Question-202882

Question Number 202882 by dimentri last updated on 05/Jan/24

$$\:\:\:\:\downharpoonleft\underline{\:} \\ $$

Answered by cortano12 last updated on 05/Jan/24

{ ((5∫_3 ^6 f(x)dx=10)),((5∫_1 ^6 f(x)dx=2)) :} ⇒∫_1 ^6 f(x)dx = ∫_1 ^3 f(x)dx+∫_3 ^6 f(x)dx ⇒5∫_1 ^6 f(x)dx = 5∫_1 ^3 f(x)dx+5∫_3 ^6 f(x)dx ⇒2 = 5∫_1 ^3 f(x)dx+10 ⇒5∫_1 ^3 f(x)dx = −8

$$\:\:\begin{cases}{\mathrm{5}\underset{\mathrm{3}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{10}}\\{\mathrm{5}\underset{\mathrm{1}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{2}}\end{cases} \\ $$$$\:\:\Rightarrow\underset{\mathrm{1}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\:=\:\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\underset{\mathrm{3}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:\Rightarrow\mathrm{5}\underset{\mathrm{1}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\:=\:\mathrm{5}\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\mathrm{5}\underset{\mathrm{3}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:\Rightarrow\mathrm{2}\:=\:\mathrm{5}\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\mathrm{10} \\ $$$$\:\Rightarrow\mathrm{5}\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\:=\:−\mathrm{8}\: \\ $$

Answered by MM42 last updated on 05/Jan/24

$$\int_{\mathrm{1}} ^{\mathrm{3}} \mathrm{5}{f}=\int_{\mathrm{1}} ^{\mathrm{6}} \mathrm{5}{f}+\int_{\mathrm{6}} ^{\mathrm{3}} \mathrm{5}{f}=\mathrm{2}−\mathrm{10}=−\mathrm{8}\:\checkmark \\ $$

Answered by aba last updated on 05/Jan/24

∫_1 ^6 f(x)dx=∫_1 ^3 f(x)dx+∫_3 ^6 f(x)dx ⇒∫_1 ^3 f(x)dx=∫_1 ^6 f(x)−∫_3 ^6 f(3)dx ⇒ 5∫_1 ^3 f(x)dx=5.(2/5)−5.2 ⇒ ∫_1 ^3 5f(x)dx=−8✓✓

$$\int_{\mathrm{1}} ^{\mathrm{6}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{1}} ^{\mathrm{3}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\int_{\mathrm{3}} ^{\mathrm{6}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{\mathrm{3}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{1}} ^{\mathrm{6}} \mathrm{f}\left(\mathrm{x}\right)−\int_{\mathrm{3}} ^{\mathrm{6}} \mathrm{f}\left(\mathrm{3}\right)\mathrm{dx} \\ $$$$\Rightarrow\:\mathrm{5}\int_{\mathrm{1}} ^{\mathrm{3}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{5}.\frac{\mathrm{2}}{\mathrm{5}}−\mathrm{5}.\mathrm{2} \\ $$$$\Rightarrow\:\int_{\mathrm{1}} ^{\mathrm{3}} \mathrm{5f}\left(\mathrm{x}\right)\mathrm{dx}=−\mathrm{8}\checkmark\checkmark \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply