Question

Another Calc II question.
Find y'(x) if y= integral of sin(2e^t) dt *limits of integration are from 0 to ln(4x)*
I don't understand the y'(x) part really.

Answer 1

\[\frac d{dx}~(\int_{a(x)}^{b(x)}f(x)~dx)=f(b)b'-f(a)a'\]

Answer 2

we can follow this using the definitions of integration and derivatives

Answer 3

\[\int_{a(x)}^{b(x)}f(t)dt=F(b(x))-F(a(x))\]
take the derivative of that, noting the chain rule
\[Dx[F(b(x))-F(a(x))]=f(b(x))b'-f(a(x))a'\]

Answer 4

I understand how to integrate but why does it say y'(x) because to integrate you are finding the original function.

Answer 5

yes, notice that the integration creates a function of "x"

Answer 6

but when i integrate as well I cannot use the U substitution because setting u=(2e^t) would be du/dx = 2e^t and there isn't anywhere in the integral containing that

Answer 7

notice the form from the definitions; dont try to do all the work .. you cant

Answer 8

what is f(t) ?

Answer 9

-cos(2e^t)?

Answer 10

according to the post: f(t) = sin(2e^t)

Answer 11

ohhh im so confused!

Answer 12

what then is f(b(x)) ?
and f(a(x)) ?

Answer 13

sin(2e^ln(4x))-sin(2e^0)?

Answer 14

good, now lets fill in the missing parts; the b' and a'


sin(2e^ln(4x)) * (ln(4x))' - sin(2e^0)*0'

Answer 15

the key is to know that by definitions of integration and derivative, we have all the parts we need. It doesnt matter what the solution of the integration is, becasue we end up taking the derivative back down again

Answer 16

sin(2e^ln(4x)) * (1/4x)?

Answer 17

close, your ln(4x) derivative needs work i think

Answer 18

\[D_x(ln(u))=\frac{u'}{u}\]
\[D_x(ln(4x))=\frac{4}{4x}\to\ \frac{1}{x}\]

Answer 19

right i just did that can and can i simplify sin(2e^ln(4x))

Answer 20

im sure you can
\[e^{ln(4x)}=e^{ln4+lnx}=e^{ln4}*e^{lnx}\to\ 4x\]

Answer 21

right i got sin(8x) (1/x)

Answer 22

correct

Answer 23

and then i just derive it from there?

Answer 24

derive what? all youre doing is simplifying what youve already derived

Answer 25

so that's y'(x)?

Answer 26

yes

Answer 27

okay...im still a bit confused with that though

Answer 28

i understand the process but the concept is a bit confusing

Answer 29

\[y(x)=\int_{a(x)}^{b(x)}f(t)~dt\]
\[y(x)=F(b(x))-F(a(x))\]

you agree with this by definition of integration?

Answer 30

yes

Answer 31

then y'(x) is the derivative\[y'(x)=[F(b(x))−F(a(x))]'\]
then y'(x) is the derivative\[y'(x)=[F(b(x))]'−[F(a(x))]'\]
then using the chain rule; y'(x) is the derivative\[y'(x)=F'(b(x))b'−F'(a(x))a'\]
agreed?

Answer 32

ohh yess i see now

Answer 33

and we already now what F' is.... its just the f that we started with

Answer 34

koayy thanks you so much I understand now. Much clearer

Answer 35

youre welcome :)