# integral

• Nov 11th 2006, 09:17 PM
viet
integral
consider the function f(x) whose second derivative is f''(x) = 3x+2sin(x)
if f(0)=4, f'(x)=4. what is f(2) = ?

heres what i did
$\displaystyle f'(x) = \int (3x+2sin(x)$
$\displaystyle = \frac{3x^2}{2} - 2cos(x) + C1$

$\displaystyle f(x) = \int \frac{3x^2}{2} - 2cos(x) + C1$
$\displaystyle = \frac{x^3}{2} + C2 - 2sin(x) + C2$
• Nov 11th 2006, 09:54 PM
srinivas
i thik something is wrong how can a function which has sin o cos i it give the derivative as 4 that is a constant?
it ought to be a straight line for that
• Nov 12th 2006, 12:14 AM
CaptainBlack
Quote:

Originally Posted by srinivas
i thik something is wrong how can a function which has sin o cos i it give the derivative as 4 that is a constant?
it ought to be a straight line for that

Because it is a typo in the intial conditions. Presumably they shout be:

f(0)=4, f'(0)=4.

RonL
• Nov 12th 2006, 12:19 AM
CaptainBlack
Quote:

Originally Posted by viet
consider the function f(x) whose second derivative is f''(x) = 3x+2sin(x)
if f(0)=4, f'(x)=4. what is f(2) = ?

heres what i did
$\displaystyle f'(x) = \int (3x+2sin(x)$
$\displaystyle = \frac{3x^2}{2} - 2cos(x) + C1$

$\displaystyle f(x) = \int \frac{3x^2}{2} - 2cos(x) + C1$
$\displaystyle = \frac{x^3}{2} + C2 - 2sin(x) + C2$

$\displaystyle = \frac{x^3}{2} - 2\sin(x) + C1\ x + C2$

RonL
• Nov 12th 2006, 07:14 AM
Soroban
Hello, viet!

I agree with the Captain: there must be a typo in the problem.
I'll assume it is supposed to be: $\displaystyle f'(0) = 4$

Quote:

Consider the function $\displaystyle f(x)$ whose second derivative is: $\displaystyle f''(x) \:= \:3x+2\sin x$
If $\displaystyle f(0) = 4,\;f'(0) = 4$, find $\displaystyle f(2)$

We have: .$\displaystyle f'(x)\;=\;\int \left(3x + 2\sin x\right)\,dx \;= \;\frac{3}{2}x^2 - 2\cos x + C_1$

Since $\displaystyle f'(0) = 4$, we have: .$\displaystyle \frac{3}{2}(0^2) - 2\cos0 + C_1\:=\:4\quad\Rightarrow\quad C_1 = 6$

. . Hence: .$\displaystyle f'(x)\:=\:\frac{3}{2}x^2 - 2\cos x + 6$

We have: .$\displaystyle f(x)\;=\;\int\left(\frac{3}{2}x^2 - 2\cos x + 6\right)\,dx \;=\;\frac{1}{2}x^3 - 2\sin x + 6x + C_2$

Since $\displaystyle f(0) = 4$, we have: .$\displaystyle \frac{1}{2}(0^3) - 2\sin0 + 6(0) + C_2\:=\:4\quad\Rightarrow\quad C_2 = 4$

. . Hence: .$\displaystyle \boxed{f(x) \;= \;\frac{1}{2}x^3 - 2\sin x + 6x + 4}$

Therefore: .$\displaystyle f(2) \;= \;\frac{1}{2}(2^3) = 2\sin2 + 6(2) + 4$ . . . etc.