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Question Number 191507 by Spillover last updated on 24/Apr/23

Answered by cortano12 last updated on 25/Apr/23

$$\left(\mathrm{a}\right)\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=1$$$$\mathrm{k}\underset{0}{\stackrel{4}{\int }}({\mathrm{x}}^2-3\mathrm{x}+4)\mathrm{dx}+\underset{4}{\stackrel{5}{\int }}8\mathrm{k}\mathrm{dx}=1$$$$\mathrm{k}(\frac{64}{3}-\frac{3.16}{2}+16)+8\mathrm{k}(5-4)=1$$$$\frac{40\mathrm{k}}{3}+8\mathrm{k}=1\Rightarrow \mathrm{k}=\frac{3}{64}$$

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