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Question Number 113668 by mohammad17 last updated on 14/Sep/20

Answered by john santu last updated on 14/Sep/20

If ∫_(−2) ^6 (f(x)+3)dx = 32 →∫_(−2) ^6 f(x)dx+3(8)=32   ∫_(−2) ^6 f(x)dx = 8...(i)   ∫_1 ^6 f(x) dx = 6 ...(ii)  we want to find ∫_(−2) ^1 f(x) dx  → ∫_(−2) ^6 f(x)dx = ∫_(−2) ^1 f(x)dx+∫_1 ^6 f(x)dx  →  8 = ∫_(−2) ^1 f(x)dx + 6   ∴ ∫_(−2) ^1 f(x)dx = 2

$${If}\:\int_{−\mathrm{2}} ^{\mathrm{6}} \left({f}\left({x}\right)+\mathrm{3}\right){dx}\:=\:\mathrm{32}\:\rightarrow\int_{−\mathrm{2}} ^{\mathrm{6}} {f}\left({x}\right){dx}+\mathrm{3}\left(\mathrm{8}\right)=\mathrm{32} \\ $$$$\:\int_{−\mathrm{2}} ^{\mathrm{6}} {f}\left({x}\right){dx}\:=\:\mathrm{8}...\left({i}\right) \\ $$$$\:\int_{\mathrm{1}} ^{\mathrm{6}} {f}\left({x}\right)\:{dx}\:=\:\mathrm{6}\:...\left({ii}\right) \\ $$$${we}\:{want}\:{to}\:{find}\:\int_{−\mathrm{2}} ^{\mathrm{1}} {f}\left({x}\right)\:{dx} \\ $$$$\rightarrow\:\underset{−\mathrm{2}} {\overset{\mathrm{6}} {\int}}{f}\left({x}\right){dx}\:=\:\underset{−\mathrm{2}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}+\overset{\mathrm{6}} {\int}_{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$\rightarrow\:\:\mathrm{8}\:=\:\underset{−\mathrm{2}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}\:+\:\mathrm{6}\: \\ $$$$\therefore\:\underset{−\mathrm{2}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}\:=\:\mathrm{2} \\ $$

Commented by mohammad17 last updated on 14/Sep/20

thank you sir very thank you

$${thank}\:{you}\:{sir}\:{very}\:{thank}\:{you} \\ $$

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