Integrals

[tkvictim]

The directions read:
Evaluate

by interpreting the integral in terms of area.


So, I started off with putting 1 over the sq.rt. of 1-x^2, so that when I take the integral, it would be 1/sin^-1x but I know this is far from right. I tried u substitution but the resulting du didn't match with dx. Is u substitution even necessary?

Also. I have other homework problems that I completed, and I feel sort-of confident in my answers but I am not always right. Is anyone willing to double-check them? You would be life savers!

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ \int_{-1}^{1}[2-\sqrt{1-x^{2}}]dx \ = \ \int_{-1}^{1}2dx \ -\int_{-1}^{1}\sqrt{1-x^{2}}dx$

$\displaystyle = \ 2x]_{-1}^{1} \ -\int_{-1}^{1}\sqrt{1-x^{2}}dx \ = \ 4-\int_{-1}^{1}\sqrt{1-x^{2}}dx$

$\displaystyle Let \ x \ = \ sin(t), \ then \ dx \ = \ cos(t)dt$

$\displaystyle Hence, \ f(x) \ = \ 4 \ -\int_{-\pi/2}^{\pi/2}\sqrt{1-sin^{2}(t)}cos(t)dt \ = \ 4-\int_{-\pi/2}^{\pi/2}cos^{2}(t)dt$

\(\displaystyle = \ 4-1/2\int_{-\pi/2}^{\pi/2}[1+cos(2t)]dt \ = \ 4-\int_{0}^{\pi/2}[1+cos(2t)}]dt\)

$\displaystyle = \ 4 \ -\bigg[t+\frac{sin(2t)}{2}\bigg]_{0}^{\pi/2} \ = \ 4-[t+sin(t)cos(t)]_{0}^{\pi/2} \ = \ 4-\frac{\pi}{2}$

 

[tkvictim]

Wow, thank you. I would have never thought of doing that.

Should I approach this problem the same way?

Evaluate

.

I started off with U substitution, but my resulting du had an x variable in it and it just looked a mess. My final answer came out to be
(1/u) (du/2x^3)=
2x^3 ln | 1+x^6 | + C

 

[JeffWest]

I will give you a hint.

use as your substitiution

$\displaystyle u=x^3$

Then

$\displaystyle du=3x^2$

Then check out a standard table of integrals..that should help

 

[BigGlenntheHeavy]

$\displaystyle \int\frac{3x^{2}}{1+x^{6}}dx \ = \ 3\int\frac{x^{2}}{1+(x^{3})^{2}}dx$

$\displaystyle Let \ u \ = \ x^{3}, \ then \ du \ = \ 3x^{2}dx, \ ergo$

$\displaystyle we \ have \ \int\frac{du}{1+u^{2}} \ = \ arctan(u)+C \ = \ arctan(x^{3})+C$

 

[tkvictim]

Ah! Thank you!

 

[tkvictim]

Alright, and I have one last question. I just need to know one thing.

One problem just gave me a chart, as follows:

Directions read:
The following table shows the speedometer reading of a truck, taken at ten-minute intervals during one hour of a trip.
Use the table and the indicated technique to estimate the distance that the truck traveled in the hour. Give exact answers.

The techniques are trapezoid rule and midpoint rule. I got the trapezoid rule, however the midpoint rule is confusing since I wasn't given an equation to plug-in those values. For the trapezoid rule, I got 170/3 mi. The midpoints that I've chosen are
M(6) = 10[f(5) + f(15) + f(25) + f(35) + f(45) + f(55)]
I think I'm on the right track, but I'm not sure how to complete it after this.

 

[JeffWest]

M(6) = 10[f(5) + f(15) + f(25) + f(35) + f(45) + f(55)]
I think I'm on the right track, but I'm not sure how to complete it after this.[/quote]

Yes you are on the right track... Check out the midpoint formula.

$\displaystyle x_{mid}=\frac{x_2+x_1}{2}$

$\displaystyle f(x_{mid})=\frac{y_2+y_1}{2}$

you got your x mid point just fine. now to find your $\displaystyle f(x)$ just find the midpoints your speed values.

I will give you one to start and then you can pick it up from there.

$\displaystyle f(x_{mid})=f(5)=\frac{45+40}{2}=42.5$

 

[tkvictim]

Ah, thank you. I thought it would be the average of the two points, but I wasn't sure. =)

 

[JeffWest]

I am glad that we could help.