Area Between Curves — Question 7

Question 7 – Solution

1. Identify the upper curve on $[1,2]$.

For $x\ge1$, $$x-\frac1x=\frac{x^2-1}{x}\ge0.$$ Thus the line $y=x$ lies above $y=1/x$. They meet at $(1,1)$, and the boundary $x=2$ closes the region.

2. Set up the area.

$$A=\int_1^2\left(x-\frac1x\right)\,dx.$$

3. Integrate each term.

Since $x>0$ on this interval, an antiderivative of $1/x$ is $\ln x$. $$\int x\,dx=\frac{x^2}{2},\qquad\int\frac1x\,dx=\ln x.$$

4. Substitute the bounds and simplify.

$$\begin{align*} A&=\left[\frac{x^2}{2}-\ln x\right]_1^2\\ &=\left(2-\ln2\right)-\left(\frac12-\ln1\right)\\ &=\boxed{\frac32-\ln2}. \end{align*}$$

5. Check the result.

Using $\ln2\approx0.69315$, the area is $1.5-0.69315\approx0.80685$ square units. The result is positive, as required.