Question

Find the definite integral

Answer 1

\[\int\limits_{0}^{12} \frac{8x}{\sqrt{x^2+25}} dx\]

Answer 2

hmm....

Answer 3

trig sub is the first thing that comes to mind

Answer 4

I was thinking of tan, but that doesn't work in this case tho?

Answer 5

I do remember there were 3 additional ones for cot, csc, and cos but we didn't do those.

Answer 6

sin is a-u

Answer 7

you usually use tan when you have + under the radical

Answer 8

no radical :(

Answer 9

tan is just a+u with no sqrt().

Answer 10

?

Answer 11

are you saying you miss-typed the Q ?

Answer 12

the common trig sub forms are as follows:\[\sqrt{a^2x^2+b^2}\implies x=\frac ba\tan u\]\[\sqrt{a^2x^2-b^2}\implies x=\frac ba\sec u\]\[\sqrt{b^2-a^2x}\implies x=\frac ba\sin u\]

Answer 13

wait a minute, I'm over-thinking
u=x^2+25 is all you need here

Answer 14

hmm

Answer 15

in my book it just says \[\frac{du}{u^2+a^2} is \arctan\]

Answer 16

Yes you only need an u substitution in here
\[u = x^2+25\]
and therefore
\[\frac{du}{dx}= 2x \rightarrow 4\frac{du}{dx}=8x\]

Answer 17

where does that 4 come from?

Answer 18

I multiply it times 4 to match it with the numerator.

Answer 19

then I can substitute that entire stuff into the numerator.

Answer 20

oh okay... so it's \[4\int\limits \frac{du}{\sqrt{u}} ?\]

Answer 21

I am left with:
\[4\int\limits_{}^{}\frac{1}{\sqrt{u}} du\]

Answer 22

perfect

Answer 23

Makes sense... I always am learning cool new tricks :P.

Answer 24

It's "magic" !

Answer 25

so now I can do the integration which would be \[4 * \frac{1}{2\sqrt{u}} +c?\]

Answer 26

yes that is the answer

Answer 27

or \[ \frac{2}{\sqrt{u}} +c?\]

Answer 28

\[ \frac{2}{\sqrt{x^2+25}} +c?\]

Answer 29

ah poo it's a definite integral okay...

Answer 30

\[[\frac{2}{\sqrt{12^2+25}} ] - [\frac{2}{\sqrt{0+25}} ]\]

Answer 31

wait wait wait. you are going too fast, remember 4/2 = 2

Answer 32

\[[\frac{2}{\sqrt{49}} ] - [\frac{2}{\sqrt{25}} ]\]

Answer 33

somewhere we are going wrong here (x

Answer 34

I already did 4/2...?

Answer 35

I think you guys did the integral wrong

Answer 36

yes Turing is right

Answer 37

maybe I thought of the derivative of a sqrt?

Answer 38

moving too fast lol

Answer 39

it should be
\[\frac{1}{2}\sqrt{u}\]

Answer 40

which is what I did...

Answer 41

that one step where your square term ended up in the denominator, that was wrong, I said yes to that, my fault.

Answer 42

then we multiplied by 4.

Answer 43

no, for you it's in the enumerator.

Answer 44

oh woops.

Answer 45

yes as I said, my bad, it looked so right to me, because it's a natural derivative, but not the integral hehe.

Answer 46

yeah I confused them :P

Answer 47

\[2{\sqrt{49}} ] - [2\sqrt{25} ]\]

Answer 48

\[[2*7 ] - [2 * 5]\]

Answer 49

14-10 = 4

Answer 50

at least these are nice numbers, always a good indication hehe.

Answer 51

:)

Answer 52

actually I think I'm wrong again...

Answer 53

since it should be 4*2

Answer 54

8*7 -8*5 = 56-40 = 16?

Answer 55

let me write it down, I have only followed via screen so far, never works well for me obviously (-;

Answer 56

:p

Answer 57

yes you are right 4* 2sqrt(u) = 8 sqrt(u)

Answer 58

the good thing is, now we have made all the possible mistakes that should be avoided in an exam in just one single exercise (-:

Answer 59

and I messed up again it seems...

Answer 60

12^2 = 144

Answer 61

sqrt169 = 13

Answer 62

but our integral is right, I just checked with WolframAlpha:
\[8\sqrt{x^2+25}\]
so the rest will be F(a)-F(b) without making careless mistakes (-:

Answer 63

8 * 13 = 104 - 40 = 64

Answer 64

which are the endpoints again? from ? to ?

Answer 65

0-12

Answer 66

yes 64

Answer 67

:)