Question

QUESTION:
Find the area of the first arch of the curve y = e^(-x) sin x
TOPIC: PLANE AREAS IN RECTANGULAR COORDINATES IN CALCULUS

Answer 1

\[y = e^{-x} sinx\]

Answer 2

@kittiwitti1

Answer 3

@NoelGreco

Answer 4

I can't help, sorry. I'm also working on a problem of my own at the minute --
@ganeshie8 ?

Answer 5

1. Find the zeros. Remember that the "e" factor is always ppositive

2. Integrate by parts twice

3. Evaluate

Answer 6

are the limits is from 0 to 0?

Answer 7

can you explain it one by one so i can understand please

Answer 8

you want \(\int\limits_0^{\pi / 2} \sin x \; e^{-x} \; dx\), right?!

so in an IBP the \(\sin\) will flip back and forward from \(\sin \to \cos \to \sin \to \cos \dots\), and the exponential just changes from \(+ \to - \to + \to - \dots\)

so you can pretty much do what you fancy, quite a luxury really

:p

Answer 9

so by ibp i will get

Answer 10

so by ibp i will get
\[\int\limits e ^{-x} sinx dx \]

Answer 11

\[u = sinx, du = \cos x dx ,dv = \int\limits e ^{-x}dx , v =e ^{x}\]

Answer 12

\[=e ^{-x}sinx dx - \int\limits e ^{-x}\cos xdx\]

Answer 13

\[u^`= cosx, du^` = `sinx dx , dv = e ^{-x} \cos xdx , v = e^^{-x}\]

Answer 14

then i will get a equation of \[1/2 e ^{-x} (sinx - \cos x)+ c \]

Answer 15

@SolomonZelman

Answer 16

As you were directed, an integration by parts twice will allow you to solve the integral. You will integrate it without completely integrating it.

Want an example of such?

Answer 17

yes

Answer 18

thats my answer

Answer 19

is it right?

Answer 20

Suppose you have the following integral:
\(\color{#000000}{ \displaystyle \int e^x\cos x~dx }\)

Integrate by parts.

Set,
\(\color{#000000}{ \displaystyle u=\cos x \quad\quad\quad dv=e^x~dx }\)

And from that setup you have,
\(\color{#000000}{ \displaystyle du=-\sin x~dx \quad\quad\quad v=e^x }\)

then, you get:
\(\color{#000000}{ \displaystyle \int e^x\cos x~dx=e^x\cos x-\left(\int(e^x)(-\sin x~dx)\right) }\)

\(\color{#000000}{ \displaystyle \int e^x\cos x~dx=e^x\cos x+\int e^x\sin x~dx }\)

Then, you will apply integration by parts to the second integral with e^x•sin(x):

\(\color{#000000}{ \displaystyle \int e^x\sin x~dx=? }\)

Set,
\(\color{#000000}{ \displaystyle u=\sin x \quad\quad\quad dv=e^x~dx }\)

from there,
\(\color{#000000}{ \displaystyle du=\cos x~dx \quad\quad\quad v=e^x }\)

\(\color{#000000}{ \displaystyle \int e^x\sin x~dx=e^x\sin x-\int e^x\cos x~dx }\)

Going back to the previous integral you have:
\(\color{#000000}{ \displaystyle \int e^x\cos x~dx=e^x\cos x+\int e^x\sin x~dx }\)

\(\color{#000000}{ \displaystyle \int e^x\cos x~dx=e^x\cos x+\left(e^x\sin x-\int e^x\cos x~dx\right) }\)

Then, I will rely on your algebra:

\(\color{#000000}{ \displaystyle \int e^x\cos x~dx+\int e^x\cos x~dx=e^x\cos x+e^x\sin x }\)

\(\color{#000000}{ \displaystyle 2\int e^x\cos x~dx=e^x\cos x+e^x\sin x }\)

\(\color{#000000}{ \displaystyle \int e^x\cos x~dx=\frac{e^x\cos x+e^x\sin x}{2} }\)

Answer 21

I think no right.
v = \(\color{red}{-}\)e^(-x).

