Question

Find the arc length of the function \(f(x)=\int_{e}^{x} \sqrt{\ln ^{2} t-1} d t\) on \(\left[e, e^{3}\right]\).

Short answer

Based on the given solution above, the arc length of the function \(f(x)=\int_{e}^{x} \sqrt{\ln^{2} t-1} dt\) on the interval \(\left[e, e^{3}\right]\) is approximately \(40.172\).

Step-by-step solution

Find the derivative of f(x)

Using the Fundamental Theorem of Calculus, we can find the derivative of f(x) with respect to x: \(f'(x)=\frac{d}{dx}\left(\int_{e}^{x}\sqrt{\ln^2 t-1}\right)dt = \sqrt{\ln^2 x-1}\).

Write down the arc length formula for the given interval

Now we'll use the arc length formula with the given derivative: \(L=\int_{e}^{e^3} \sqrt{1+\left(\sqrt{\ln^2 x-1}\right)^2} dx\).

Simplify the expression inside the square root

We can now simplify the expression inside the square root: \(L=\int_{e}^{e^3} \sqrt{1+\ln^2 x-1} dx\). This simplifies to: \(L=\int_{e}^{e^3} \sqrt{\ln^2 x} dx\).

Integrate the function on the given interval

Now, we will integrate the function on the given interval: \(L=\int_{e}^{e^3} \ln|x| dx\). To solve this integral, we can use integration by parts: Let \(u=\ln|x|\) and \(dv=dx\). Differentiate u to get \(du = \frac{1}{x} dx\) and integrate dv to get \(v=x\). Using integration by parts formula, we have: \(L=\left[uv\right]_e^{e^3} - \int_{e}^{e^3} vu' dx\) Plugging in u, du, v and dv, we get: \(L=\left[x\ln|x|\right]_e^{e^3} - \int_{e}^{e^3} x\left(\frac{1}{x} dx\right)\). Simplify the equation: \(L= \left[x\ln|x|\right]_e^{e^3} - \int_{e}^{e^3} dx\). Now, integrate and evaluate over the interval: \(L=\left[\left(x\ln|x|-x\right) \right]_{e}^{e^3}\).

Evaluate the integral on the given interval

Finally, substitute the limits of the interval into the expression found in step 4: \(L=\left[\left(e^3\ln{e^3}-e^3\right)-\left(e\ln{e}-e\right)\right]\). This simplifies to: \(L=(e^3 (3\ln{e} -1)) - (e (1-1))\) Which gives us the arc length: \(L = e^3 (3 -1) = 2 e^3\approx 2(20.086)\). Therefore, the arc length of the function f(x) on the interval \(\left[e, e^{3}\right]\) is approximately \(40.172\).

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key concept in connecting differentiation and integration, two of the main subjects within calculus. It provides a way to easily evaluate the integral of a function provided we know its antiderivative. This theorem states: if a function is continuous over a certain interval, then it possesses an antiderivative over that interval. Here's what it entails:

• The first part of the theorem tells us that if we have an integral function, the derivative of that function is, in fact, the original function inside the integral.
• The second part shows that if you want to evaluate the definite integral of a function over an interval, you find the antiderivative of the function and compute the difference between its values at the endpoints of the interval.

In our exercise, we used the first part of this theorem to determine the derivative of \( f(x) \), which was essential for finding the arc length of the function.

Integration by Parts

Integration by Parts is a technique derived from the product rule for differentiation and is extremely helpful for integrating products of functions. It is particularly useful when faced with an integral for which standard methods are inadequate. The formula is given by:\[\int u \, dv = uv - \int v \, du\]Here’s how it works:

• You choose parts of the integral to be \(u\) and \(dv\).
• Differentiate \(u\) to find \(du\), and integrate \(dv\) to get \(v\).
• Then substitute into the integration by parts formula.

In the exercise, we used integration by parts to handle the integral \(\int_{e}^{e^3} \ln|x| \, dx\). By setting \(u = \ln|x|\) and \(dv = dx\), we effectively simplified the computation by transforming it into another integral that was easier to solve.

Definite Integrals

Definite Integrals are fundamental in calculus as they provide the net area under a curve bound by a certain interval. It is expressed by the symbol \( \int_a^b f(x) \, dx \), where \(a\) and \(b\) are the limits of integration. Key characteristics include:

• Unlike indefinite integrals, definite integrals result in a number representing the area.
• The process involves finding an antiderivative and using the Fundamental Theorem of Calculus.
• Endpoints of the interval are crucial as they determine the limits for evaluating the area.

In the context of our problem, the definite integral was crucial to calculate the arc length over the interval \([e, e^3]\). After finding the necessary antiderivatives, we plugged the limits into the expression to get the final result of the arc length.