Find the area under a logarithmic curve using its inverse

To use the inverse-area method for this logarithmic rule and interval, what should be done first?

Move u and compare P with Q on equal axis scales. The diagonal is y=x. Reflecting a point exchanges its input and output; it does not itself prove an area formula

When P has coordinates (eᵘ,u), what are the coordinates of the matching inverse point Q?

Use the inverse-area link

Transpose the input interval to the output interval and subtract the inverse integral from x₂y₂−x₁y₁. In the displayed ln(x) example the lower output is zero, so the boundary term is the enclosing rectangle

1. 1Restrict to a one-to-one domain Use an interval on which the logarithmic function has an inverse function \(\displaystyle f(x)=\log_e(x),\quad x>0\)
2. 2Form the inverse rule Swap x and y, then solve for the new output in terms of the new input \(\displaystyle y=\log_e(x)\Rightarrow x=e^{y}\Rightarrow f^{-1}(y)=e^{y}\)
3. 3Transpose the interval Evaluate the original function at both input endpoints and use those output values as inverse limits \(\displaystyle x=1\Rightarrow y=0,\quad x=e^{2}\Rightarrow y=2\)
4. 4Subtract the inverse area The required area is the enclosing rectangle less the area under the inverse curve \(\displaystyle A=2e^{2}-\int_0^2e^{y}\,dy=2e^{2}-\left(e^{2}-1\right)=e^{2}+1\)

Function and area data

\(\displaystyle \log_e\left(e^{2}\right)=\square\)

Find the upper output value of the logarithmic function

Function and area data

\(\displaystyle 2e^{2}-\left(e^{2}-1\right)=e^{2}+\square\)

Collect the constant term after applying the rectangle and inverse-integral terms

Both expressions give the same area. Which one explicitly uses the inverse-area relationship?

Keep the shifted endpoint grouped

Use the inverse-area relationship for this increasing, non-negative curve on the stated interval. First find the maximal logarithmic domain and the restricted output interval

Function and interval data

\(\displaystyle f(x)=2\ln(3x-2),\quad 3x-2>0\)

The maximal domain is x>d. Find the boundary d; it is excluded

Function and interval data

\(\displaystyle y=2\ln(3x-2)\)

Solve y=2 ln(3x−2). Find k in the exponential term exp(y/k)

Function and interval data

\(\displaystyle f(x)=2\ln(3x-2),\quad f^{-1}(y)=\frac{e^{y/2}+2}{m}\)

Find the denominator m in the inverse rule

Distinguish maximal and interval-restricted inverses

The full logarithmic rule maps (2/3,∞) onto all real outputs, so its maximal inverse has domain ℝ. For this area task only, restrict the inverse input to the function values at the two given endpoints

Function and interval data

\(\displaystyle f(x)=2\ln(3x-2),\quad x_2=\frac{e^2+2}{3}\)

Find the upper limit for the inverse integral on the stated area interval

Integrate the inverse over the output interval

Integrating exp(y/2) supplies a factor of 2. Evaluate the constant term and both endpoints before collecting terms

Function and interval data

\(\displaystyle I=\frac23e^2+c=\left[\frac23e^{y/2}+\frac23y\right]_0^4\)

Collect the constant term c in the inverse integral

Which endpoint-product expression keeps the whole upper input multiplied by its output?

Function and interval data

\(\displaystyle A=\frac{4(e^2+2)}3-\left(\frac23e^2+2\right)\)

Subtract the inverse integral and find K in A=K e²+c.

Function and interval data

\(\displaystyle A=\frac{4(e^2+2)}3-\left(\frac23e^2+2\right)\)

Find the constant c in the exact area form A=(2/3)e²+c.

Check the exact area before rounding

The domain and output interval are valid, the curve is non-negative, and the endpoint product exceeds the inverse integral. Keep this exact value until a decimal answer is requested

Exact area

\(\displaystyle A=\frac23(e^2+1)\)

Give the area to two decimal places

Do not round intermediate terms

Rounding the endpoint product and inverse integral separately can change the final rounded result. Retain the exact expressions through the subtraction

Keep the lower endpoint product when it is non-zero

For ln(x) on [e,e²], the output interval is [1,2]. The lower product is e×1, not zero. This identity gives a signed integral in general; here the function is positive, so it also gives geometric area

For ln(x) on [e,e²], why is 2e² minus the inverse integral insufficient?

Construct the complete solution, ungraded

On paper, give the maximal domain, form the inverse, find its restricted limits, sketch the paired curves and derive the exact area using both endpoint products. The component entries that follow do not grade the complete sketch, inverse or justification

Function and interval data

\(\displaystyle f(x)=3\ln(2x-1),\quad [1,(e^3+1)/2]\)

After completing the full derivation, enter K in A=K e³+c.

Function and interval data

\(\displaystyle f(x)=3\ln(2x-1),\quad [1,(e^3+1)/2]\)

Enter c in A=3e³+c from your full derivation

Check every part of the written argument

The curve is increasing and non-negative on this interval. Compare your complete argument, not only its two final coefficients

1. 1Domain and inverse The maximal domain is shown below. On the area interval the inverse input runs from 0 to 9 \(\displaystyle x>\frac12,\quad f^{-1}(y)=\frac{e^{y/3}+1}{2},\quad 0\le y\le9\)
2. 2Sketch check Sketch these restricted curves as reflections across the diagonal, with matching endpoints exchanged \(\displaystyle \begin{aligned}f(x)&=3\ln(2x-1),\quad 1\le x\le\frac{e^3+1}{2}\\f^{-1}(y)&=\frac{e^{y/3}+1}{2},\quad 0\le y\le9\end{aligned}\)
3. 3Inverse integral Integrate exp(y/3) with its factor 3 and evaluate both limits \(\displaystyle I=\left[\frac32e^{y/3}+\frac y2\right]_0^9=\frac32e^3+3\)
4. 4Subtract with grouped endpoint The lower endpoint product is 1×0. Keep the whole compound upper input multiplied by 9. \(\displaystyle A=\frac{9(e^3+1)}2-\left(\frac32e^3+3\right)=3e^3+\frac32\)

Transpose the interval, not just the rule

State the logarithmic domain, find the inverse, use the original output values as its limits and retain both endpoint products. Group compound endpoints before multiplying and round only when requested. Direct integration can be mathematically correct even when an inverse-area method is requested