Answer 22

since the integral of e^(-x) is -e^(-x)

Answer 23

yes, same thing that i integrate, did u see the jpeg that i attached?

Answer 24

Yes, I did, and that is exactly I am saying wha I said just now.

Answer 25

exactly **why** I am saying ...

Answer 26

so in your first integration by parts, you would be actually adding the integral, not subtracting (because there are two negatives, (1) you subtract the integral when integrating by parts, (2) the -e^(-x), which is the integral of dv. So --=+)

Answer 27

why? the formula of ibp is uv - integral of vdu

Answer 28

yes, and asides from the negative in the formula, you have another negative coming from -e^(-x).

Answer 29

So you have two negatives, and that is plus.

Answer 30

You error is that the integral of e^(-x) is -e^(-x), and you missed that negative.

Answer 31

you missed the minus in front of the e^(-x).

Answer 32

but the negative is in exponent so nothing will do with that? @SolomonZelman

Answer 33

Ok, what is the integral of e^(-x)?

Answer 34

1/ dx

Answer 35

That is the answer to my question?

Answer 36

no, -e^(-x)

Answer 37

Yes, exactly.

Answer 38

can you help me understand it by steps,

Answer 39

So, you have:
\(\color{#000000}{ \displaystyle \int e^{-x}\sin x dx }\)

Then you set,
\(\color{#000000}{ \displaystyle u=\sin x }\), and so \(\color{#000000}{ \displaystyle du=\cos x~dx }\)
\(\color{#000000}{ \displaystyle dv=e^{-x}~dx }\), and so \(\color{#000000}{ \displaystyle v=-e^{-x} }\)

The formula is:
\(\color{#000000}{ \displaystyle \int u~dv=uv-\int v~du }\)

And here you get:
\(\color{#000000}{ \displaystyle \int(\sin x)( e^{-x} ~dx)=\sin x(-e^{-x})-\int (-e^{-x})~(\cos x~dx) }\)

Answer 40

So, you should have had the following integral:
\(\color{#000000}{ \displaystyle \int(\sin x)( e^{-x} ~dx)=-e^{-x}\sin x+\int e^{-x}\cos x~dx }\)

Answer 41

ah yes, so because we need 2 integration i will next integrate the \[\int\limits e ^{-x} \cos x\]

Answer 42

Yes, indeed!

Answer 43

So, now, YOU need to integrate this integral....

Answer 44

well, you should be consistent.
So, you will need to be integrating e^(-x) again.
(So, you can't choose cosine as dv, because then you are just going to return to your initial integral.)

Answer 45

So you must set dv=e^(-x). and thus u=cos(x).

Answer 46

dv=e^(-x) dx I meant.

Answer 47

here

Answer 48

is that right

Answer 49

@SolomonZelman

Answer 50

yes, I am there...
Please check your sines.

Answer 51

I mean signs

Answer 52

oops its should b positive at \[2 \int\limits e ^{-x} \sin x dx = -e ^{-x} sinx + e ^{-x} \cos x\]

Answer 53

so it would five a answer of \[\int\limits e ^{-x} \sin x dx = -e ^{-x} (sinx + cosx)/ 2 + c\]

Answer 54

Yes, very good!

\(\color{#000000}{ \displaystyle \int(\sin x)( e^{-x} ~dx)=\frac{-e^{-x}}{2}(\sin x+ \cos x)+C }\)

Answer 55

yes, it is will be divide by 2?

Answer 56

?

Answer 57

elaborate

Answer 58

I just didn't quite understand what you asked.

Answer 59

In any case, the first arc, from 0 to π/2. That is just plugging in the limits of integration and preforming the algebraic task. I think you can do this without me.

