Question

∫ 1 between 0 of : t/√(4-t²) dt

Answer 1

sub u = 4-t^2

Answer 2

that's what I did but it doesn't work ...

Answer 3

lets do it again and see...

Answer 4

the answer is : 0,27 and that's why this doesnt work :(

Answer 5

u = 4-t^2
du = -2t dt
-du/2 = tdt

as t->0
u -> 4-0^2 = 4

as t->1
u-> 4-1^2 = 3

Answer 6

^corrected the typo above

Answer 7

\(\large \mathbb{\int_1^0 \frac{t}{\sqrt{4-t^2}}dt}\)

\(\large \mathbb{\int_3^4 \frac{1}{\sqrt{u}}(-du/2)}\)

Answer 8

evaluate the integral ?

Answer 9

that's what I try to do :P

Answer 10

\(\large \mathbb{\int_3^4 \frac{1}{\sqrt{u}}(-du/2)}\)

\(\large \mathbb{\frac{-1}{2}\int_3^4 u^{\frac{-1}{2}}du}\)

Answer 11

try now :)

Answer 12

but where did you took your 4/3? Cuz I have the same equation if it where not this :o

Answer 13

(not good in english, sorry :) )

Answer 14

i get exactly what u mean lol, ur english is good ok

Answer 15

u are asking how did i get 3 and 4
in place of
1 and 0

right ?

Answer 16

Umm kind of yea

Answer 17

we substituted u = 4-t^2
that means, we have introduced a new 'u' variable in place of 't'
so the bounds also will change.

when t = 0, u need to find whats the value of u
when t = 1, u need to find whats the value of u

Answer 18

basically when u do substitutions, u need to carry out below steps :-
1) substitute u
2) find du in terms of dt
3) find how bounds will change
4) evaluate the new integral in u's
5) convert back to t's

Answer 19

if that confuses u, forget it.
just see that u are substituting u = 4-t^2

so u need to find the new bounds.

Answer 20

u = 4-t^2

when t = 0, u = ?

Answer 21

4`:) haha

Answer 22

yes,
when t = 1, u = ?

Answer 23

3

Answer 24

good :)

Answer 25

so, bounds from 0->1
will change to
4->3

Answer 26

fine ? are u at peace wid the bounds ha ? :)

Answer 27

so when you have the new bounds, you do the soustraction?

Answer 28

u mean,
when u substitute, u need to change the bounds accordingly ?
then yes :)

Answer 29

when u substitute,
u need to change both 'bounds' and 'differential'

Answer 30

here is the final answer :
http://www.wolframalpha.com/input/?i=%5Cfrac%7B-1%7D%7B2%7D%5Cint_4%5E3++%5Cfrac%7B1%7D%7B%5Csqrt%7Bu%7D%7Ddu

Answer 31

just u need to practice more... i see u have the concept already :)

Answer 32

Umm not sur finally

Answer 33

ya

Answer 34

do you put ''3'' in the equation - (4 in the equation) ?

Answer 35

do u have skype ?

Answer 36

no :(

Answer 37

np, i dint get ur q.... can u elaborate a bit ha :)

Answer 38

\[\int\limits_{a}^{b} f(x)dx = F(b) - F(a)\]

Answer 39

yes

Answer 40

so you do this with the new bounds?

Answer 41

yup, lets do it step by step again ok ?

Answer 42

\(\large \mathbb{\int_0^1 \frac{t}{\sqrt{4-t^2}}dt}\)

Answer 43

ur original question is like that right ?

Answer 44

Ok but when you pu the 3 in the equation, the interior of the radical is negative :(

Answer 45

Yes!

Answer 46

thats a good observation actually !
lets see whats actually happening while solving it :)

Answer 47

\(\large \mathbb{\int_0^1 \frac{t}{\sqrt{4-t^2}}dt}\)

substitute \(\large u = 4-t^2\)

Answer 48

\(\large u = 4-t^2\)

differentiate both sides

Answer 49

wat do u get ?

Answer 50

-du/2 = t dt

Answer 51

yes, so we can replace tdt wid -du/2
and 4-t^2 wid u

Answer 52

lets do that first,
we can wry about bounds afterwards :)

Answer 53

\(\large \mathbb{\int_0^1 \frac{t}{\sqrt{4-t^2}}dt}\)

substitute \(\large u = 4-t^2\)

\(\large du = -2t dt\)
\(\large -du/2 = tdt\)

so the itegrand becomes :

\(\large \int_{...}^{....} \frac{1}{\sqrt{u}} (-du/2)\)

Answer 54

let me knw if ur fine so far

Answer 55

ok thanks :)

Answer 56

next, setup the bounds

Answer 57

u = 4-t^2

when t=0, u = ?
when t = 1, u = ?

Answer 58

4 and 3

Answer 59

\(\large \mathbb{\int_0^1 \frac{t}{\sqrt{4-t^2}}dt}\)

substitute \(\large u = 4-t^2\)

\(\large du = -2t dt\)
\(\large -du/2 = tdt\)

so the itegrand becomes :

\(\large \int_{...}^{....} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \int_{4}^{3} \frac{1}{\sqrt{u}} (-du/2)\)

Answer 60

evaluate the new integral now

Answer 61

factor out the constant -1/2

Answer 62

\(\large \mathbb{\int_0^1 \frac{t}{\sqrt{4-t^2}}dt}\)

substitute \(\large u = 4-t^2\)

\(\large du = -2t dt\)
\(\large -du/2 = tdt\)

so the itegrand becomes :

\(\large \int_{...}^{....} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \int_{4}^{3} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \frac{-1}{2}\int_{4}^{3} \frac{1}{\sqrt{u}} du\)

Answer 63

fine so far ?
can u do the rest ha

Answer 64

you + 1 the exposant of the u ( that you put in the nominator) / the answer of the addition? And after you do the thing with the soustraction ?

