A few Webwork Integral problems have me stuck.

• May 29th 2007, 01:43 PM
Whitewolfblue
A few Webwork Integral problems have me stuck.
I am working on substitution and I get far with each question but then I get near the end of evaluating the integrals with differentiation (or antiderivatives, can't remember off the top off my head) and I get stuck or get the wrong answer.

here are the ones I am stuck on(I will try multiple forms of writing it):

1.)

Evaluate the indefinite integral:

Integral bar --cos(x)-- dx/1 = _______________ + C
-----------4sin(x)+8

(I started by substituting "4sin(x) + 8" with "u".

I got in the end (1/4)(1/(-cos(x)+8)) + C .

2.)

Evaluate the definite integral

Integral bar High# e^6, Low#1 or [1,e^6] if you prefer.----dx-----
------------------------------------------------------x(1+ln(x))

= _______________

I substituted "u" for 1+ln(x) because 1/x can be used to replace the 1/x in the main integral.

I worked this one so much I got lost.

u = 1 + ln(x)
du = 1/x dx or dx/x

= integral bar [1,7] (<= replacing the old values since I changed it.) du 1/u
=-(1+1/x) and a couple other versions as well.

3.) This one stumped me entirely.

Find the following indefinite integral

integral bar ----x----- dx
--------------sqrt(x+8)

The last isn't too hard to get the first part but then I tried the second part the same way and I got a wrong answer.

4.) If f(x) = integral bar [x,x^2] (t^2)dt , then

f '(-3)= _________.
• May 29th 2007, 02:09 PM
Jhevon
Quote:

Originally Posted by Whitewolfblue
I am working on substitution and I get far with each question but then I get near the end of evaluating the integrals with differentiation (or antiderivatives, can't remember off the top off my head) and I get stuck or get the wrong answer.

here are the ones I am stuck on(I will try multiple forms of writing it):

1.)

Evaluate the indefinite integral:

Integral bar --cos(x)-- dx/1 = _______________ + C
-----------4sin(x)+8

(I started by substituting "4sin(x) + 8" with "u".

I got in the end (1/4)(1/(-cos(x)+8)) + C .

This is incorrect.

$\int \frac { \cos x}{4 \sin x + 8}dx$

Let $u = 4 \sin x + 8$

$\Rightarrow du = 4 \cos x dx$

$\Rightarrow \frac {1}{4} du = \cos x dx$

So our integral becomes:

$\frac {1}{4} \int \frac {1}{u}du$

Now continue....

Quote:

2.)

Evaluate the definite integral

Integral bar High# e^6, Low#1 or [1,e^6] if you prefer.----dx-----
------------------------------------------------------x(1+ln(x))

= _______________

I substituted "u" for 1+ln(x) because 1/x can be used to replace the 1/x in the main integral.

I worked this one so much I got lost.

u = 1 + ln(x)
du = 1/x dx or dx/x

= integral bar [1,7] (<= replacing the old values since I changed it.) du 1/u
=-(1+1/x) and a couple other versions as well.
$\int_{1}^{e^6} \frac {1}{x(1 + \ln x)}dx$

Let $u = 1 + \ln x$

$\Rightarrow du = \frac {1}{x}dx$

Now, when $x=1$, $u = 1 + \ln 1 = 1 + 0 = 1$

when $x = e^6$, $u = 1 + \ln \left( e^6 \right) = 7$

So our integral becomes:

$\int_{1}^{7} \frac {1}{u}du$

Now continue

Quote:

3.) This one stumped me entirely.

Find the following indefinite integral

integral bar ----x----- dx
--------------sqrt(x+8)

$\int \frac {x}{ \sqrt {x + 8}}dx$

Let $u = x + 8 \mbox { } \Rightarrow x = u - 8$

$\Rightarrow du = dx$

So our integral becomes:

$\int \frac {u - 8}{ \sqrt {u}}du$

$= \int \frac {u - 8}{u^{ \frac {1}{2}}}du$

$= \int \left( u^{ \frac {1}{2}} - 8u^{- \frac {1}{2}} \right) du$

Now continue...

Quote:

The last isn't too hard to get the first part but then I tried the second part the same way and I got a wrong answer.

4.) If f(x) = integral bar [x,x^2] (t^2)dt , then

f '(-3)= _________.
We need the second fundamental theorem of calculus here, i'm not that good with it.

$f(x) = \int_{x}^{x^2} t^2 dt = \int_{x}^{0} t^2 dt + \int_{0}^{x^2} t^2 dt$

$= - \int_{0}^{x} t^2 dt + \int_{0}^{x^2} t^2 dt$

Apply the second fundamental theorem of calculus:

$\Rightarrow f'(x) = \frac {d}{dt} \left( \int_{0}^{x^2} t^2 dt - \int_{0}^{x} t^2 dt \right)$

$= \left( x^2 \right)^2 \cdot 2x - x^2$

$= 2x^5 - x^2$

Now just find $f'(-3)$ .......i did pretty much the whole problem, i leave this last bit to you

I guess you could check this manually, it won't be too much work in this case to actually find the integral, evaluate between the limits and then differentiate again. good luck

if you get stuck with any again, say so

EDIT: I verified the last one myself, it checks out
• May 29th 2007, 03:22 PM
Soroban
Hello, Whitewolfblue!

Another approach to #3 . . .

Quote:

$3)\;\int\frac{x}{\sqrt{x+8}}\,dx$

Let $\sqrt{x+5} \,= \,u\quad\Rightarrow\quad x \,= \,u^2 - 5\quad\Rightarrow\quad dx \,= \,2u\,du$

Substitute: . $\int\frac{u^2-5}{u}(2u\,du)\;=\;2\int(u^2-5)\,du \;=\;2\left(\frac{u^3}{3} - 5u\right) + C$

Factor: . $\frac{2}{3}u(u^2 - 15) + C$

Back-substitute: . $\frac{2}{3}\sqrt{x+5}\left([\sqrt{x+5}]^2 - 15\right) + C \;=\;\frac{2}{3}\sqrt{x+5}\,(x + 5 - 15) + C$

Answer: . $\frac{2}{3}\sqrt{x+5}\,(x - 10) + C$

• May 29th 2007, 04:20 PM
Whitewolfblue
Thanks Soroban for that interesting way. It gave me an idea for a separate problem and I got it right doing so. Unfortunately trying it your way for that exact problem I was given and "Incorrect" message.

I got as far as Jhevon demonstrated on questions 2,3 and 4 but I am still stuck.

Kiba:confused:
• May 29th 2007, 04:33 PM
Jhevon
Quote:

Originally Posted by Whitewolfblue
Thanks Soroban for that interesting way. It gave me an idea for a separate problem and I got it right doing so. Unfortunately trying it your way for that exact problem I was given and "Incorrect" message.

I got as far as Jhevon demonstrated on questions 2,3 and 4 but I am still stuck.

Kiba:confused:

if you got as far as i did, that means you didn't do anything. you should have any problem picking up the problems where i left off.

2)
we left off here:

$\int_{1}^{7} \frac {1}{u}du$ .....the integral of $\frac{1}{u}$ is $\ln u$

$\Rightarrow \int_{1}^{7} \frac {1}{u}du = \left[ \ln u \right]_{1}^{7}$

$= \ln 7 - \ln 1 = \ln 7$

3)
we left off at

$\int \left( u^{ \frac {1}{2}} - 8u^{- \frac {1}{2}} \right) du$ ......we integrate using the power rule

$= \frac {2}{3}u^{ \frac {3}{2}} - 16u^{ \frac {1}{2}} + C$

4)
we left off at

$f'(x) = 2x^5 - x^2$

$\Rightarrow f'(-3) = 2(-3)^5 - (-3)^2$

$= -486 - 9$

$= -495$ .......you especially shouldn't have any problems with this one
• May 30th 2007, 04:06 PM
Whitewolfblue
A new one of integration by parts
I am having trouble with another integral problem.

Use integration by parts to evaluate the integral:

integral bar [1,4] sqrt(t) ln(t) dt =______________

I did it this way:

u = ln(t) | dv = t^.5
du = 1/x dx | v = (t^1.5)/1.5

Using : uv - integral bar v du

(ln(t))((t^1.5)/1.5)-(ln(t))((t^1.5)/1.5)-integral bar[1,4] (t^1.5)/1.5 (1/x) dx

This is what I got
((ln(4))(4^1.5/1.5)-(ln(1))(1^1.5/1.5))-((4^2.5/2.5)(ln(4))-(1^2.5/2.5)(ln(1)))

or -10.351

I am thinking maybe I am supposed to use the product rule but I am unsure exactly.

Kiba
• May 30th 2007, 04:28 PM
Krizalid
You mean $\int_1^4\sqrt t\ln t\,dt$ ??

• May 30th 2007, 04:31 PM
Jhevon
Quote:

Originally Posted by Whitewolfblue
I am having trouble with another integral problem.

Use integration by parts to evaluate the integral:

integral bar [1,4] sqrt(t) ln(t) dt =______________

I did it this way:

u = ln(t) | dv = t^.5
du = 1/x dx | v = (t^1.5)/1.5

Using : uv - integral bar v du

(ln(t))((t^1.5)/1.5)-(ln(t))((t^1.5)/1.5)-integral bar[1,4] (t^1.5)/1.5 (1/x) dx

This is what I got
((ln(4))(4^1.5/1.5)-(ln(1))(1^1.5/1.5))-((4^2.5/2.5)(ln(4))-(1^2.5/2.5)(ln(1)))

or -10.351

I am thinking maybe I am supposed to use the product rule but I am unsure exactly.

Kiba

now that i can use LaTex, i find it hard to read stuff like this, so i will do it

You should post new questions in a new thread. the formula you used is incorrect.

$\int_{1}^{4} \sqrt {t} \ln t \mbox { } dt = \frac {2}{3}t^{ \frac {3}{2}} \ln t - \frac {2}{3} \int_{1}^{4} t^{ \frac {3}{2}} \cdot \frac {1}{t} \mbox { } dt$

$= \frac {2}{3}t^{ \frac {3}{2}} \ln t - \frac {2}{3} \int_{1}^{4} t^{ \frac {1}{2}} \mbox { } dt$

$= \left[ \frac {2}{3}t^{ \frac {3}{2}} \ln t - \frac {4}{9}t^{ \frac {3}{2}} \right]_{1}^{4}$

$= \frac {2}{3}4^{ \frac {3}{2}} \ln 4 - \frac {4}{9} 4^{ \frac {3}{2}} + \frac {4}{9}$

$\approx 4.2825$

If you are unclear about anything, please ask. i don't like using the formula, so i just did it directly.

PS. i know Krizalid told me how to align things, but i still haven't learnt the code yet, i'll have to review it
• May 30th 2007, 04:33 PM
Krizalid
Quote:

Originally Posted by Jhevon
now that i can use LaTex, i find it hard to read stuff like this, so i will do it

Did you know that MathType works as a LaTeX translator?
• May 30th 2007, 04:36 PM
Whitewolfblue
I apologize I am new to Mathtype I was wondering how everyone was able to create pictures.

I will try and use IT next time.

When I tried it a seperate way I got a negative version of what you got.

thanks again

Kiba
• May 30th 2007, 04:47 PM
Jhevon
Quote:

Originally Posted by Krizalid
Did you know that MathType works as a LaTeX translator?

i didn't know that, how do you access that feature?
• May 30th 2007, 04:49 PM
Jhevon
Quote:

Originally Posted by Whitewolfblue
I apologize I am new to Mathtype I was wondering how everyone was able to create pictures.

I will try and use IT next time.

When I tried it a seperate way I got a negative version of what you got.

thanks again

Kiba

chances are you used the wrong formula again. look up the formula and make sure you're using it properly. on the interval [1,4] the graph $\sqrt {t} \ln t$ is above the x-axis, any negative answer is wrong

the formula is:

$\int u \cdot dv = uv - \int v \cdot du$

that is the version i see used most often

i like this version better

$\int uv' = uv - \int u'v$
• May 30th 2007, 04:54 PM
Krizalid
OK

I assume you have full MathType version.

.- Open it
.- Preferences
.- Translators...
.- Select "translation to other language (text)"
.- Select the translator TeX--AMS-LaTeX
.- Deactivate "include MathType data in translation"

Then, for example, write something, and you paste it into the forum

$\int {(x + 1)dx}$

Remove the $and$, and enclose the rest on math /math; finally, you'll have $\int {(x + 1)dx}$
• May 30th 2007, 05:05 PM
Jhevon
Quote:

Originally Posted by Krizalid
OK

I assume you have full MathType version.

.- Open it
.- Preferences
.- Translators...
.- Select "translation to other language (text)"
.- Select the translator TeX--AMS-LaTeX
.- Deactivate "include MathType data in translation"

Then, for example, write something, and you paste it into the forum

$\int {(x + 1)dx}$

Remove the $and$, and enclose the rest on math /math; finally, you'll have $\int {(x + 1)dx}$

ah! I see! thanks!