Question Number 134152 by mohammad17 last updated on 28/Feb/21
Commented by mohammad17 last updated on 28/Feb/21
$${help}\:{me}\:{sir} \\ $$
Answered by bobhans last updated on 28/Feb/21
![Area = ∫_1 ^( 9) (1+(√x) )−(1+(x/3)) dx = ∫_1 ^( 9) ((√x)−(x/3))dx = [(2/3)x^(3/2) −(x^2 /6) ]_1 ^9 = ((2/3).(27−1)−(1/6)(80)) =((52)/3) − ((40)/3) = 4](Q134163.png)
$$\mathrm{Area}\:=\:\int_{\mathrm{1}} ^{\:\mathrm{9}} \left(\mathrm{1}+\sqrt{\mathrm{x}}\:\right)−\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{3}}\right)\:\mathrm{dx} \\ $$$$\:=\:\int_{\mathrm{1}} ^{\:\mathrm{9}} \:\left(\sqrt{\mathrm{x}}−\frac{\mathrm{x}}{\mathrm{3}}\right)\mathrm{dx} \\ $$$$\:=\:\left[\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}/\mathrm{2}} −\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\:\right]_{\mathrm{1}} ^{\mathrm{9}} \\ $$$$\:=\:\left(\frac{\mathrm{2}}{\mathrm{3}}.\left(\mathrm{27}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{80}\right)\right) \\ $$$$\:=\frac{\mathrm{52}}{\mathrm{3}}\:−\:\frac{\mathrm{40}}{\mathrm{3}}\:=\:\mathrm{4} \\ $$