Help on some calculus problems!!!

[nicotrial]

Hi guys im new arround here in i am in serach of some urgent help.. i have an exame in a few days and still have many things i dont know how to solve yet..
In the attached file are some problems i could not solve....
if posible could some one solve these for me (step by step) so i can go through them real fast and learn how they are solved..

Thank you...and dont worry i wont dissapear i will also try to help others (and post many more questions) after i finish my exams..

 

[tkhunny]

This looks like the actual exam. You had better get started! It may help if you suggest exactly what course you are in.

 

[nicotrial]

This looks like the actual exam. You had better get started! It may help if you suggest exactly what course you are in.

i know im trying!!!

no its different exersices from different previous exams from different years..
im an engineer student in first year university..

 

[mmm4444bot]

solve these for me (step by step) so i can go through them real fast and learn how they are solved

I'm pretty sure that's not how engineers learn. :wink:

Do you know how to integrate? That is, for example, are you able to determine the following antiderivatives?

\(\displaystyle \displaystyle \int \dfrac{4x}{x^2+1} dx\)

\(\displaystyle \displaystyle \int \dfrac{1}{x^2} dx\)

\(\displaystyle \displaystyle \int \dfrac{1}{x} dx\)

\(\displaystyle \displaystyle \int \dfrac{3}{x-1} dx\)

Show me what you get, and I'll have a suggestion for your first exercise.

Also, please check out the summary of posting guidelines page. Cheers :cool:

 

[nicotrial]

I'm pretty sure that's not how engineers learn. :wink:

Do you know how to integrate? That is, for example, are you able to determine the following antiderivatives?

\(\displaystyle \displaystyle \int \dfrac{4x}{x^2+1}\)

\(\displaystyle \displaystyle \int \dfrac{1}{x^2}\)

\(\displaystyle \displaystyle \int \dfrac{1}{x}\)

\(\displaystyle \displaystyle \int \dfrac{3}{x-1}\)

Show me what you get, and I'll have a suggestion for your first exercise.

Also, please check out the summary of posting guidelines page. Cheers :cool:

of course these are easy..
--------

i solved the one with the squared root i do not know if it is correct..

the first integral i do not know where to start am i soposed to use the method where you calculate parameters (A/(x-1)+ /(x^2+1)...) and get easy to solve integrals??? if so isint it soposed to be that the top power of the devision is sopposed to be lower then the bottom?? what do i do with the x^2..??

 

[lookagain]

Do you know how to integrate? That is, for example, are you able to determine the following antiderivatives?

\(\displaystyle \displaystyle \int \dfrac{4x}{x^2+1}\)dx

\(\displaystyle \displaystyle \int \dfrac{1}{x^2}\)dx

\(\displaystyle \displaystyle \int \dfrac{1}{x}\)dx

\(\displaystyle \displaystyle \int \dfrac{3}{x-1}\)dx

Important: Make sure the "dx" is present.

 

[nicotrial]

this is the first integral this is how i think it is solved but i am not sure if this s correct..

 

[DrPhil]

this is the first integral this is how i think it is solved but i am not sure if this s correct..

When the denominator includes an irreducible quadratic, such as (x^2+1), then for that factor you need two terms:

\(\displaystyle \dfrac{D + Ex}{x^2 + 1}\)

There are five parameters to match the 5 terms in the numerator. Note that the degrees of the numerator (4) is less than the degree of the denominator (5), so this rational expression is a proper fraction.

\(\displaystyle \dfrac{8x^4 -6x^3 +5x^2 -2x + 1}{x^2(x-1)(x^2+1)} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x-1} + \dfrac{D + Ex}{x^2 + 1}\)

\(\displaystyle \displaystyle 8x^4 -6x^3 +5x^2 -2x + 1 = (A+Bx)(x-1)(x^2+1) + C(x^2)(x^2+1) + (D+Ex)(x^2)(x-1) \)

Find A by letting x=0. I think you missed the sign because A is multiplied by (x-1)=-1

Find C by letting x=1. I think you missed a factor of 2 because C is multiplied by (x^2+1)=2

Then you have a system of equations to solve by your favorite method for B, D, and E.

 

[nicotrial]

When the denominator includes an irreducible quadratic, such as (x^2+1), then for that factor you need two terms:

\(\displaystyle \dfrac{D + Ex}{x^2 + 1}\)

There are five parameters to match the 5 terms in the numerator. Note that the degrees of the numerator (4) is less than the degree of the denominator (5), so this rational expression is a proper fraction.

\(\displaystyle \dfrac{8x^4 -6x^3 +5x^2 -2x + 1}{x^2(x-1)(x^2+1)} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x-1} + \dfrac{D + Ex}{x^2 + 1}\)

\(\displaystyle \displaystyle 8x^4 -6x^3 +5x^2 -2x + 1 = (A+Bx)(x-1)(x^2+1) + C(x^2)(x^2+1) + (D+Ex)(x^2)(x-1) \)

Find A by letting x=0. I think you missed the sign because A is multiplied by (x-1)=-1

Find C by letting x=1. I think you missed a factor of 2 because C is multiplied by (x^2+1)=2

Then you have a system of equations to solve by your favorite method for B, D, and E.

OHHHH ok thank you very much i totaly forgot about that....

 

[DrPhil]

i solved the one with the squared root i do not know if it is correct..

Checking - looks well done, but I will see if I get the same result.

\(\displaystyle \displaystyle \int \dfrac{x^3}{\sqrt{1 - x^2}}\ dx \)

Let \(\displaystyle t = 1-x^2\)..........\(\displaystyle dt = -2x\ dx\)

The numerator is \(\displaystyle x^3\ dx = -\frac{1}{2}(1-t)\ dt \) and the integral is

\(\displaystyle \displaystyle -\frac{1}{2}\int \dfrac{1-t}{\sqrt{t}}\ dt = \frac{1}{2}\int t^{1/2} dt - \frac{1}{2}\int t^{-1/2} dt\)

...........................\(\displaystyle \displaystyle = \frac{1}{3}t^{3/2} - t^{1/2} + C\)

...........................\(\displaystyle \displaystyle = \sqrt{1-x^2}\ \left[ \frac{1}{3}(1 - x^2) - 1\right] + C\)

...........................\(\displaystyle \displaystyle = -\sqrt{1-x^2}\ \left[ \dfrac{x^2 + 2}{3}\right] + C\)

Looks like we have a sign difference - please check your work (or find where I am wrong!).
Also, I carried out another step of simplification.
Thanks for showing your work!

 

[mmm4444bot]

Make sure the "dx" is present.

Yes. (Fixed.)

Thanks.