Integrating 1/x^3: -2/x^2 or -1/2x^2?

In summary, the general process for integrating 1/x^3 is to use the power rule, which gives the integral as -1/2x^2 + C. Similarly, the power rule can be used to integrate -2/x^2 and -1/2x^2, which give integrals of 2/x + C and 1/6x^3 + C, respectively. The negative sign in front of the integral for 1/x^3 is due to the integration process, where the exponent of x is increased by 1 and the coefficient becomes the reciprocal of the new exponent. The constant of integration, denoted as C, can be any real number and is added at the end of the integration process
  • #1
angel
18
0
Hi,

ive got a really simple question, but I am just slightly confused,
im trying to integrate 1/x^3 , now i take x^3 to the top and i get x^-3, and now when i intergrate that i get:

x^-2/-2 (i think)
now when i put it back into fractional form do i get:

-2/x^2 or -1/2x^2 ?
 
Physics news on Phys.org
  • #2
(x^-2)/-2 = -1/(2x^2)
 
  • #3


Thanks for your question! The correct answer is -1/2x^2. This is because when you integrate x^-3, you bring the power up by 1 to get x^-2 and then divide by the new exponent, which is -2. This results in -1/2x^2. Hope this helps!
 

1. What is the general process for integrating 1/x^3?

The general process for integrating 1/x^3 is to use the power rule, which states that the integral of x^n is equal to (x^(n+1))/(n+1) + C. In the case of 1/x^3, the integral would be (x^(-2))/-2 + C, which simplifies to -1/2x^2 + C.

2. Can I use the power rule to integrate -2/x^2?

Yes, the power rule can be used to integrate -2/x^2. The integral would be (-2x^(-1))/(-1) + C, which simplifies to 2/x + C.

3. Why is there a negative sign in front of the integral for 1/x^3?

The negative sign in front of the integral for 1/x^3 is a result of the integration process. When using the power rule, the exponent of x is increased by 1 and the coefficient becomes the reciprocal of the new exponent. In this case, the original exponent was -2, so the new exponent becomes -1 and the coefficient becomes -1/2.

4. Can I use the power rule to integrate -1/2x^2?

Yes, the power rule can be used to integrate -1/2x^2. The integral would be (-1/2x^3)/(-3) + C, which simplifies to 1/6x^3 + C.

5. How do I know which constant of integration to use?

The constant of integration, denoted as C, can be any real number. It is added at the end of the integration process to account for any unknown constant values that may have been lost during differentiation. To determine the specific value of C, you would need to know the initial condition or boundary conditions of the problem.

Similar threads

I Negative area above x-axis from integrating x^2?
    • Calculus
Replies
4
Views
1K
MHB Integrate 3/(x(2-sqrt(x))) - No Partial Fractions
    • Calculus
Replies
5
Views
1K
B Question about the limits of integration in a change of variables
    • Calculus
Replies
2
Views
252
B Calculating shaded area: I'm getting discrepancies between methods
    • Calculus
Replies
3
Views
289
B I need to check if I am right solving this integral
    • Calculus
Replies
2
Views
995
MHB Finding k for x is undefined at $x=2$
    • Calculus
Replies
3
Views
733
I Substitution in a definite integral
    • Calculus
Replies
6
Views
1K
A Non solvable integral? (dx/dt)^2 dt
    • Calculus
Replies
3
Views
2K
I How Do You Integrate 1/√(x^3 + x^2 + x + 1) dx?
    • Calculus
Replies
8
Views
1K
B Having trouble deriving the volume of an elliptical ring toroid
    • Calculus
Replies
4
Views
301

Hot Threads

Recent Insights

Back
Top