Good luck with your maths:)

Answer 60

i mean the whole equation of sinx + cos x will be divided by 2

Answer 61

no i can`t still @SolomonZelman

Answer 62

Yes, it will be. (You can use my answer, where I essentially factored the exponential with the 1/2)

Answer 63

We had twice the integral on the left, so to solve for the integral we needed to divide by 2, correct ?

Answer 64

so what about the first arc thingy in the question?

Answer 65

yep

Answer 66

Well, that is this integral from 0 to \(\pi/2\).

Answer 67

So you just plug in the limits of integration....
that on its own is just algebra.

Answer 68

i dont know how to compute e

Answer 69

Well, show me your work to the point where you are stock, please.

Answer 70

here

Answer 71

i dont know how to compute in calcu it gives me a decimal answer

Answer 72

You can leave the answer in exact form.
You can be very fancy and say: \(e^{-\pi/2}=i^i\)
You can approximate it as a decimal.

it all depends on what your teacher would want.

Answer 73

I assume that you know cosine and sine of π and of π/2, tho.
Can I assume so?

Answer 74

I would say that just getting rid of negative exponents, but to leave the answer in exact-simplified form is the best thing to do.

Answer 75

but my teacher want a not decimal answer

Answer 76

So, then let's get the exact answer.

Answer 77

I am lagging this whole time. Sorry

Answer 78

okay so the answer is 0?

Answer 79

It is not

Answer 80

i got 0 because pi/2 is equivalent to 180 degree, so by replacing it
0-1+0+1 = 0

Answer 81

You are missing something.

Answer 82

the first (0-1) would be multiplied by e^(-π/2), and the second (0+1) would be multiplied by 1/2, since e^(0)=1.

Answer 83

At first, lets agree that we are evaluating the following:

\(\color{#000000}{ \displaystyle -\frac{e^{-\pi/2}}{2} (\sin{\tiny~} \pi/2+\cos{\tiny~}\pi/2) -\left\{-\frac{e^{-0}}{2} (\sin{\tiny~} 0+\cos{\tiny~}0) \right\} }\)

(result of plugging the limits of integration.

Answer 84

i cant compute the e^ (-pi/2) caus ethe calcu says 0.20

Answer 85

Your teacher wants a non-decimal answer, so why not give him/her the exact answer.

Answer 86

We will enchance the answer by applying some algebra, but please try to follow with me as far as you can.

Answer 87

We are really evaluating the following:

\(\color{#000000}{ \displaystyle -\frac{e^{-\pi/2}}{2} (\sin{\tiny~} \pi/2+\cos{\tiny~}\pi/2) +\frac{e^{-0}}{2} (\sin{\tiny~} 0+\cos{\tiny~}0) }\)

do you understand why?

Answer 88

okay i will follow you

Answer 89

so 2 will be cancel out and e^pi will remain

Answer 90

Can you tell me: \(e^{-0}=e^0=?\)

Answer 91

is equals to 1

Answer 92

So, let's re-write the expression:

\(\color{#000000}{ \displaystyle -\frac{e^{-\pi/2}}{2} (\sin{\tiny~} \pi/2+\cos{\tiny~}\pi/2) +\frac{1}{2} (\sin{\tiny~} 0+\cos{\tiny~}0) }\)

YOu can evaluate step by step, if quicker wrking results in small errors.

Answer 93

okay then what to do with the first e how will going to eliminate it

Answer 94

And as you said:
\(\sin 0=\cos \pi/2 =0\)
\(\sin \pi/2=\cos 0 =1\)

So you would then end up with:
\(\color{#000000}{ \displaystyle -\frac{e^{-\pi/2}}{2} (1+0) +\frac{1}{2} (0+1) }\)

\(\color{#000000}{ \displaystyle -\frac{e^{-\pi/2}}{2} +\frac{1}{2} }\)

\(\color{#000000}{ \displaystyle \frac{1}{2}-\frac{e^{-\pi/2}}{2} }\)

Answer 95

And then what you so much desired.
But, the bad news is that you can't eliminate it, really.
THo, you could re-write it without negative exponent.