Limit Problem with Intergral

[Guest]

Can someone help me on number 4 amd 5? I am stuck, and for number 4, I tried the intergral from 0 to e of (ln(x+1)+1)^2dx.

THANKS. I apperciate your help, and again, thanks for your time in advanced. lol. =]. Yeah. lol k. : ]. Yep, Thanks. lol k. : ]. And again, I apperciate it. lol k. : ]. = ].

 

[pka]

For #5: \(\displaystyle \L
\begin{array}{l}
f(t) = \int_0^t {\sqrt {x^3 + 8} dx} \\
\frac{{df}}{{dt}}(1) = \ {lim}\limits_{h \to 0} \frac{{f(1 + h) - f(1)}}{h} = \ {lim}\limits_{h \to 0} \frac{{\int_1^{1 + h} {\sqrt {x^3 + 8} dx} }}{h} \\
\end{array}\)

 

[soroban]

Hello, atse1900!

I found an approach . . .

#4. The intergral from 0 to e of [ln(x+1)+1]^2 dx

Let: \(\displaystyle \,\ln(x+1)\,+\,1\:=\;w\;\;\Rightarrow\;\;\ln(x+1)\:=\:w\,-\,1\;\;\Rightarrow\;\;x\,+\,1\:=\:e^{w-1}\)

\(\displaystyle \;\;\;\Rightarrow\;\;x\:=\:e^{w-1}\,-\,1\;\;\Rightarrow\;\;dx\:=\:e^{w-1}\,dw\)

Substitute: \(\displaystyle \L\,\int\)\(\displaystyle w^2\cdot\,e^{w-1}\,dw\)

Integrate by parts (tabular method) and we get:

\(\displaystyle \;\;\;w^2\cdot e^{w-1}\,-\,2w\cdot e^{w-1}\,+\,e^{w-1}\,+\,C \;=\;e^{w-1}(w^2\,-\,2w\,+\,2)\,+\,C\)


Back-substitute: \(\displaystyle \:e^{w-1}\;=\;e^{\ln(x+1)+1\,-\,1} \;=\;e^{\ln(x+1)}\;=\;x\,+\,1\)

And \(\displaystyle \,w^2\,-\,2w\,+\,2\;=\;[\ln(x+1)\,+\,1]^2\,-\,2[\ln(x+1)\,+\,1] \,+\,2\)

\(\displaystyle \;\;\;=\;[\ln(x+1)]^2\,+\,2\cdot\ln(x+1)\,+\,1\,-\,2\cdot\ln(x+1)\,-\,2\,+\,2\)

\(\displaystyle \;\;\;=\;\ln^2(x+1)\,+\,1\)


And we have: \(\displaystyle \x\,+\,1)\left[\ln^2(x+1)\,+\,1\right]\,+\,C\)

\(\displaystyle \;\;\)I'll let you evaluate the definite integral . . .

 

[skeeter]

I have a feeling #4 is to be done using a calculator ... (cuz I've seen it before)

f(x) = ln(x+1)

using the method of washers ...

R = f(x) + 1
r = 1

V = pi*INT{a to b} (R^2 - r^2)dx

V = pi*INT{0 to e} {[f(x)+1]^2 - 1}dx

V = 20.146

#5 ... use L'Hopital's rule

d/dh[INT{1 to 1+h} sqrt(x^3 + 8)dx] = sqrt[(1+h)^3 + 8]

d/dh[h] = 1

limit as h->0 ... sqrt(9)/1 = 3

 

[galactus]

Here's the method I used, but I thought I made a boo-boo somewhere because the method the poster used ends up being answer E. It seems viable.

Solving \(\displaystyle \L\\y=ln(x+1)\) for x, \(\displaystyle \L\\x=e^{y}-1\)

\(\displaystyle \L\\2{\pi}\int_{0}^{ln(e+1)}(y+1)(e-(e^{y}-1))dy=20.146\)

This ends up being D.

\(\displaystyle \L\\{\pi}\int_{0}^{ln(e+1)}((e^{y}-1)+1)^{2}dy={\pi}\int_{0}^{ln(e+1)}(e^{y})^{2}dy=20.146\)