Question

bantu jawab dong........​

Answer 1

pembahasan :

∫₀¹ (5x⁴ - 6x³ + 4x² + x) dx =

( x⁵ -6/4 x⁴ + 4/3 x³ + ½x² + C) |₀¹ =

( 1⁵ -6/4 (1)⁴ + 4/3 (1)³ + ½(1)² ) - 0 =

4/3

jawaban OPSI B

Answer 2

Verified answer

[tex]\sf \int\limits^{ \: \: 1}_{ \: \: 0} \: (5 {x}^{4} - 6 {x}^{3} + {4x}^{2} + x)dx[/tex]

[tex]\small\sf \int\limits^{ \: \: 1}_{ \: \: 0} \: ( \frac{5}{4 + 1} {x}^{4 + 1} - \frac{6}{3 + 1} {x}^{3 + 1} + \frac{4}{2 + 1} {x}^{2 + 1} + \frac{1}{1 + 1} {x}^{1 + 1} )dx[/tex]

[tex]\sf \int\limits^{ \: \: 1}_{ \: \: 0} \: ( \frac{5}{5} {x}^{5} - \frac{6}{4} {x}^{4} + \frac{4}{3} {x}^{3} + \frac{1}{2} {x}^{2} )dx[/tex]

[tex]\sf \int\limits^{ \: \: 1}_{ \: \: 0} \: ( {x}^{5} - \frac{3}{2} {x}^{4} + \frac{4}{3} {x}^{3} + \frac{1}{2} {x}^{2} )dx[/tex]

[tex]\sf {x}^{5} - \frac{3}{2} {x}^{4} + \frac{4}{3} {x}^{3} + \frac{1}{2} {x}^{2} \huge| ^{\small 1}_{\small 0}[/tex]

[tex]\small\sf (1 {}^{5} - \frac{3}{2} (1 {)}^{4} + \frac{4}{3} (1 {)}^{3} + \frac{1}{2} (1 {)}^{2} ) - 0[/tex]

[tex]\sf (1 - \frac{3}{2} + \frac{4}{3} + \frac{1}{2} ) - 0[/tex]

[tex]\sf ( \frac{6}{6} - \frac{9}{6} + \frac{8}{6} + \frac{3}{6} ) - 0[/tex]

[tex]\sf \frac{8}{6} - 0[/tex]

[tex] \tt \frac{8}{6} = \boxed{ \frac{4}{3} }[/tex]