# Integration Problems

1. Dec 5, 2004

### Astronomer107

Integration Problems!!

Help! I have a problem with the following:
1. Sketch the graph of the function y^2 = x^4(x+4) and find the area enclosed by the loop. -- I don't see any loop!

2. Find the area of the region enclosed by th curves y= x^2 and y^2=8x. -- The answer given in the back is 8/3, but I got 120/3. I don't know if the book is wrong or I am wrong.

3. Find the integral of Sin^-1 x (by dx) with the boundaries 0 and 1.-- I tried substitution and that didn't work. I need to get cos x in the numerator, but I couldn't figure out how.

4. Find the area of the region bounded by the curve y = 1 - (2/x), the x-axis and the line x=1.-- I got 1 as my answer, but the back of the book said it was 1-2ln2. The graph was a little odd as far as how the boundaries worked was concerned.

Any help would be GREAT! Thanks!

2. Dec 5, 2004

### dextercioby

1.Try Maple to show you the graphics.There has to be a loop since the problem tells you so and in addition,as a clue,the function is multiform.The integral should not be difficult.You can make the substitution $x+4\rightarrow t^2$
2.The intersections are at x=0 and x=2.Evaluate the integral properly knowing that on this interval the second parabola is on top of the first and you can take the positive sign when expressing "y" as function of "x" for the second parabola.
$$\int_{0}^{1} \frac{dx}{\sin x}$$...It's a messy integral,but it works the result exactly:
Hint:
$$\int_{0}^{1} \frac{dx}{\sin x} = \int_{0}^{1} \frac{\sin x dx}{\sin^{2} x} =\int_{1}^{0} \frac{d(\cos x)}{1-cos^{2}x}$$.I'll let u take it from here.Be careful with the substitutions and the integration limits.
4.Be careful,since the the area is under the "x" axis,the result should be negative.The book's answer seems correct (it should involve logarithms and be less than zero) and yours is wrong.

Good luck!!

Last edited: Dec 5, 2004
3. Dec 5, 2004

### Astronomer107

Thanks for your help! On the 2nd one, I still got the same answer, so the book is probably wrong (it frequently is) or I still am incorrect. Thanks for reminding me that sin^2x = 1-cos^2x. I'm still working on getting an answer for that one though. For the last one, I don't know how to find the lower boundary. I know the upper boundary is 1, but is the lower boundary 0 or does it extend beyond the y-axis? Thanks!

4. Dec 5, 2004

### dextercioby

Some books are,other aren't.Your result is definitely wrong.It should be less than the area of the rectangle between 0 and 2 and with the height along "y" axis of 4,which is obviously 8.
8/3<8.
For the last one,the upper boundary is 2 (i.e.the intersection between the graphic and "x" axis) and the lower is 1 (the vertical bar x=1 which appears in the text).

Better luck this time!!

5. Dec 5, 2004

### Astronomer107

ok... here's what I'm doing.

the integral of 8x^(1/2) - x^2) by dx, boundaried 0 and 2, which is:

2/3 * 8x^(3/2) - 1/3 x^3. When I plug in the 2, I still get 120/3, which is wrong.

On the Sin^-1 x problem, I stopped at sin x/ (1-cos^2 x) because I couldn't figure out how to cancel out 1- cos x to make it so that f ' (x) / f (x) would work. I am unfamiliar with the d(cos x) that you used.

I did manage to figure out the last one, thank you! I'm sorry to keep asking questions, but I really want to understand.

6. Dec 5, 2004

### dextercioby

The first one is ridiculously simple:
$$\int_{0}^{2} (\sqrt{8}\sqrt{x}-x^2) dx=\frac{2}{3} \sqrt{8} x^{\frac{3}{2}} \mid_{0}^{2} -\frac{x^3}{3}\mid_{0}^{2} = \frac{16}{3}-\frac{8}{3}=\frac{8}{3}$$.

For the second,make the substitution $\cos x\rightarrow u$,transform the integration limits properly and you'll get something about
$$\int_{\cos{1}}^{1}\frac{du}{1-u^2}$$ .The rest is obvious,the integral is given in a reference table,if u don't kow know to calculate it.Calculating it means expressing the integrand in terms of simpler fractions,using the identity $(1-u^2)=(1-u)(1+u)$.

Last edited: Dec 5, 2004