A few integral questions i need help with!

  • #1
Evaluate the integrals:

(integral sign) {2x^3 - 7x^2 + 3x^1/2 -x}dx
So when i evaluated this integral, i got:

1/2X^4 - 7/3X^3 + 2*(3)^1/2 / 3 X^3/2 - X^2 +C

oookay, so that looks a bit confusing...so if you don't understand that, can someone post me the answer to this question?

My other question is:

Solve the following differential equation:

dy/dx = sinx-sec^2x

For this question, i got
-cosx-tanx+C
is that one correct?

Lastly, i would like to know if i got this answer right..
A pebble is tossed 6m/s down a cliff 72m high. how long will it take to hit the ground?
my final answer was 3.27 seconds..

any help would work you guys! thanks so much!
 

Answers and Replies

  • #2
VietDao29
Homework Helper
1,424
3
Uhm, number 1 is wrong.
2 and 3 look fine to me.
Viet Dao,
 
  • #3
so what's the right answer for number 1..? i dont know how else to do it..
please help!
 
  • #4
551
2
laker_gurl3 said:
so what's the right answer for number 1..? i dont know how else to do it..
please help!

Look back at your answer and see. Some of the terms are wrong. Edit: the first two terms are ok.
 
Last edited:
  • #5
okay, so how do i find the integral of square root 3X? cuz i have a feeling that's the wrong term in my answer..i dont know how to go about finding it..
 
  • #6
551
2
laker_gurl3 said:
okay, so how do i find the integral of square root 3X? cuz i have a feeling that's the wrong term in my answer..i dont know how to go about finding it..

The way you had it written wasn't clear, it looked like it was [itex]3x^\frac{1}{2}[/itex]. If you did mean [itex](3x)^\frac{1}{2}[/itex] then that term is right too.The last one is still wrong though.
 
  • #7
ooh okay, is the last term - (1/2)X^2
 
  • #8
551
2
Yes, it is.
 
  • #9
okay thanks alot! another question i had was

Solve the following differential equation:
dy/dx = sec(y)sin(x) y= (pi/2) when x = pi

i changed the sec(y) to 1/(cos(y)) then multiplied to the dy side so this is what it looks like now

sec(y) dy = sin(x) dx + C

then i antideriv of the both side

sin(y) dy = -cos(x) + C

subbed in the pi/2 and pi then i got:

C = 0

Is that a good enough answer or do i have to write it in a different form?
thanks a bunch!
 
  • #10
bump... thanks a lot you guys
 
  • #11
siddharth
Homework Helper
Gold Member
1,130
0
laker_gurl3 said:
okay thanks alot! another question i had was

Solve the following differential equation:
dy/dx = sec(y)sin(x) y= (pi/2) when x = pi

i changed the sec(y) to 1/(cos(y)) then multiplied to the dy side so this is what it looks like now

sec(y) dy = sin(x) dx + C

How did you get that equation.
Shouldn't it be
cos(y)dy=sin(x)dx?

laker_gurl3 said:
then i antideriv of the both side

sin(y) dy = -cos(x) + C

What's the 'dy' term doing after you find the anti-derivative on both sides?
 

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