Another homework problem: Calculus

• May 30th 2006, 01:46 PM
Susie38
Another homework problem: Calculus
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.
• May 30th 2006, 05:11 PM
Soroban
Hello, Susie!

Quote:

Suppose $\displaystyle f,\;f',\'f''$ are continuous on $\displaystyle [0,\,\ln2]$

and that $\displaystyle f(0) = 0,\;f'(0) = 3,\;f(\ln2) = 6,\;f'(\ln2) = 4$

and: $\displaystyle \int^{\ln2}_0\!\!e^{-2x}f(x)\,dx\;= \;3$

Find the value of: $\displaystyle I \;= \;\int^{\ln 2}_0 e^{-2x}\cdot f''(x)\,dx$
We will integrate by parts . . .

. . Let: $\displaystyle u = e^{-2x}\qquad\qquad dv = f''(x)\,dx$

. . Then: $\displaystyle du = -2e^{-2x}dx\quad v = f'(x)$

And we have: $\displaystyle I \;= \;e^{-2x}\!\cdot\!f'(x) + 2\int e^{-2x}\!\cdot\!f'(x)\,dx$

Do by-parts again:

Let: $\displaystyle u = e^{-2x}\qquad\qquad dv = f'(x)\,dx$

Then: $\displaystyle du = -2e^{-2x}dx\quad v = f(x)$

So we have: $\displaystyle I \;= \;e^{-2x}\!\cdot\!f'(x) + 2\left[e^{-2x}\!\cdot\!f(x) + 2\int e^{-2x}\!\cdot\!f(x)\,dx\right]$

and: $\displaystyle I \;= \;e^{-2x}\!\cdot\!f'(x) + 2e^{-2x}\!\cdot\!f(x) + 4\int e^{-2x}\!\cdot\!f(x)\,dx$

Now we evaluate from $\displaystyle x = 0$ to $\displaystyle x = \ln 2$
. . Note that: $\displaystyle e^{-2\cdot\ln 2}\:=\:e^{\ln(2^{-2})} \:= \:e^{\ln\frac{1}{4}} \;=\;\frac{1}{4}$

We have: $\displaystyle I \;= \;\frac{1}{4}\!\cdot\underbrace{f'(\ln 2)} \;+ \;2\left(\frac{1}{4}\right)\!\cdot\underbrace{f(\l n\,2)} \;+ \;4\underbrace{\int^{\ln 2}_0\!\!\!e^{-2x}\!\cdot\!f(x)\,dx}$
. . . . . . . . . . . . . . . .$\displaystyle \downarrow$ . . . . . . . . . . . .. $\displaystyle \downarrow$ . . . . . . . . . $\displaystyle \downarrow$
Hence: . . $\displaystyle I \;= \quad\;\frac{1}{4}(4) \quad\;+ \quad\;2\left(\frac{1}{4}\right)(6) \quad + \qquad4\;(3)$

. . . . . . . $\displaystyle I \;= \; 1 + 3 + 12 \;= \;16$

But someone check my work . . . please!
• May 30th 2006, 05:20 PM
malaygoel
Quote:

Originally Posted by Soroban
Hello, Susie!

We will integrate by parts . . .

. . Let: $\displaystyle u = e^{-2x}\qquad\qquad dv = f''(x)\,dx$

. . Then: $\displaystyle du = -2e^{-2x}dx\quad v = f'(x)$

And we have: $\displaystyle I \;= \;e^{-2x}\!\cdot\!f'(x) + 2\int e^{-2x}\!\cdot\!f'(x)\,dx$

Do by-parts again:

Let: $\displaystyle u = e^{-2x}\qquad\qquad dv = f'(x)\,dx$

Then: $\displaystyle du = -2e^{-2x}dx\quad v = f(x)$

So we have: $\displaystyle I \;= \;e^{-2x}\!\cdot\!f'(x) + 2\left[e^{-2x}\!\cdot\!f(x) + 2\int e^{-2x}\!\cdot\!f(x)\,dx\right]$

and: $\displaystyle I \;= \;e^{-2x}\!\cdot\!f'(x) + 2e^{-2x}\!\cdot\!f(x) + 4\int e^{-2x}\!\cdot\!f(x)\,dx$

