evaluate-f-x-dx-where-f-x-e-x-x-0-1-x-0-lt-x-1-1-x-2-1-lt-x-2-5-2-lt-x-5-5-1-x-5-2-x-gt-5-

Question Number 216 by 123456 last updated on 25/Jan/15

evaluate

\underset{- \infty}{\int^{+ \infty}} f (x) d x

where

f (x) = {\begin{cases} e^{x} & x ⩽ 0 \\ 1 + x & 0 < x ⩽ 1 \\ 1 + x^{2} & 1 < x ⩽ 2 \\ 5 & 2 < x ⩽ 5 \\ \frac{5}{1 + {(x - 5)}^{2}} & x > 5 \end{cases}

Answered by prakash jain last updated on 16/Dec/14

= \underset{- \infty}{\int^{0}} e^{x} d x + \underset{0}{\int^{1}} (1 + x) d x + \underset{1}{\int^{2}} (1 + x^{2}) d x + \underset{2}{\int^{5}} 5 d x

+ \underset{5}{\int^{\infty}} \frac{5}{1 + {(x - 5)}^{2}} d x

= 1 + {[x + \frac{x^{2}}{2}]}_{0}^{1} + {[x + \frac{x^{3}}{3}]}_{1}^{2} + {[5 x]}_{2}^{5} + {[{5 \tan}^{- 1} \frac{x - 5}{1}]}_{5}^{\infty}

= 1 + \frac{3}{2} + [2 + \frac{8}{3} - 1 - \frac{1}{3}] + [25 - 10] + [\frac{5 π}{2}]

= \frac{5}{2} + \frac{10}{3} + 15 + \frac{π}{2}

= \frac{15 + 20 + 90}{6} + \frac{π}{2} = \frac{125}{2} + \frac{5 π}{2}