Suppose f, f', f", are continuous on [0, ln2] and that f(0) = 0, f'(0) = 3, f(ln2) = 6, f'(ln2) = 4 and the integral of (e^-2x * f(x) dx from 0 to ln 2) = 3.
Find the vlaue of the integral of (e^-2x *f"(x) dx) from 0 to ln 2.
Suppose f, f', f", are continuous on [0, ln2] and that f(0) = 0, f'(0) = 3, f(ln2) = 6, f'(ln2) = 4 and the integral of (e^-2x * f(x) dx from 0 to ln 2) = 3.
Find the vlaue of the integral of (e^-2x *f"(x) dx) from 0 to ln 2.
Hello, Susie!
We will integrate by parts . . .Suppose are continuous on
and that
and:
Find the value of:
. . Let:
. . Then:
And we have:
Do by-parts again:
Let:
Then:
So we have:
and:
Now we evaluate from to
. . Note that:
We have:
. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . .
Hence: . .
. . . . . . .
But someone check my work . . . please!
sorobsn has substituted only ln2Originally Posted by Soroban
and got the answer 16
can you follow upto here
if we substitute 0 we will get 1/3
result of integration is difference of the resulting function at upper limit and lower limit
tell me if I am not clear to you