Question

Let f be twice differentiable with f (0) = 6, f (1) = 5, and f ′(1) = 2. Evaluate the integral...

Answer 1

\[\int\limits_{0}^{1}x f \prime \prime(x) dx\]
which is
\[f \prime(1) - f \prime(0)\]
and my logic is that because there is an x in front of the prime, then
\[f \prime(0) = 0\]

Answer 2

am i in the right path?

Answer 3

Not entirely - only partially (pun intended) You need integration by parts where u = x
dv = f''

Answer 4

i see...

Answer 5

so i have
\[x f \prime(x) - f \prime\]

Answer 6

\[d(uv)=udv+vdu \rightarrow \int\limits d(uv)=uv=intudv+intvdu. \]
Rearranging gives
intudv=uv-intvdu

Answer 7

\[ \int\limits udv=uv-\int\limits vdu\]

Answer 8

oh poop, i made a mistake there

Answer 9

y u dd

Answer 10

so i would have
\[x f \prime(x) - f(x) \]?

Answer 11

xf' - VALUES_DIFFRENCE of f'

Answer 12

VALUES_DIFFRENCE of xf' - VALUES_DIFFRENCE of f'

Answer 13

is this correct?

\[[1 f \prime(1) - f(1)] - [0 f \prime(0) - f(0)]\]

Answer 14

or am i arranging them incorrectly?

Answer 15

Second expr wrong change to f'(0} to f'(0)

Answer 16

Change fourth to f'(0)

Answer 17

how come?

if
\[dv= f \prime \prime\]
then
\[v= f \prime\]
thus the antiderivative of \[v= f \prime\] would be f wouldn't it?

Answer 18

u = x
du = dx

v = f'
dv = f'' dx

Answer 19

\[\int x~f''~dx=x~f'-\int f' dx\]

Answer 20

so it is
\[x f \prime(x) - f(x)\]

Answer 21

yes, evaled at 0 and 1

Answer 22

thank you, mikael was telling me that it was

\[x f \prime - f \prime\]
which didn't make sense to me

Answer 23

\[(x f \prime(x) - f(x))^1-(x f \prime(x) - f(x))^0\]
\[(1 f \prime(1) - f(1))-(0 f \prime(0) - f(0))\]
\[f \prime(1) - f(1)+ f(0)\]

Answer 24

awesome, thats what i was getting.
thank you so much, you're the man... or woman ;)

Answer 25

lol, good luck ;)