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.