Now we evaluate from $\displaystyle x = 0$ to $\displaystyle x = \ln 2$
. . Note that: $\displaystyle e^{-2\cdot\ln 2}\:=\:e^{\ln(2^{-2})} \:= \:e^{\ln\frac{1}{4}} \;=\;\frac{1}{4}$

We have: $\displaystyle I \;= \;\frac{1}{4}\!\cdot\underbrace{f'(\ln 2)} \;+ \;2\left(\frac{1}{4}\right)\!\cdot\underbrace{f(\l n\,2)} \;+ \;4\underbrace{\int^{\ln 2}_0\!\!\!e^{-2x}\!\cdot\!f(x)\,dx}$
. . . . . . . . . . . . . . . .$\displaystyle \downarrow$ . . . . . . . . . . . .. $\displaystyle \downarrow$ . . . . . . . . . $\displaystyle \downarrow$
Hence: . . $\displaystyle I \;= \quad\;\frac{1}{4}(4) \quad\;+ \quad\;2\left(\frac{1}{4}\right)(6) \quad + \qquad4\;(3)$

. . . . . . . $\displaystyle I \;= \; 1 + 3 + 12 \;= \;16$

But someone check my work . . . please!

I think that you have substitutedx=ln2 but forgor to substitute x=0
• May 30th 2006, 05:22 PM
malaygoel
Quote:

Originally Posted by malaygoel
I think that you have substitutedx=ln2 but forgor to substitute x=0

I think the answer may be 16-3/4=61/4
• May 30th 2006, 06:27 PM
Soroban
Hello, malaygoel!

Quote:

I think that you have substituted $\displaystyle x = \ln2$ but forgot to substitute $\displaystyle x = 0$
You're right . . . *blush*
• May 31st 2006, 09:46 AM
malaygoel
Quote:

Originally Posted by Soroban
Hello, malaygoel!

You're right . . . *blush*

Could you please explain the solution given in the thread "Calculus Problem"
• Jun 3rd 2006, 03:10 PM
Nichelle14
how did you get 16- 3/4? I got 16 - 3.

• Jun 3rd 2006, 05:25 PM
malaygoel
Quote:

Originally Posted by Soroban
$\displaystyle I \;= \;e^{-2x}\!\cdot\!f'(x) + 2e^{-2x}\!\cdot\!f(x) + 4\int e^{-2x}\!\cdot\!f(x)\,dx$

We have: $\displaystyle I \;= \;\frac{1}{4}\!\cdot\underbrace{f'(\ln 2)} \;+ \;2\left(\frac{1}{4}\right)\!\cdot\underbrace{f(\l n\,2)} \;+ \;4\underbrace{\int^{\ln 2}_0\!\!\!e^{-2x}\!\cdot\!f(x)\,dx}$

sorobsn has substituted only ln2
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
• Jun 3rd 2006, 05:39 PM
Nichelle14
sorry I don't see how still.
• Jun 4th 2006, 06:56 AM
malaygoel
Quote:

Originally Posted by Nichelle14
sorry I don't see how still.

you got 16 right
explain me how you got 3.
• Jun 4th 2006, 08:25 AM
Nichelle14
I substituted 0 in for x, this is after substituting the ln 2

-(e^-2(0)f'(0) + 2e^-2(0)f(0))

-(1(3) + 2(1)(0)) = -3
• Jun 4th 2006, 09:46 AM
malaygoel
Quote:

Originally Posted by Nichelle14
I substituted 0 in for x, this is after substituting the ln 2

-(e^-2(0)f'(0) + 2e^-2(0)f(0))

-(1(3) + 2(1)(0)) = -3

thank you for correcting me
I was wrong
it is 16 -3 =13 :rolleyes: