Question

Evaluate the integrals below.

Answer 1

\[\int\limits_{}^{} (3x \sqrt[5]{x^3 +1}) dx\]

Answer 2

\[\int\limits_{}^{}(8 \theta \sqrt[3]{\theta^2 -1} ) d \theta\]

Answer 3

\[y=\sqrt[5]{x ^{3}+1} \rightarrow x=\sqrt[3]{y ^{5}-1} \rightarrow dx = \frac{ 5y ^{4} }{ \sqrt[3]{(y ^{5}-1)^{2}} }\]
thus the integral becomes :\[\int\limits_{}^{} 3\sqrt[3]{y ^{5}-1}* y * \frac{ 5y ^{4} }{ 3\sqrt[3]{(y ^{5}-1)^{2}} }\]\[5 \int\limits\limits_{}^{} \frac{ y ^{4} }{\sqrt[3]{y ^{5}-1}}*y dy\]

Answer 4

how do you end up evaluating that integral, it looks crazy! hah

Answer 5

what if we simplify it first. to \[\int\limits_{}^{} 3x(x^3+1)^{\frac{ 1 }{ 5 }}\]

Answer 6

now you can apply the formula of integration by parts to get:
\[y \sqrt[3]{y ^{5}-1}-\int\limits_{}^{} \sqrt[3]{y ^{5}-1}\]
\[\frac{ 1 }{2 } \int\limits 2y \sqrt[3]{y ^{^{2}}-1} = \frac{ 1 }{2 } (\sqrt[3]{(y ^{2}-1)^{4}}\]

Answer 7

dude can you please check the simplifications carefully that's really gonna take time :( , it's only an algebraic operation
no, only you can simplify after substituting..
and also I am sorry if anything went wrong

Answer 8

@zepdrix Help? pls. haha

Answer 9

Have you learned Integration by Parts yet?

Answer 10

No I haven't.

Answer 11

Ok, I didn't think so.
The second problem won't give us any trouble.

But the first one.. hmm..
Did you paste it correctly? Or is it suppose to be x^2+1 under the root?
Just checking.

Answer 12

it's x^3+1 under the root.

Answer 13

Is it supposed to be 3x^2 outside of the root? :D lol just checking again XD

Answer 14

for the first problem it wan't us to evaluate
\[\int\limits_{}^{}(3x^5 \sqrt{x^3+1} )dx\]

Answer 15

Oh I see...... the 5 is suppose to be an exponent.
You wrote the 5 on the root.. that's where all the confusion is :(((

Answer 16

Whooops just noticed that now ._. lol

Answer 17

Ok ok ok now this makes more sense D:

Answer 18

\[\Large \int\limits 3x^5 \sqrt{\color{orangered}{x^3+1}}\;dx\]
Ok we'll want this orange term to be our `u`.

Answer 19

We'll have to do a little work to fill in all the pieces.

Answer 20

okay(: haha

Answer 21

What do you get for du?

Answer 22

3x^2

Answer 23

\[\Large \color{orangered}{u=x^3+1}\]\[\Large \color{royalblue}{du=3x^2\;dx}\]Ok good.
We'll have to do something clever and break up the powers of x that are outside of the root.\[\Large \int\limits\limits 3x^5 \sqrt{\color{orangered}{x^3+1}}\;dx \qquad=\qquad \int\limits\limits 3x^2\cdot x^3 \sqrt{\color{orangered}{x^3+1}}\;dx\]

So that allows us to write our integral like this,\[\Large \int\limits x^3\sqrt{\color{orangered}{x^3+1}}(\color{royalblue}{3x^2\;dx})\]Understand why I grouped it up like that?

Answer 24

Pay attention to the colors ^^ hehe

Answer 25

okay. hah(:

Answer 26

Brb gonna make a sammich real quick +_+
Do you understand how to plug in the u and du?

Answer 27

mmkays and sorta.

Answer 28

\[\int\limits_{}^{} x^3 \sqrt{u} (du)\]

Answer 29

Ok good!
So our goal is to completely replace the x's with u's.
Looks like we still have a little bit of work to do.

How can we write x^3 in terms of u?
Let's look back at our original substitution we made,\[\Large u=x^3+1\]Could we maybe solve for x^3 here?

Answer 30

3x^2?

Answer 31

mmmmmmmmm no -_-
If we subtract 1 from each side it gives us,\[\Large u=x^3+1 \qquad\qquad\to\qquad\qquad x^3=u-1\]Right? :o

Answer 32

hahah, sorry. I seriously suck at this stuff.. but okay. makes sense! :p

Answer 33

\[\Large \color{purple}{x^3=u-1}\]
\[\Large \int\limits\limits\limits \color{purple}{x^3}\sqrt{\color{orangered}{u}}(\color{royalblue}{du}) \qquad=\qquad \int\limits\limits\limits \color{purple}{(u-1)}\sqrt{\color{orangered}{u}}(\color{royalblue}{du})\]

So I think that takes care of the last piece we needed to fix.

