Areas with logarithmic functions

Finding areas under the hyperbola y = 1/x, between a hyperbola and a line, and areas involving y = ln x using the y-axis.

Areas under the hyperbola

Example 1●

Find the area under $y = \dfrac1x$ from (a) $x = 1$ to $x = e^2$, and (b) $x = \tfrac12$ to $x = 2$.

$\Big[\ln x\Big]_1^{e^2} = 2$ square units.

$\ln 2 - \ln\tfrac12 = 2\ln 2$ square units.

Between a hyperbola and a line

Between y = 1/x and x + y = 5/2

The line is above the hyperbola between their intersections at x = 1/2 and x = 2.

Example 2●●●

Find the area enclosed between $y = \dfrac1x$ and the line $x + y = \tfrac52$.

$\dfrac1x = \tfrac52 - x$ gives $2x^2 - 5x + 2 = 0$, so $x = \tfrac12$ or $x = 2$.

Intersections

$\displaystyle\int_{\frac12}^{2} \left(\tfrac52 - x - \tfrac1x\right)dx = \Big[\tfrac52x - \tfrac{x^2}{2} - \ln x\Big]_{\frac12}^{2}$.

Integrate

$= (3 - \ln 2) - \left(\tfrac98 + \ln 2\right) = \tfrac{15}{8} - 2\ln 2$ square units.

Areas involving $y = \ln x$

Example 3●●

Find the area bounded by $y = \ln x$, the $x$-axis and the line $x = e$.

Using $\displaystyle\int \ln x\,dx = x\ln x - x$: $\Big[x\ln x - x\Big]_1^e = 0 - (-1) = 1$.

Method 1

Use the $y$-axis: $x = e^y$, and the region lies between $x = e^y$ and $x = e$ for $0 \le y \le 1$. Area $= \displaystyle\int_0^1 (e - e^y)\,dy = e - (e - 1) = 1$ square unit.

Method 2

Flashcards

Area under $y = dfrac1x$ from $1$ to $t$

$\ln t$

Area under $y = ln x$ from $1$ to $e$

$1$

$y = ln x$ rewritten for the $y$-axis

$x = e^y$

Practice

Find the area under $y = \dfrac1x$ from $x = 1$ to $x = e^2$.
Find the area under $y = \dfrac1x$ from $x = \tfrac12$ to $x = 2$. Type $\ln 2$ as ln2.
Find the area enclosed between $y = \dfrac1x$ and $x + y = \tfrac52$.
Find the area bounded by $y = \ln x$, the $x$-axis and $x = e$.
Find the area under $y = \dfrac{1}{x + 1}$ from $x = 0$ to $x = 3$.
Find the area under $y = \dfrac2x$ from $x = 1$ to $x = 4$.
Find the area under $y = \dfrac1x$ from $x = 2$ to $x = 6$.
Find the area under $y = \dfrac{1}{2x}$ from $x = 1$ to $x = e$.
The area under $y = \dfrac1x$ from $x = 1$ to $x = t$ (where $t > 1$) is: