antiderivative 3x^3 - 5x w/ value 6 when x = 1

[needurhelp]

Find the antiderivative of 3x^3-5x that has the value 6 when x = 1

Here is what I did:

. . .3x^3 - 5x
. . .=
. . .(3x^4/4) - 5x
. . .=
. . .(3/4)x^4 - 5x + C

Plug 1 in for x:

. . .(3/4) - 5 + C
. . .=
. . .C = 6 - (3/4)+5

But my answer is wrong. Can someone please explain? Thank you!

 

[soroban]

Re: antiderivative 3x^3-5x

Hello, needurhelp!

Find the antiderivative of \(\displaystyle 3x^3\,-\,5x\) that has the value 6 when \(\displaystyle x\,=\,1\)

Here's what i did: \(\displaystyle \L\,\int(3x^3\,-\,5x)\,dx \:=\:\frac{3x^4}{4}\,-\,5x\;\) . . . no

You want: \(\displaystyle \L\,\frac{3x^4}{4}\,-\,\frac{5x^2}{2}\,+\,C\)

 

[needurhelp]

hey soroban,
i plug 1 for (3x^4/4) - (5x^2/2) + C and solved for C, got the answer as 11/4 is that right?

 

[needurhelp]

not sure what im doing is right but i got;
(3(1)^4/4)-(5(1)^2/2)+C
C = 7/4

so the antiderivative is:
(3x^4/4)-(5x^2/2)+(7/4)

 

[tkhunny]

You're guessing, but you're really close.

Try solving this for C: 3/4 - 5/2 + C = 6

Whatever you were doing before, you didn't seem to have an actual equation. How did you manage to solve it for anything?

 

[needurhelp]

i think i got it, heres what i did:
after i solve for C = 7.75 or 31/4
the antiderivative is 3x^4/4)-(5x^2/2)+7.75
thanks for every1s help