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1. ∫(3x^2-1/2)dx

2. ∫(2-3x+4x^3)dx

3. ∫(xy+2x)dx

4. ∫(2-4/x^3+1/x^2)dx

5. ∫(1-x-x^2-2/x^2)dx

 Jan 21, 2016

Best Answer 

 #1
avatar
+5

1-

 

Solve for x over the real numbers:
3 x^2-1/2 = 0

Add 1/2 to both sides:
3 x^2 = 1/2

Divide both sides by 3:
x^2 = 1/6

Take the square root of both sides:
Answer: | x = 1/sqrt(6) or x = -1/sqrt(6)

 

2-

 

Take the integral:
integral (4 x^3-3 x+2) dx
Integrate the sum term by term and factor out constants:
  =  4 integral x^3  dx-3 integral x dx+2 integral 1 dx
The integral of x^3 is x^4/4:
  =  x^4-3 integral x dx+2 integral 1 dx
The integral of x is x^2/2:
  =  -(3 x^2)/2+x^4+2 integral 1 dx
The integral of 1 is x:
Answer: | = x^4-(3 x^2)/2+2 x+constant

 

3-

Take the integral:
integral (x y+2 x) dx
Integrate the sum term by term and factor out constants:
  =  (y+2) integral x dx
The integral of x is x^2/2:
Answer: | =  1/2 x^2 (y+2)+constant

 

4-

Take the integral:
integral (-4/x^3+1/x^2+2) dx
Integrate the sum term by term and factor out constants:
  =   integral 1/x^2 dx-4 integral 1/x^3 dx+2 integral 1 dx
The integral of 1/x^2 is -1/x:
  =  -1/x-4 integral 1/x^3 dx+2 integral 1 dx
The integral of 1/x^3 is -1/(2 x^2):
  =  2/x^2-1/x+2 integral 1 dx
The integral of 1 is x:
  =  2/x^2+2 x-1/x+constant
Which is equal to:
Answer: | =  (2 x^3-x+2)/x^2+constant

 

5-

Take the integral:
integral (-x^2-2/x^2-x+1) dx
Integrate the sum term by term and factor out constants:
  =  - integral x^2  dx- integral x dx-2 integral 1/x^2 dx+ integral 1 dx
The integral of x^2 is x^3/3:
  =  -x^3/3- integral x dx-2 integral 1/x^2 dx+ integral 1 dx
The integral of x is x^2/2:
  =  -x^2/2-x^3/3-2 integral 1/x^2 dx+ integral 1 dx
The integral of 1/x^2 is -1/x:
  =  2/x-x^2/2-x^3/3+ integral 1 dx
The integral of 1 is x:
Answer: | =  -x^3/3-x^2/2+x+2/x+constant

 Jan 21, 2016
 #1
avatar
+5
Best Answer

1-

 

Solve for x over the real numbers:
3 x^2-1/2 = 0

Add 1/2 to both sides:
3 x^2 = 1/2

Divide both sides by 3:
x^2 = 1/6

Take the square root of both sides:
Answer: | x = 1/sqrt(6) or x = -1/sqrt(6)

 

2-

 

Take the integral:
integral (4 x^3-3 x+2) dx
Integrate the sum term by term and factor out constants:
  =  4 integral x^3  dx-3 integral x dx+2 integral 1 dx
The integral of x^3 is x^4/4:
  =  x^4-3 integral x dx+2 integral 1 dx
The integral of x is x^2/2:
  =  -(3 x^2)/2+x^4+2 integral 1 dx
The integral of 1 is x:
Answer: | = x^4-(3 x^2)/2+2 x+constant

 

3-

Take the integral:
integral (x y+2 x) dx
Integrate the sum term by term and factor out constants:
  =  (y+2) integral x dx
The integral of x is x^2/2:
Answer: | =  1/2 x^2 (y+2)+constant

 

4-

Take the integral:
integral (-4/x^3+1/x^2+2) dx
Integrate the sum term by term and factor out constants:
  =   integral 1/x^2 dx-4 integral 1/x^3 dx+2 integral 1 dx
The integral of 1/x^2 is -1/x:
  =  -1/x-4 integral 1/x^3 dx+2 integral 1 dx
The integral of 1/x^3 is -1/(2 x^2):
  =  2/x^2-1/x+2 integral 1 dx
The integral of 1 is x:
  =  2/x^2+2 x-1/x+constant
Which is equal to:
Answer: | =  (2 x^3-x+2)/x^2+constant

 

5-

Take the integral:
integral (-x^2-2/x^2-x+1) dx
Integrate the sum term by term and factor out constants:
  =  - integral x^2  dx- integral x dx-2 integral 1/x^2 dx+ integral 1 dx
The integral of x^2 is x^3/3:
  =  -x^3/3- integral x dx-2 integral 1/x^2 dx+ integral 1 dx
The integral of x is x^2/2:
  =  -x^2/2-x^3/3-2 integral 1/x^2 dx+ integral 1 dx
The integral of 1/x^2 is -1/x:
  =  2/x-x^2/2-x^3/3+ integral 1 dx
The integral of 1 is x:
Answer: | =  -x^3/3-x^2/2+x+2/x+constant

Guest Jan 21, 2016
 #2
avatar+8581 
0

Great Job, Jc! ( if that is you)!!

* Gives a medal* Nice Job :)

 Jan 21, 2016
 #3
avatar
+5

1-

 

Take the integral:
integral (3 x^2-1/2) dx
Integrate the sum term by term and factor out constants:
  =  3 integral x^2  dx-1/2 integral 1 dx
The integral of x^2 is x^3/3:
  =  x^3-1/2 integral 1 dx
The integral of 1 is x:
Answer: | =  x^3-x/2+constant

 

SORRY FOR ANSWER 1 ABOVE. I DIDN'T SEE THE INTEGRAL SIGN!!.

 Jan 21, 2016
 #4
avatar+8581 
0

it's okay, we all make mistakes. :)

 Jan 21, 2016

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