Question Number 168665 by mokys last updated on 15/Apr/22
Commented by cortano1 last updated on 15/Apr/22
![A_1 = ∫_(−2) ^0 (8−x^3 )dx= [8x−(x^4 /4)]_(−2) ^0 = 8(2)−(1/4)(−16)=16+4=20 A_2 =∫_0 ^2 (8−2x^2 )dx=[8x−((2x^3 )/3)]_0 ^2 = 8(2)−(2/3)(8)=8×((4/3))=((32)/3) ∴ A= 20+((32)/3)= ((92)/3)](https://www.tinkutara.com/question/Q168667.png)
$$\:{A}_{\mathrm{1}} =\:\underset{−\mathrm{2}} {\overset{\mathrm{0}} {\int}}\left(\mathrm{8}−{x}^{\mathrm{3}} \right){dx}=\:\left[\mathrm{8}{x}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right]_{−\mathrm{2}} ^{\mathrm{0}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{8}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{16}\right)=\mathrm{16}+\mathrm{4}=\mathrm{20} \\ $$$$\:{A}_{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\left(\mathrm{8}−\mathrm{2}{x}^{\mathrm{2}} \right){dx}=\left[\mathrm{8}{x}−\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{8}\left(\mathrm{2}\right)−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{8}\right)=\mathrm{8}×\left(\frac{\mathrm{4}}{\mathrm{3}}\right)=\frac{\mathrm{32}}{\mathrm{3}} \\ $$$$\:\therefore\:{A}=\:\mathrm{20}+\frac{\mathrm{32}}{\mathrm{3}}=\:\frac{\mathrm{92}}{\mathrm{3}} \\ $$
Answered by mr W last updated on 15/Apr/22
$${A}=\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)×\mathrm{8}×\mathrm{2}=\frac{\mathrm{92}}{\mathrm{3}}=\mathrm{30}.\mathrm{7} \\ $$