Question

Find the following improper integrals:

f(infinity-negativeinfinity) xe^-x2dx

f(3-2) dx/rad3-x

Answer 1

@cruffo

Answer 2

I suspect the first one will be 0. Take a look at the graph:
http://www.wolframalpha.com/input/?i=f%28x%29+%3D+xe%5E%7B-x%5E2%7D

Answer 3

do you know how you get 0?

Answer 4

To work it out you'll need the "by parts" integration technique.
http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx

Answer 5

it's not loading !

Answer 6

By parts OR substitution. Are you familiar with either method?

Answer 7

No show me please

Answer 8

What level math are you in. The method we use depends on what class you are in.

Answer 9

Calculus 2

Answer 10

So you should be familiar with both "by parts" and "substitution". Let's use substitution, it will be the easiest to explain.

Answer 11

\[\int_{-\infty}^{\infty} xe^{x^2} dx\]

We start by making a guess at what to let out "u" substitution be. Let's let u = x^2.

Answer 12

kk

Answer 13

then differentiating both sides we have du = 2x dx

Answer 14

Right

Answer 15

so \(\dfrac{1}{2x} du = dx\). Now make the appropriate substitutions for x^2 and dx:
\[\int xe^{x^2} dx = \int \left(x e^{\color{red}u}\right)\color{red}{\frac{1}{2x} du } =\int \frac{1}{2} e^u du \]

Answer 16

ok

Answer 17

You remember what \(\int e^u \;du \) is?

Answer 18

i think so

Answer 19

It's one of the basic integrals from cal 1

Answer 20

dude i haven't taken calc 1 for a yr

Answer 21

pls refresh my memory

Answer 22

cal II must not be much fun...

Answer 23

it's horrible

Answer 24

ok \(\int e^u du = e^u\)

Answer 25

i'm barely passing my class with a C

Answer 26

ok

Answer 27

whats next?

Answer 28

So...
\[\int xe^{x^2} dx = \int \left(x e^{\color{red}u}\right)\color{red}{\frac{1}{2x} du } =\int \frac{1}{2} e^u du = \frac{1}{2}e^u\]

Now for a bit of "back substitution". We started out letting u = x^2, soooo
\[\frac{1}{2}e^u = \frac{1}{2}e^{x^2}\]


The last bit is to apply the limits of integration.

Answer 29

\[\lim_{x \rightarrow -\infty}\frac{1}{2}e^{x^2} = 0 \]
\[\lim_{x \rightarrow \infty}\frac{1}{2}e^{x^2} = \infty \]


damn.... made a mistake. Forgot the negative sign ... should have been

Answer 30

wait why is it pie?

Answer 31

\[\int xe^{-x^2} dx = \int \left(x e^{-\color{red}u}\right)\color{red}{\frac{1}{2x} du } =\int \frac{1}{2} e^{-u} du = -\frac{1}{2} e^{-u} \]

Answer 32

oops

Answer 33

ok

Answer 34

so..
\[\lim_{x \rightarrow -\infty}\frac{1}{2}e^{-x^2} = 0\]
\[\lim_{x \rightarrow \infty}\frac{1}{2}e^{-x^2} = 0\]

so
\[\int_{-\infty}^{\infty} xe^{x^2} dx = 0 - 0 = 0\]

Answer 35

http://www.wolframalpha.com/input/?i=integral+from+-infty+to+infty+%28xe%5E%7B-x%5E2%7D%29

Answer 36

interesting

Answer 37

what about the second one?

Answer 38

working on that, give me a sec...

Answer 39

kk

Answer 40

It's another substitution integration technique.

Answer 41

\[\int \frac{dx}{\sqrt{3-x}}\]
Let u = 3 - x
Then du = -dx
so -du = dx.

Answer 42

ok

Answer 43

right

Answer 44

what is the next step?

Answer 45

So it should look like:
\[\int \frac{dx}{\sqrt{3-x}} = \int \frac{-du}{\sqrt{u}}\]

This can be done with power rule for integration, with a little trick on rewritting \(\sqrt{u} = u^{1/2}\)

Answer 46

Power Rule: \(\int u^n \; du = \dfrac{u^{n+1}}{n+1}\)

Answer 47

\[\int \frac{dx}{\sqrt{3-x}} = \int \frac{-du}{\sqrt{u}} = \int -u^{-1/2} du = -\frac{u^{-1/2 + 1}}{-1/2 + 1}\]

Answer 48

is that the final form?

Answer 49

Not yet, this is still in terms of u, not x. Let's clean that up and back substitute.

Answer 50

\[-\frac{u^{-1/2 + 1}}{-1/2 + 1} = -2\sqrt{u} = -2\sqrt{3-x}\]

Answer 51

Then apply the limits of integration:
When x = 3:
\[-2\sqrt{3 - 3} = 0\]
When x = 2:
\[-2\sqrt{3 - 2} = -2\]

So 0 - (-2) = 2

http://www.wolframalpha.com/input/?i=integral+from+2+to+3+%281%2Fsqrt%7B3-x%7D%29

Answer 52

so basically you just substitute and integrate for all of these

Answer 53

Yes, these problems involve "integration by substitution"

Answer 54

okay got it . i'm gona go practice these

Answer 55

cruffo are you a math teacher?

Answer 56

or math major?

Answer 57

I teach - college.

Answer 58

how often are you on this website? you seem very good at explaining calc 2 problems

Answer 59

I haven't been here in a while.

Answer 60

:(

Answer 61

i could use your help in the future

Answer 62

Maybe....

Answer 63

will you be on here tomorrow as well?