) is \(10\ln|v|\). Note that the constant of integration is not considered because we are calculating a definite integral.

Applying the Fundamental Theorem of Calculus

Using the Fundamental Theorem of Calculus, we can now find \( F(x) \) by subtracting the value of the antiderivative at the lower limit from its value at the upper limit. So, \( F(x) = 10\ln|x| - 10\ln|1| = 10(\ln|x| - \ln|1|) = 10\ln|x|\). This is the function of \( x \).

Evaluating F(x)

Now that we have F(x), we can substitute in the desired values of x. For \(x = 2\), \(F(2) = 10\ln|2|\), for \(x = 5\), \(F(5) = 10\ln|5|\) and for \(x = 8\), \(F(8) = 10\ln|8|\).

Key concepts

Antiderivative

The concept of an antiderivative is foundational to understanding calculus. In essence, finding an antiderivative is the reverse process of differentiation. If you're given a function, the antiderivative is a function whose derivative gives you back the original function.

For example, if you have the function \( f(x) = x^2 \), an antiderivative would be \( F(x) = \frac{1}{3}x^3 \), since the derivative of \( \frac{1}{3}x^3 \) with respect to \( x \) is \( x^2 \). This process is also known as integration. The important thing to remember is that antiderivatives are not unique; they include a constant of integration, \( C \), because differentiating a constant always gives zero.

When calculating definite integrals, as in the textbook exercise, we are concerned with the antiderivative evaluated at specific points. The constant of integration cancels out when we subtract the antiderivative value at the upper limit from the value at the lower limit. This is why we do not consider the constant of integration in this context.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) links the process of differentiation with that of integration, providing a powerful connection between the two main concepts of calculus. The FTC states that if \( f \) is continuous on \( [a, b] \) and \( F \) is an antiderivative of \( f \) on \( [a, b] \) then:
\[\int_{a}^{b} f(x) dx = F(b) - F(a)\]
The exercise given involves the application of FTC to find the value of a definite integral by taking the antiderivative of the function and evaluating it at the upper and lower bounds.

In simpler terms, to calculate the area under the curve of the continuous function \( f(x) \) between \( a \) and \( b \) on the x-axis, you can simply find the antiderivative of \( f(x) \) and evaluate the difference at \( b \) and \( a \). This makes calculating areas and solving many other practical problems much more manageable.

Natural Logarithm

The natural logarithm is a mathematical function commonly denoted by \( \ln(x) \) and is the inverse operation of raising \( e \), the base of the natural logarithm, to a power. For example, if \( e^y = x \) then \( \ln(x) = y \). The logarithm tells us the power to which the base must be raised to obtain a certain number.

In calculus, the natural logarithm is particularly important because it is the inverse of the exponential function, and it appears frequently when dealing with integration.

Antiderivative of \(1/x\)

In the exercise we're discussing, the function \( 10v^{-1} \) or equivalently \( \frac{10}{v} \) has an antiderivative of \( 10\ln|v| \). It's especially notable because the natural log of 1, \( \ln(1) \) is 0, which simplifies the evaluation of definite integrals involving natural logarithms, as seen when finding the function \( F(x) \) in the step by step solution.

This property allows us to easily find the values of the function \( F(x) \) for different values of \( x \), providing a clear and practical illustration of how the natural logarithm operates within the realm of calculus.