Area — Question 9

Question 9 - Solution

On the interval $0\le x\le 1$, we compare the curves. At $x=0$, $$\ln(1)=0,\qquad \frac{0}{1}=0.$$ At $x=1$, $$\ln 2>\frac12,$$ so $$\ln(1+x)\ge \frac{x}{1+x} \quad\text{on }[0,1].$$

Thus the enclosed area is $$A=\int_{0}^{1}\left(\ln(1+x)-\frac{x}{1+x}\right)\,dx.$$

Split the integral: $$A=\int_{0}^{1}\ln(1+x)\,dx-\int_{0}^{1}\frac{x}{1+x}\,dx.$$

Evaluate the first integral: $$\int \ln(1+x)\,dx=(1+x)\ln(1+x)-x,$$ so $$\int_{0}^{1}\ln(1+x)\,dx =\bigl[(1+x)\ln(1+x)-x\bigr]_{0}^{1} =2\ln 2-1.$$

For the second integral, rewrite: $$\frac{x}{1+x}=1-\frac{1}{1+x}.$$ Then $$\int_{0}^{1}\frac{x}{1+x}\,dx =\int_{0}^{1}\left(1-\frac{1}{1+x}\right)\,dx =1-\ln 2.$$

Subtract: $$A=(2\ln 2-1)-(1-\ln 2)=3\ln 2-2.$$

$$\boxed{3\ln 2-2}$$