Exponential and Logarithmic Graph-Length Examples

Why These Graphs Have a Useful Length Pattern

The arc-length formula for a graph turns a derivative into a square root: a continuously differentiable function \(y=f(x)\) on a finite interval \([a,b]\) has length

L=\int_a^b\sqrt{1+\bigl(f'(x)\bigr)^2}\,dx.

For an exponential graph, the derivative introduces another exponential inside this square root. For a logarithmic graph, the derivative introduces a reciprocal. In both cases, a carefully chosen substitution reduces the integral to the same antiderivative. The endpoint conditions matter: the logarithm is defined only for positive \(x\), and the simplification of its integrand uses that positivity.

Key idea: For \(y=e^x\), use \(u=e^x\). For \(y=\ln x\), retain \(x>0\) when simplifying the square root. Both calculations lead to an integral with integrand \(\sqrt{1+u^2}/u\), with a positive integration variable.

The Shared Antiderivative

For \(u>0\), define

F(u)=\sqrt{1+u^2}+\ln\left(\frac{u}{1+\sqrt{1+u^2}}\right).

The logarithm is defined here because both the numerator and denominator inside its argument are positive. To check that this function is an antiderivative, differentiate each term. Write \(s=\sqrt{1+u^2}\), so \(s'=u/s\). Then

\begin{aligned} F'(u) &=\frac{u}{s}+\frac{1}{u}-\frac{u}{s(1+s)}\\ &=\frac{u}{1+s}+\frac{1}{u}\\ &=\frac{u^2+1+s}{u(1+s)}\\ &=\frac{s^2+s}{u(1+s)}\\ &=\frac{s}{u} =\frac{\sqrt{1+u^2}}{u}. \end{aligned}

In the second line, the first and third terms of the first line have been combined. The fourth line uses \(s^2=1+u^2\). Thus, for positive \(u\),

\int \frac{\sqrt{1+u^2}}{u}\,du=F(u)+C.

The restriction \(u>0\) is convenient for the logarithm as written. It will hold automatically in the exponential substitution, because \(e^x>0\), and throughout the logarithmic examples, because their \(x\)-intervals lie in the positive domain.

Exponential Graphs: Transforming the Integral

For \(y=e^x\), the derivative is \(dy/dx=e^x\). Therefore, on any finite interval \([\alpha,\beta]\),

L=\int_\alpha^\beta\sqrt{1+e^{2x}}\,dx.

Set \(u=e^x\). Then \(du=e^x\,dx=u\,dx\), so \(dx=du/u\). Since the exponential is increasing, the endpoints transform in order: \(x=\alpha\) gives \(u=e^\alpha\), and \(x=\beta\) gives \(u=e^\beta\). Both transformed endpoints are positive. The length becomes

L=\int_{e^\alpha}^{e^\beta}\frac{\sqrt{1+u^2}}{u}\,du =F(e^\beta)-F(e^\alpha).

The substitution changes both the integrand and the limits. Keeping the transformed limits is essential: after changing from \(x\) to \(u\), the integral must be evaluated at \(u=e^\alpha\) and \(u=e^\beta\), not at the original \(x\)-values.

Worked Example: Length of \(y=e^x\) from \(x=0\) to \(x=\ln 2\)

The interval is finite, and \(e^x\) is continuously differentiable there, so the graph-length formula applies. Since \(dy/dx=e^x\), the length is

L=\int_0^{\ln 2}\sqrt{1+e^{2x}}\,dx.

Use \(u=e^x\), with \(dx=du/u\). The lower endpoint becomes \(u=e^0=1\); the upper endpoint becomes \(u=e^{\ln 2}=2\). Thus

\begin{aligned} L &=\int_1^2\frac{\sqrt{1+u^2}}{u}\,du\\ &=F(2)-F(1)\\ &=\left(\sqrt{5}+\ln\frac{2}{1+\sqrt{5}}\right) -\left(\sqrt{2}+\ln\frac{1}{1+\sqrt{2}}\right)\\ &=\sqrt{5}-\sqrt{2} +\ln\left(\frac{2(1+\sqrt{2})}{1+\sqrt{5}}\right). \end{aligned}

The last line combines the logarithms using \(\ln A-\ln B=\ln(A/B)\), with both arguments positive. To check the result, differentiate the antiderivative \(F(e^x)\):

\frac{d}{dx}F(e^x) =F'(e^x)e^x =\frac{\sqrt{1+e^{2x}}}{e^x}e^x =\sqrt{1+e^{2x}}.

This is exactly the length integrand. Evaluating at \(x=\ln 2\) and \(x=0\) gives the endpoint difference above.

Logarithmic Graphs: Why Positivity Matters

For \(y=\ln x\), the domain requires \(x>0\), and the derivative is \(dy/dx=1/x\). On a finite interval \([a,b]\) with \(0<a<b\), the length is

L=\int_a^b\sqrt{1+\frac{1}{x^2}}\,dx.

Because \(x>0\), we can write \(\sqrt{x^2}=x\). Consequently,

\sqrt{1+\frac{1}{x^2}} =\sqrt{\frac{x^2+1}{x^2}} =\frac{\sqrt{x^2+1}}{x}.

The integrand is now \(F'(x)\), so the logarithmic graph has length

$$ L=F(b)-F(a). $$

This formula requires \(0<a<b\): the graph must be defined throughout the interval, and the antiderivative above is being used with positive arguments. The interval does not include \(x=0\), which is not in the domain of \(\ln x\).