Answer 65

yes, here is the formula :
\(\large \int x^n dx = \frac{x^{n+1}}{n+1}\)

Answer 66

\(\large \mathbb{\int_0^1 \frac{t}{\sqrt{4-t^2}}dt}\)

substitute \(\large u = 4-t^2\)

\(\large du = -2t dt\)
\(\large -du/2 = tdt\)

so the itegrand becomes :

\(\large \int_{...}^{....} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \int_{4}^{3} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \frac{-1}{2}\int_{4}^{3} \frac{1}{\sqrt{u}} du)\)

\(\large \frac{-1}{2}\int_{4}^{3} u^{\frac{-1}{2} } du\)

Answer 67

ok thnak you very mucch!! I'M okay with the rest :)

Answer 68

\(\large \mathbb{\int_0^1 \frac{t}{\sqrt{4-t^2}}dt}\)

substitute \(\large u = 4-t^2\)

\(\large du = -2t dt\)
\(\large -du/2 = tdt\)

so the itegrand becomes :

\(\large \int_{...}^{....} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \int_{4}^{3} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \frac{-1}{2}\int_{4}^{3} \frac{1}{\sqrt{u}} du)\)

\(\large \frac{-1}{2}\int_{4}^{3} u^{\frac{-1}{2} } du\)


\(\large \frac{-1}{2} \frac{u^{\frac{-1}{2}+1 }}{\frac{-1}{2}+1} \Big|_4^3\)

Answer 69

good luck wid the rest :)

Answer 70

doesnt workkkkk uhuhuhh

Answer 71

it gaves me : -1 - 0

Answer 72

\(\large \mathbb{\int_0^1 \frac{t}{\sqrt{4-t^2}}dt}\)

substitute \(\large u = 4-t^2\)

\(\large du = -2t dt\)
\(\large -du/2 = tdt\)

so the itegrand becomes :

\(\large \int_{...}^{....} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \int_{4}^{3} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \frac{-1}{2}\int_{4}^{3} \frac{1}{\sqrt{u}} du)\)

\(\large \frac{-1}{2}\int_{4}^{3} u^{\frac{-1}{2} } du\)


\(\large \frac{-1}{2} \frac{u^{\frac{-1}{2}+1 }}{\frac{-1}{2}+1} \Big|_4^3 \)

\(\large \frac{-1}{2} \frac{u^{\frac{1}{2} }}{\frac{1}{2}} \Big|_4^3 \)

Answer 73

\(\large \mathbb{\int_0^1 \frac{t}{\sqrt{4-t^2}}dt}\)

substitute \(\large u = 4-t^2\)

\(\large du = -2t dt\)
\(\large -du/2 = tdt\)

so the itegrand becomes :

\(\large \int_{...}^{....} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \int_{4}^{3} \frac{1}{\sqrt{u}} (-du/2)\)

\(\large \frac{-1}{2}\int_{4}^{3} \frac{1}{\sqrt{u}} du)\)

\(\large \frac{-1}{2}\int_{4}^{3} u^{\frac{-1}{2} } du\)


\(\large \frac{-1}{2} \frac{u^{\frac{-1}{2}+1 }}{\frac{-1}{2}+1} \Big|_4^3 \)

\(\large \frac{-1}{2} \frac{u^{\frac{1}{2} }}{\frac{1}{2}} \Big|_4^3 \)

\(\large - \sqrt{u} \Big|_4^3 \)

Answer 74

fine so far ?

Answer 75

\[\frac{ -1 }{ 2 } \frac{ u^{1/2} }{ 1/2

Answer 76

oops
but yes thats what I tap

Answer 77

wrote/

Answer 78

yahh :)

Answer 79

\(\large - \sqrt{u} \Big|_4^3 \)

u just need to evaluate this

Answer 80

4-9= -5......

Answer 81

\(\large - \sqrt{u} \Big|_4^3 \)


\(\large - [\sqrt{3} - \sqrt{4}] \)

Answer 82

\(\large - \sqrt{u} \Big|_4^3 \)


\(\large - [\sqrt{3} - \sqrt{4}] \)

\(\large - [\sqrt{3} - 2] \)

Answer 83

use ur calculator :P

Answer 84

u = 4-t^2 ?

Answer 85

we can forget about u, cuz we changed the bounds also

Answer 86

\(\large - \sqrt{u} \Big|_4^3 \)


\(\large - [\sqrt{3} - \sqrt{4}] \)

\(\large - [\sqrt{3} - 2] \)

\(\large 2 - \sqrt{3}\)

Answer 87

thats the final answer

Answer 88

umm okay

Answer 89

were suppose to change the bounds everytimes?

Answer 90

yes, everytime we do substitution, we need to change bounds also

Answer 91

im going for dinner... brb
u go ahead wid other problems.. .the more u do these, the more u wil feel confident :)

Answer 92

Okay thank you :p didn't know I was this much noob in those integrals..... haha i'm shy!! good dinner :)

Answer 93

got problems with another question ... can you help me :)

Answer 94

I can send you a picture of what I did