Question

See attached (integrals).

Answer 1

75, 81, 87

Answer 2

what do you get when you integrate 75?

Answer 3

sorry i understand 75 now.

Answer 4

i have to rotation my laptop to looking for it

Answer 5

i just need to know how to do 81 and 87.
@RadEn sorry about that. you could copy and paste in onto a word document and then rotate it...

Answer 6

ok, well, the second ftc is a consequence of the following\[\frac d{dx}\int_{a(x)}^{b(x)}f(t)~dt=\frac d{dx}~F(b(x)-F(a(x))\]
\[\frac d{dx}~F(b(x)-F(a(x))=F'(b(x))b'(x)-F'(a(x))a'(x)\]

and since F' = f, and to rewrite it more readable

\[\frac d{dx}~\int_{a}^{b}f(t)~dt~=~f(b)b'-f(a)a'\]

Answer 7

so for 81, you're just basically differentiating the integral?

Answer 8

yes

Answer 9

but without having to integrate first and then derive ... just use the consequences

Answer 10

thanks. i'm just really bad with understanding directions.
and for 87, you're integrating then plugging it into the bounds?

Answer 11

not really integrating, they give you the derivative already as the function that is integrating.

rewite the function, and plug in your limits and thier derivatives to solve

Answer 12

if given f(t), with limits a and b

f(b) b' - f(a) a' is the solution

Answer 13

85 for example

f(t) = t cos(t) ; a=0, b=x

(x cos(x))x' - (0 cos(0))0' = x cos(x)

Answer 14

yes ... but without working it up and then down again

Answer 15

ok thanks

Answer 16

ok sorry, i'm just confused on 87 now.

Answer 17

ok, you tell me how you think it should go, and ill correct you along the way :)

Answer 18

i'm not really sure. since you're supposed to find F'(x), isn't it just (4x+1) for #87? and i guess plug in the bounds?

Answer 19

@amistre64 should that be chain rule 87

Answer 20

close

(4b+1)b' - (4a+1)a'

Answer 21

chain rule is a result of the workings yes

Answer 22

(4(x+2)+1)x' - (4x+1)x'

and x'=1 soo

4(x+2)+1 - 4x-1

Answer 23

i got (8x+10) is that right?

Answer 24

without chain rule

Answer 25

(8x + 10) is not right as a final result, no

Answer 26

hmm how to do

Answer 27

would you agree that \[\int_{a}^{b}f(t)~dt=F(b)-F(a)\]???

Answer 28

yup

Answer 29

and that the derivative of F is f?

Answer 30

yup

Answer 31

ok

then all thats needed is to take the derivative of F(b)-F(a)

\[\frac d{dx}[F(b)-F(a)]\]
\[\frac d{dx}[F(b)]-\frac d{dx}[F(a)]\]
\[f(b)b'-f(a)a'\]

we agree?

Answer 32

ok

Answer 33

so the only thing to do is take the f(t) that is sitting inside the integral; and work in the limits and their derivatives ...

Answer 34

this is the work that they have.
but i don't understand why they integrate it then take the derivative.

Answer 35

f(t) = (4t+1) ; a=x, b=x+2
a'=1, b'=1

f(b)b' - f(a)a'

(4(x+2)+1)*1 - (4x+1) * 1

Answer 36

ohh wow that's simple last question @amistre64 how do we know when to derivative i mean in what form

Answer 37

first integrate it and then derivative

Answer 38

you never need to integrate first and then derive .... the rule itself proves that the extra steps are not needed

there are some on the page there that simply cant be integrated ...

Answer 39

if you want to integrate up first, and then derive down again thats totally your own personal call, but problems 75-80 were used to establish the truth of it to begin with

Answer 40

ok...

Answer 41

i got u

Answer 42

:) good luck

Answer 43

thanks.