Question Number 132123 by mohammad17 last updated on 11/Feb/21
Answered by liberty last updated on 11/Feb/21
![(1)c ∫_(−1) ^( 0) 0.2 dy + ∫_0 ^1 (0.2+cy)dy=1 0.2y ]_(−1) ^0 +(0.2y+(1/2)cy^2 )]_0 ^1 =1 0.2 +0.2+0.5 c=1 0.5 c = 0.6 ⇒ c = (6/5)=1.2](https://www.tinkutara.com/question/Q132129.png)
$$\left(\mathrm{1}\right)\mathrm{c}\: \\ $$$$\:\int_{−\mathrm{1}} ^{\:\mathrm{0}} \mathrm{0}.\mathrm{2}\:\mathrm{dy}\:+\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{0}.\mathrm{2}+\mathrm{cy}\right)\mathrm{dy}=\mathrm{1} \\ $$$$\left.\:\left.\mathrm{0}.\mathrm{2y}\:\right]_{−\mathrm{1}} ^{\mathrm{0}} +\left(\mathrm{0}.\mathrm{2y}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cy}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{1} \\ $$$$\:\mathrm{0}.\mathrm{2}\:+\mathrm{0}.\mathrm{2}+\mathrm{0}.\mathrm{5}\:\mathrm{c}=\mathrm{1} \\ $$$$\:\mathrm{0}.\mathrm{5}\:\mathrm{c}\:=\:\mathrm{0}.\mathrm{6}\:\Rightarrow\:\mathrm{c}\:=\:\frac{\mathrm{6}}{\mathrm{5}}=\mathrm{1}.\mathrm{2} \\ $$