Question

Evaluate the following:

${\displaystyle {\int }_1^4}$ f(x) dx where f(x) = ${\displaystyle \{\begin{array}{ll}4x+3\text{,}& 1\le x\le 2\\ 3x+5\text{,}& 2<x\le 4\end{array}}$

Answer 1

${\displaystyle {\int }_1^4}$ f(x) dx = ${\displaystyle {\int }_1^2\text{f}\left(x\right) \text{d}x+{\int }_2^4\text{f}\left(x\right) \text{d}x}$

= ${\displaystyle {\int }_1^2(4x+3) \text{d}x+{\int }_2^4(3x+5) \text{d}x}$

= ${\displaystyle {[4\left(\frac{x^2}{2}\right)+3x]}_1^2+{[3\left(\frac{x^2}{2}\right)+5x]}_2^4}$

= ${\displaystyle {[2x^2+3x]}_1^2+{[\frac{3}{2}{x}^2+5x]}_2^4}$

= ${\displaystyle [2{\left(2\right)}^2+3\left(2\right)]-[2{\left(1\right)}^2+3\left(1\right)]+[\frac{3}{2}{\left(4\right)}^2+5\left(4\right)]-[\frac{3}{2}{\left(2\right)}^2+5\left(2\right)]}$

= ${\displaystyle [2\left(4\right)+6]-[2+3]+[\frac{3}{16}+20]-[\frac{3}{2}\left(4\right)+10]}$

= [8 + 6] – [5] + [24 + 20] – [6 + 10]

= 14 – 5 + 44 – 16

= 58 – 21

= 37