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Question Number 127704 by NATTAPONG4359 last updated on 01/Jan/21

$$$$$$iff\left(x\right)=\{\begin{array}{l}x-n;2n⩽x⩽2n+1\\ n+1;2n+1⩽x⩽2n+2\end{array}wheren=0,1,2,3,..,9$$$$find{\int }_0^{20}f\left(x\right)dx$$

Answered by mahdipoor last updated on 01/Jan/21

$${\int }_{2n}^{2n+1}f\left(x\right)dx={\int }_{2n}^{2n+1}(x-n)dx={[\frac{x^2}{2}-nx]}_{2n}^{2n+1}$$$$=(\frac{{(2n+1)}^2}{2}-n(2n+1))-(\frac{{\left(2n\right)}^2}{2}-2\left(2n\right))$$$$=n-\frac{1}{2}$$$${\int }_{2n+1}^{2n+2}f\left(x\right)dx={\int }_{2n+1}^{2n+2}(1+n)dx={\left[(1+n)x\right]}_{2n+1}^{2n+2}$$$$=(1+n)(2n+2-(2n+1))=1+n$$$$$$$${\int }_0^{20}f\left(x\right)dx=({\int }_0^{1}f\left(x\right)dx+{\int }_2^{3}f\left(x\right)dx+...+{\int }_{18}^{19}f\left(x\right)dx)$$$$+({\int }_1^{2}f\left(x\right)dx+{\int }_3^{4}f\left(x\right)dx+...+{\int }_{19}^{20}f\left(x\right)dx)=$$$$\underset{n=0}{\sum ^9}(n-\frac{1}{2})+\underset{n=0}{\sum ^9}(1+n)=\underset{n=0}{\sum ^9}(2n+\frac{1}{2})=$$$$2\times \left(\frac{9\times 10}{2}\right)+10\times \frac{1}{2}=95$$

Commented by mehedi last updated on 22/Dec/21

$$4$$

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