# Thread: Intergration questions

1. ## Intergration questions

Greetings. I could really use some assistance on some homework questions. I've been working on them, but can't seem to make it far on them. Any help is appreciated. Thanks!

The first and second are improper integrals.
The first is:

∫(lower limit 3, upper limit ∞) dx/√(x + 1)

The second is:

∫(lower limit 0, upper limit 4) dx/√(4 - x)

The third is:

∫ 1/√(9 + x squared)dx

2. Originally Posted by Sukasa
Greetings. I could really use some assistance on some homework questions. I've been working on them, but can't seem to make it far on them. Any help is appreciated. Thanks!

The first and second are improper integrals.
The first is:

∫(lower limit 3, upper limit ∞) dx/√(x + 1)

The second is:

∫(lower limit 0, upper limit 4) dx/√(4 - x)

The third is:

∫ 1/√(9 + x squared)dx
For the first integral, try a substitution y = x + 1. Then write the integral as:
Int(y^{-1/2}dy)

The second integral works with a similar substitution.

For the third consider a trig substitution: x = 3*tan(y). then the denominator of the integrand becomes:
sqrt(9 + 9*[tan(y)]^2) = 3*sqrt(1 + [tan(y)}^2) = 3*sec(y)

-Dan

3. Originally Posted by Sukasa
Greetings. I could really use some assistance on some homework questions. I've been working on them, but can't seem to make it far on them. Any help is appreciated. Thanks!

The first and second are improper integrals.
The first is:

∫(lower limit 3, upper limit ∞) dx/√(x + 1)
Find the antiderivative of 1/sqrt(x+1)
That is,
INTEGRAL (x+1)^{-1/2} dx
Use the substitution u=x+1 to get,
2(x+1)^{1/2}+C
Thus, we need to consider, limit of N---> oo
2(N+1)^(1/2)-4
It clearly diverges.
The second is:

∫(lower limit 0, upper limit 4) dx/√(4 - x)
This is a different type of improper integral.
It has an vertical asymptote at x=4
First find the anti-derivative,
INTEGRAL (4-x)^{-1/2} dx use u=4-x thus,
-(2)(4-x)^{1/2}+C
We have,
-2(4-N)^{1/2}+4
Next find limit N----> 4 - (from the left).
That is equal to 4

I am having problems with this one, my graphing software is showing it diverges but my results show it is 4.

The third is:

∫ 1/√(9 + x squared)dx
Express as,
1/3sqrt[1+(x/3)^2]
Use substitute u=x/3 then we have,
1/sqrt(1+u^2)
This is a tabular integral, iverse hyperbolic sine function,
asinh(u)+C
Thus,
asinh(x/3)+C

4. ## Thanks!

Thanks guys for the help!! I think I see where I was getting held up on those 3 now and should be ok on them. I got further along in my work and came upon one more that I seem to be having issues with or maybe just not understanding.

From 0 ≤ x ≤ ∞, and 0 ≤ y ≤ 1

∫(from lower 0 to upper ∞) ℮^(-x) *dx

In essence, the question is to find the thickness of the concrete of a road at x satisfy h = ℮^(-x)

O, could appreciate help on this before 3:50 central, before I head off to one of my classes. Thanks again

Edit: Nevermind! I finally figurered it out.

5. Hello, Sukasa!

Here's the third one . . .

$\displaystyle \int\frac{dx}{\sqrt{x^2+9}}$

Let $\displaystyle x = 3\tan\theta\quad\Rightarrow\quad dx = 3\sec^2\theta\,d\theta$

Substitute: .$\displaystyle \int\frac{3\sec^2\theta\,d\theta}{3\sec\theta} \;=\;\int\sec\theta\,d\theta\;=\;\ln|\sec\theta + \tan\theta| + C$

Back-substitute: .$\displaystyle \tan\theta = \frac{x}{3}\quad\sec\theta = \frac{\sqrt{x^2+9}}{3}$

And we have: .$\displaystyle \ln\left|\frac{\sqrt{x^2+9}}{3} + \frac{x}{3}\right| + C \;= \;\ln\left|\frac{x + \sqrt{x^2+9}}{3}\right| + C$

This can be simplfied further:

. . $\displaystyle \ln\left|x + \sqrt{x^2 + 9}\right| - \underbrace{\ln(3) + C}$
. . . . . . . . . . . . . . .
a constant

. . $\displaystyle =\:\ln\left|x + \sqrt{x^2+9}\right| + C$

LaTeX is back . . . $\displaystyle Yippee! \;\hdots\;\circlearrowleft\;\;\heartsuit\;\;\right squigarrow\;\;\ddots\;\;\bigstar$

6. Originally Posted by Sukasa

∫(from lower 0 to upper ∞) ℮^(-x) *dx
.
You have,
$\displaystyle \int_0^{+\infty} e^{-x} dx$
You need to consider,
$\displaystyle \lim_{n\to \infty} \int_0^n e^{-x} dx$
Which is,
$\displaystyle \lim_{n\to \infty} -e^{-n}+e^{0}=1$