What is the final area of the curvilinear trapezoid bounded by $y=1/x$ on [1,2]?

Conditions

• The curve is $y = \frac{1}{x}$.
• The interval is $[1, 2]$.
• The area is defined as the limit of the Riemann sum.

Reasoning, step by step

1. Express the Riemann sum as the difference of two harmonic series.
2. Apply the asymptotic expansion to get $(\ln 2n + \gamma + \sigma_{2n}) - (\ln n + \gamma + \sigma_n)$.
3. Cancel $\gamma$ and simplify $\ln 2n - \ln n$ to $\ln 2$.
4. Take the limit as $n \to \infty$, noting that $\sigma_{2n} \to 0$ and $\sigma_n \to 0$.
5. Conclude the area is $\ln 2 + 0 - 0 = \ln 2$.

Example

The video concludes: 'Therefore, the final result for the area of the curvilinear trapezoid is $\ln 2$.'

Common misconceptions

• Thinking the area is $\ln n$ or depends on $n$ in the final answer.
• Forgetting that the remainder terms $\sigma_n$ vanish in the limit.

Watch the explanation

Explore next

• How does this relate to the definite integral of $1/x$?
• What is the numerical value of ln 2?