Area Between Curves — Question 4

Question 4 – Solution

1. Identify the upper curve on $[1,e]$.

Let $h(x)=x-1-\ln x$. Then $h(1)=0$ and $$h'(x)=1-\frac1x\ge0\qquad(1\le x\le e).$$ Thus $x-1\ge\ln x$ throughout the interval.

2. Set up the area.

$$A=\int_1^e(x-1-\ln x)\,dx.$$

3. Find the logarithm antiderivative.

Integration by parts, with $u=\ln x$ and $dv=dx$, gives $$\int\ln x\,dx=x\ln x-\int1\,dx=x\ln x-x.$$ Here $du=dx/x$ and $v=x$. Combine the terms: $x^2/2-x-(x\ln x-x)=x^2/2-x\ln x$.

4. Apply the upper and lower bounds.

$$\begin{align*} A&=\left[\frac{x^2}{2}-x-(x\ln x-x)\right]_1^e\\ &=\left[\frac{x^2}{2}-x\ln x\right]_1^e\\ &=\left(\frac{e^2}{2}-e\right)-\frac12 =\boxed{\frac{e^2}{2}-e-\frac12}. \end{align*}$$

5. Check the result.

Since $\ln e=1$ and $\ln1=0$, the endpoint terms above give an area of approximately $0.4762$ square units, which is positive.