Question

$\left.\begin{array}{c c c} \displaystyle{f(x) ={\begin{cases} \sf {x}^{2} + 2x + 3 \: \: when \: 0 \le \: x \le 1 \\ \\ \sf\sqrt{4x} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{when \: 1 < x \le 4}\end{cases}}}\end{array}\right \}$
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Answer 1

Step-by-step explanation:

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Answer 2

$\tt\huge{Answer}$

$\displaystyle{ \int \limits_ {0}^{4} f( \sf{x})dx} = \int \limits_{0}^{1}( \sf {x}^{2} + 2x + 3)dx + \int \limits_{1}^{4}( \sqrt{4x})dx$

$\displaystyle{ = \int \limits_{0}^{1}( \sf {x}^{2} + 2x + 3)dx + \int \limits_{1}^{4} \bigg(2 \sqrt{x} \bigg)dx}$

$= \left[\begin{array}{c c c} \\ \sf{ \dfrac{ {x}^{2 + 1} }{(2 + 1) } + 2 \dfrac{ {x}^{1 + 1} }{(1 + 1)} + 3x} \\ \\ \end{array} \right]_ {0}^{1} + \\ \\ \left[\begin{array}{c c c} \\ \sf{2 \dfrac{ {x}^{ \frac{1}{2} + 1} }{ \bigg( \dfrac{1}{2} + 1 \bigg)} } \\ \\ \end{array}\right]_{1}^{4}$

$= \left[\begin{array}{c c c} \\ \sf{ \dfrac{ {x}^{3} }{3} + 2 \dfrac{ {x}^{2} }{2} + 3x} \\ \\ \end{array} \right]_ {0}^{1} + \left[\begin{array}{c c c} \\ \sf{2 \dfrac{ {x}^{ \frac{3}{2} } }{ \dfrac{3}{2} } } \\ \\ \end{array}\right]_{1}^{4}$

$= \left[\begin{array}{c c c} \\ \sf{ \dfrac{ {x}^{3} }{3} + {x}^{2} + 3x} \\ \\ \end{array} \right]_ {0}^{1} + \left[\begin{array}{c c c} \\ \sf{2 \dfrac{ {4x}^{ \frac{3}{2} } }{3 }}\\ \\ \end{array}\right]_{1}^{4}$

$= \left[\begin{array}{c c c} \\ \sf{ \bigg( \dfrac{(1)}{3} + {(1)}^{2} + 3(1)} \bigg) - \bigg( \sf{ \dfrac{ {(0)}^{3} }{3} + {(0)}^{2}} + 3(0) \bigg) \\ \\ \end{array} \right]+$

$\left[\begin{array}{c c c} \\ \bigg( \sf{\dfrac{4(4)^{ \frac{3}{2} } }{3} \bigg) - \bigg( \dfrac{4(4)^{ \frac{3}{2} } }{3} \bigg)} \\ \\ \end{array}\right]$

$= \left[\begin{array}{c c c} \\ \sf{ \bigg( \dfrac{1}{3} + 1 + 3} \bigg) - \bigg( \sf{ 0 + 0 + 0} \bigg) \\ \\ \end{array} \right] + \left[\begin{array}{c c c} \\ \sf{ \bigg( \dfrac{4(8)}{3} \bigg) - \bigg( \dfrac{4(1)}{2} \bigg) } \\ \\ \end{array}\right]$

$= \left[\begin{array}{c c c} \\ \sf{ \bigg( \dfrac{1}{3} + \dfrac{12}{3} } \bigg) \\ \\ \end{array} \right] + \left[\begin{array}{c c c} \\ \sf{ \bigg( \dfrac{32}{3} \bigg) - \bigg( \dfrac{4}{2} \bigg) } \\ \\ \end{array}\right]$

$= \sf{ \dfrac{13}{3} + \dfrac{28}{3} }$

$\sf{ = \dfrac{41}{3} }$

∴The Definite integral of the piece wise defined function f(x) from 0 to 4 is equal to $\sf{\dfrac{41}{3} }$