Question Number 88210 by bagjamath last updated on 09/Apr/20
![If f(x)= { ((1−x, 0≤x≤1)),((0, 1≤x≤2 )),(((2−x)^2 , 2≤x≤3)) :} and φ(x)=∫_( 0) ^x f(t) dt. Then for any x ∈ [2, 3], φ(x) =](Q88210.png)
$$\mathrm{If}\:{f}\left({x}\right)=\begin{cases}{\mathrm{1}−{x},\:\:\:\:\:\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}}\\{\mathrm{0},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\leqslant{x}\leqslant\mathrm{2}\:\:\:\:}\\{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} ,\:\:\:\mathrm{2}\leqslant{x}\leqslant\mathrm{3}}\end{cases}\:\mathrm{and}\: \\ $$$$\phi\left({x}\right)=\underset{\:\mathrm{0}} {\overset{\mathrm{x}} {\int}}\:{f}\left({t}\right)\:{dt}.\:\mathrm{Then}\:\mathrm{for}\:\mathrm{any}\:{x}\:\in\:\left[\mathrm{2},\:\mathrm{3}\right],\: \\ $$$$\phi\left({x}\right)\:= \\ $$
Answered by MJS last updated on 09/Apr/20
![=∫_0 ^1 (1−t)dt+0∫_1 ^2 dt+∫_2 ^x (2−t)^2 dt= =[t−(1/2)t^2 ]_0 ^1 +[4t−2t^2 +(t^3 /3)]_2 ^x = =(1/3)x^3 −2x^2 +4x−((13)/6)](Q88213.png)
$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{1}−{t}\right){dt}+\mathrm{0}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{dt}+\underset{\mathrm{2}} {\overset{{x}} {\int}}\left(\mathrm{2}−{t}\right)^{\mathrm{2}} {dt}= \\ $$$$=\left[{t}−\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} +\left[\mathrm{4}{t}−\mathrm{2}{t}^{\mathrm{2}} +\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{2}} ^{{x}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}−\frac{\mathrm{13}}{\mathrm{6}} \\ $$