Area — Question 10

Question 10 - Solution

On $1\le x\le e^{2}$, we have $\sqrt{x}\ge \ln x$, so $$A=\int_{1}^{e^{2}}\bigl(\sqrt{x}-\ln x\bigr)\,dx.$$ Compute: $$\int_{1}^{e^{2}}\sqrt{x}\,dx=\left[\frac{2}{3}x^{3/2}\right]_{1}^{e^{2}} =\frac{2}{3}(e^{3}-1),$$ $$\int_{1}^{e^{2}}\ln x\,dx=\left[x\ln x-x\right]_{1}^{e^{2}} =e^{2}+1.$$ Thus $$A=\frac{2}{3}(e^{3}-1)-(e^{2}+1)=\frac{2}{3}e^{3}-e^{2}-\frac{5}{3}.$$ $$\boxed{\frac{2}{3}e^{3}-e^{2}-\frac{5}{3}}$$