Mohon bantuannya :)beserta caranya juga ya :)

Mohon bantuannya :)beserta caranya juga ya :)

${boxed{Penyelesaian : Soal}}$

$=displaystyleint_{ 2 }^{ 3 } 3x ^ { 2 } +6x+4 d x$

$=int 3x^{2}+6x+4mathrm{d}x$

$=int 3x^{2}mathrm{d}x+int 6xmathrm{d}x+int 4mathrm{d}x$

$=3int x^{2}mathrm{d}x+6int xmathrm{d}x+int 4mathrm{d}x$

$=x^{3}+6int xmathrm{d}x+int 4mathrm{d}x$

$=x^{3}+3x^{2}+int 4mathrm{d}x$

$=x^{3}+3x^{2}+4x$

$=3^{3}+3times 3^{2}+4times 3-left(2^{3}+3times 2^{2}+4times 2right)$

$= {3}^{3} + 3 times {3}^{2} + 4 times 3 - (8 + 3 times 4 + 4 times 2)$

$={3}^{3} + 3 times { 3 }^{2} + 4 times 3 - (8 + 12 + 8)$

$= {3}^{3} + 3 times {3}^{2} + 4 times 3 - (20 + 8)$

$={3}^{3} + 3 times {3}^{2} + 4 times 3 - 28$

$=27 + 3 times 9+ 4 times 3 -28$

$=27 + 27+ 12 -28$

$=54 + 12 - 28$

$=66 - 28$

$=38$

Jawabannya: B