Worked Example: Length of \(y=\ln x\) from \(x=1\) to \(x=2\)

Both endpoints are positive, and \(\ln x\) is continuously differentiable on \([1,2]\). Its derivative is \(1/x\), so

\begin{aligned} L &=\int_1^2\sqrt{1+\frac{1}{x^2}}\,dx\\ &=\int_1^2\frac{\sqrt{1+x^2}}{x}\,dx\\ &=F(2)-F(1)\\ &=\sqrt{5}-\sqrt{2} +\ln\left(\frac{2(1+\sqrt{2})}{1+\sqrt{5}}\right). \end{aligned}

To check, differentiate \(F(x)\) on \(x>0\). The derivative calculation above gives

F'(x)=\frac{\sqrt{1+x^2}}{x} =\sqrt{1+\frac{1}{x^2}},

where the final equality uses \(x>0\). This matches the graph-length integrand, and the evaluation at \(2\) minus the evaluation at \(1\) is the displayed length.

Reading the Endpoint Values Correctly

The same function \(F\) handles both graph families, but the endpoints supplied to it come from different variables. On an exponential graph, the input to \(F\) is \(e^x\); on a logarithmic graph, the input is \(x\). This distinction is especially clear when the intervals are chosen so their transformed endpoints match.

Worked Example: Length of \(y=\ln x\) from \(x=1\) to \(x=e\)

The interval \([1,e]\) lies entirely in the domain \(x>0\). The derivative is \(1/x\), and the positive-\(x\) simplification gives the integrand \(\sqrt{1+x^2}/x\). Therefore

\begin{aligned} L &=\int_1^e\frac{\sqrt{1+x^2}}{x}\,dx\\ &=F(e)-F(1)\\ &=\left(\sqrt{1+e^2}+\ln\frac{e}{1+\sqrt{1+e^2}}\right) -\left(\sqrt{2}+\ln\frac{1}{1+\sqrt{2}}\right)\\ &=\sqrt{1+e^2}-\sqrt{2} +\ln\left(\frac{e(1+\sqrt{2})}{1+\sqrt{1+e^2}}\right). \end{aligned}

The logarithm's argument is positive because \(e\), \(1+\sqrt{2}\), and \(1+\sqrt{1+e^2}\) are positive. To check the endpoint evaluation, \(F'(x)=\sqrt{1+x^2}/x\) on this interval, which is exactly the length integrand. Hence \(F(e)-F(1)\) gives the required length.

There is also a useful comparison: the exponential graph on \([0,\ln 2]\) and the logarithmic graph on \([1,2]\) have equal lengths. The exponential substitution sends its endpoints \(0,\ln 2\) to \(1,2\), respectively, so both lengths equal \(F(2)-F(1)\). This agrees with the fact that the graphs \(y=e^x\) and \(y=\ln x\) are reflections of one another across the line \(y=x\).

A Domain Pitfall to Avoid

When simplifying the logarithmic integrand, it is not enough to cancel the square in the denominator without considering its sign. The general identity is

\sqrt{x^2}=|x|,

not \(\sqrt{x^2}=x\) for every real \(x\). On the domain of \(\ln x\), however, \(x>0\), so \(|x|=x\). This is why the simplification used here is valid.

Common mistake: Writing \(\sqrt{1+1/x^2}=\sqrt{1+x^2}/x\) without stating a domain condition. The equality in that form requires \(x>0\). In a logarithmic graph-length problem, first check that the whole interval satisfies \(0<a<b\); then the denominator \(x\) is positive and the simplification is justified.

A second error is to substitute \(u=e^x\) but keep the original endpoints \(0\) and \(\ln 2\). After this substitution, the variable is \(u\), so the limits must be \(1\) and \(2\). The differential changes as well: \(dx=du/u\), not \(dx=du\). These two checks prevent a transformed integral from being evaluated in the wrong variable.

A Reliable Calculation Sequence

For these two graph families, the following sequence keeps the derivative, substitution, and domain conditions aligned.

Write the graph-length integrand.
Use \(\sqrt{1+(f'(x))^2}\) and identify the interval on which the graph is defined.

Transform or simplify carefully.
For \(e^x\), put \(u=e^x\) and replace \(dx\) by \(du/u\). For \(\ln x\), use \(x>0\) to simplify \(\sqrt{1+1/x^2}\).

Use the shared antiderivative.
Evaluate \(F\) at the transformed exponential endpoints or the original positive logarithmic endpoints.

Check the result.
Differentiate the antiderivative in the original variable and confirm that it reproduces the graph-length integrand.

These calculations illustrate how a suitable substitution can expose a shared structure in two apparently different problems. The exponential substitution converts growth in the independent variable into a positive integration variable, while the logarithmic derivative already produces the reciprocal form directly. In each case, verifying the domain and the endpoint transformation is part of the mathematical argument, not merely bookkeeping.

Check Your Understanding

Use the graph-length formula and the domain conditions in each question. Keep track of endpoint changes whenever the variable is substituted.

For \(y=e^x\) on \([0,\ln 3]\), what are the limits after setting \(u=e^x\), and what expression involving \(F\) gives the length?
For \(y=\ln x\) on \([2,4]\), write the simplified length integrand and the corresponding endpoint expression.
Why is \(\sqrt{1+1/x^2}=\sqrt{1+x^2}/x\) valid on the domain of \(y=\ln x\)?
Differentiate \(F(e^x)\) and explain why its derivative is the exponential graph-length integrand.
What must change in an integral's limits and differential when the substitution \(u=e^x\) is made?