Asked by steven
Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.)
f(x) = 1/2x+ 2/3x^2 - 5/6x^3
f(x) = 1/2x+ 2/3x^2 - 5/6x^3
Answers
Answered by
bobpursley
what is .25x^2 + (2/9)x^3- (5/24)x^4 + C
check that.
check that.
Answered by
Bosnian
x / 2 + 2 x^2 / 3 - 5 x^3 / 6 =
3 ∙ x / 3 ∙ 2 + 2 ∙ 2 x^2 / 2 ∙ 3 - 5 x^3 / 6 =
3 ∙ x / 6 + 4 x^2 / 6 - 5 x^3 / 6 =
( 1 / 6 ) ∙ ( 3 x + 4 x^2 - 5 x^3 ) =
( 1 / 6 ) ∙ ( - 5 x^3 + 3 x + 4 x^2 )
∫ ( 1 / 6 ) ∙ ( - 5 x^3 + 4 x^2 + 3 x ) dx =
( 1 / 6 ) ∙ ( - 5 ∫ x^3 dx + 4 ∫ x^2 dx + 3 ∫ x dx ) =
( 1 / 6 ) ∙ ( - 5 x^4 / 4 + 4 x^3 / 3 + 3 x^2 ) + C =
( 1 / 6 ) ∙ ( - 5 ∙ 3 x^4 / 3 ∙ 4 + 4 ∙ 4 x^3 / 4 ∙ 3 + 3 ∙ 12 x^2 / 12 ) + C =
( 1 / 6 ) ∙ ( - 15 ∙ x^4 / 12 + 16 x^3 / 12 + 36 x^2 / 12 ) + C =
( 1 / 6 ∙ 12 ) ∙ ( - 15 ∙ x^4 + 16 x^3 + 36 x^2 ) + C =
( 1 / 72 ) ∙ ( - 15 ∙ x^4 + 16 x^3 + 36 x^2 ) + C =
( - 1 / 72 ) ∙ ( 15 ∙ x^4 - 16 x^3 - 36 x^2 ) + C
The same answer as bobpursley's but written differently.
3 ∙ x / 3 ∙ 2 + 2 ∙ 2 x^2 / 2 ∙ 3 - 5 x^3 / 6 =
3 ∙ x / 6 + 4 x^2 / 6 - 5 x^3 / 6 =
( 1 / 6 ) ∙ ( 3 x + 4 x^2 - 5 x^3 ) =
( 1 / 6 ) ∙ ( - 5 x^3 + 3 x + 4 x^2 )
∫ ( 1 / 6 ) ∙ ( - 5 x^3 + 4 x^2 + 3 x ) dx =
( 1 / 6 ) ∙ ( - 5 ∫ x^3 dx + 4 ∫ x^2 dx + 3 ∫ x dx ) =
( 1 / 6 ) ∙ ( - 5 x^4 / 4 + 4 x^3 / 3 + 3 x^2 ) + C =
( 1 / 6 ) ∙ ( - 5 ∙ 3 x^4 / 3 ∙ 4 + 4 ∙ 4 x^3 / 4 ∙ 3 + 3 ∙ 12 x^2 / 12 ) + C =
( 1 / 6 ) ∙ ( - 15 ∙ x^4 / 12 + 16 x^3 / 12 + 36 x^2 / 12 ) + C =
( 1 / 6 ∙ 12 ) ∙ ( - 15 ∙ x^4 + 16 x^3 + 36 x^2 ) + C =
( 1 / 72 ) ∙ ( - 15 ∙ x^4 + 16 x^3 + 36 x^2 ) + C =
( - 1 / 72 ) ∙ ( 15 ∙ x^4 - 16 x^3 - 36 x^2 ) + C
The same answer as bobpursley's but written differently.
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