Question Number 191507 by Spillover last updated on 24/Apr/23
Answered by cortano12 last updated on 25/Apr/23
$$\left(\mathrm{a}\right)\:\underset{−\infty} {\overset{\infty} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{1} \\ $$$$\:\:\:\:\mathrm{k}\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{4}\right)\mathrm{dx}+\underset{\mathrm{4}} {\overset{\mathrm{5}} {\int}}\mathrm{8k}\:\mathrm{dx}\:=\:\mathrm{1} \\ $$$$\:\:\:\mathrm{k}\left(\frac{\mathrm{64}}{\mathrm{3}}−\frac{\mathrm{3}.\mathrm{16}}{\mathrm{2}}+\mathrm{16}\right)+\mathrm{8k}\left(\mathrm{5}−\mathrm{4}\right)=\mathrm{1} \\ $$$$\:\:\:\frac{\mathrm{40k}}{\mathrm{3}}+\mathrm{8k}=\mathrm{1}\Rightarrow\mathrm{k}=\frac{\mathrm{3}}{\mathrm{64}} \\ $$