Answer 34

What's funny about U-subs is that sometimes they may not look simpler when they really are :D lol

We'll be able to evaluate this integral a lot easier now.
Let's rewrite our sqrt(u) as a rational expression \(\large \sqrt{u}=u^{1/2}\).
Then we'll distribute it to each term inside the purple brackets.\[\Large \int\limits u\cdot u^{1/2}-u^{1/2}\;du\]

Understand how to multiply u and u^1/2?

Answer 35

2u^1/2?

Answer 36

No.
When we multiply terms of similar bases, we `add` the exponents.\[\Large u^1\cdot u^{1/2} \qquad=\qquad u^{1+1/2}\]So in our integral that gives us,\[\Large \int\limits u^{3/2}-u^{1/2}\;du\]

Answer 37

gosh.... I'm stupid. how could I have forgotten that.lol

Answer 38

:3

Answer 39

and now we reapply or subs in right?

Answer 40

Now we integrate. Looks like we can use the power rule here.

Answer 41

ohhh, right.
sorry my comp froze -,-

Answer 42

\[\frac{ u^{\frac{ 3 }{ 2 }+1} }{ \frac{ 3 }{ 2 }+1 } - \frac{ u^{\frac{ 1 }{ 2 }+1} }{ \frac{ 1 }{ 2 }+1 }\]

Answer 43

Ok.
\[\Large \frac{u^{5/2}}{5/2}\]So like with this first term, remember what to do when you divide by a fraction?

Answer 44

\[\frac{ 5u^{5/2} }{ 2 } -\frac{ 3u^{3/2} }{ 2 }\]

Answer 45

or would the front fraction flip?

Answer 46

\[\frac{2u^{5/2} }{ 5 }-\frac{ 2u^{3/2} }{ 3 }\]

Answer 47

Yes, good, we flip the fraction and write it as multiplication.
Here is another way we can write it.
A little easier to read imo :O
\[\Large \frac{2u^{5/2}}{5} \qquad=\qquad \frac{2}{5}u^{5/2}\]

Answer 48

oh okay.. hahah

Answer 49

now we apply the subs? lol

Answer 50

Undo our sub? yes

Answer 51

haha that's what I ment. so for u i plug in x^3 +1 right?

Answer 52

mhm :x

Answer 53

\[=> \frac{ 2 }{ 5 } (x^3+1)^{5/2} - \frac{ 2 }{ 3 }(x^3+1)^{3/2}\]

Answer 54

yay good job \:o/

Answer 55

is that all, or do I have to simplify it more?

Answer 56

No, that's all.

Answer 57

oh okay!(: hahah and for the second one.

Answer 58

\[\Large \int\limits 8\theta\;\sqrt[3]{\theta^2 -1}\;d\theta\]Is the 3 on the root?

Answer 59

yes it is. I double checked lol

Answer 60

\[\Large \int\limits\limits 8\theta\;\sqrt[3]{\color{orangered}{\theta^2 -1}}\;d\theta\]Ok this one will be a tad easier than the last.
So there is our u in orange, du=?

Answer 61

\[du=2\theta\]

Answer 62

btw for the last one it's +C right? cause there is no given interval

Answer 63

\[\Large \color{orangered}{u=\theta^2-1}\]\[\Large \color{royalblue}{2\theta\;d \theta}\]Hmm ok.
So we'll need to do what we did in the last problem.
We'll need to fiddle with it and pull a 2theta out of the mix.
8=4*2 right? so let's pull the 2 out of there.\[\Large \int\limits\limits\limits 8\theta\;\sqrt[3]{\color{orangered}{\theta^2 -1}}\;d\theta \qquad=\qquad \int\limits 4 \sqrt[3]{\color{orangered}{\theta^2-1}}(\color{royalblue}{2\theta\;d \theta})\]

Answer 64

Oh true. It was an `indefinite` integral. No limits, so yes we'll have to throw a +C onto the end to show it represents the whole family of solutions :U

Answer 65

okay, I got the same results so far.

Answer 66

Since 4 is just a constant, we can pull it outside of the integral.\[\Large 4\int\limits \sqrt[3]{\color{orangered}{u}}(\color{royalblue}{du})\]

Answer 67

And again, we'll want to write this as a rational expression: \(\large \sqrt[3]{u}=u^{1/3}\) and apply the power rule.

Answer 68

\[4\int\limits_{}^{}u^{1/3} du\]

Answer 69

mhm

Answer 70

then i aintiderive.

Answer 71

\[4\int\limits_{}^{}\frac{ 3 }{ 4 }u^{1/3} du\]

Answer 72

@zepdrix

Answer 73

Hmm your power didn't increase for some reason..
And remember, when you anti-differentiate, the squiggly bar and du should disappear.

Answer 74

\[4*\frac{ 3 }{ 4 }u^{4/3} du\]

Answer 75

\[=>3u^{4/3}du\]

Answer 76

D::::::::::::::
"And remember, when you anti-differentiate, the \(\bf \text{squiggly bar}\) and \(\bf\text{du}\) should disappear."

Answer 77

ahhh! lol
\[=> 3u^{4/3} \]

Answer 78

yay good job \c:/\[\large 3u^{4/3}+C\]

Answer 79

but I sub \[(\theta^2 -1)\] back in right?

Answer 80

Oh true :U

Answer 81

yay! thank you so much for all the help!(: I really appreciate it!

Answer 82

np \c:/