Calculus question...

[novasatori]

Find the function S(x), given that S'(x)=5x^2-(2/3) and that S(1)=1

I don't even know where to start with this problem, and unfortunately I missed the answers to the test review last thursday, any help with setting it up would be appreciated.


I'm sure its easier than it seems so go easy on me 🙁...

 

[henryay]

Integrate both sides, solve for the constant term. (that's when S(1)=1 comes into play)

 

[BigPoppa]

Integrate S'(x). Solve the general form for the specific form (zyx + C) by plugging in 1 for x and setting the equation = 1. Solve for C.

 

[FleshLight]

5/3x^3-(2/3x)+c

c = -10+2/3

(5/3x)^3-(2/3x)-10+(2/3)

 

[novasatori]

I (5x^2-(2/3))dx
f(x)=((5x^3)/3)-(2x/3) + C

((5(1)^3)/3)-(2(1)/3) + C = 1

1+C = 1

1-1 = C

C = 0 ?


Hope that's right, thanks a lot.

 

[novasatori]

Anyone check that ?

Please 🙂

 

[JujuFish]

Originally posted by: novasatori
I (5x^2-(2/3))dx
f(x)=((5x^3)/3)-(2x/3) + C

((5(1)^3)/3)-(2(1)/3) + C = 1

1+C = 1

1-1 = C

C = 0 ?


Hope that's right, thanks a lot.

Appears to be correct to me.

 

[RadioHead84]

yeah thats right. I just had that on my last test actually. I got an 86 🙁

 

[czech09]

Originally posted by: JujuFish

Originally posted by: novasatori
I (5x^2-(2/3))dx
f(x)=((5x^3)/3)-(2x/3) + C

((5(1)^3)/3)-(2(1)/3) + C = 1

1+C = 1

1-1 = C

C = 0 ?


Hope that's right, thanks a lot.

Appears to be correct to me.

Werd.