Intergral from -2 to 2 (-2 is at the bottom of the integral sign and 2 is at the top) x^-2dx = -x^-1] (the bracket has a 2 at the top and -2 at the bottom) = -(2)^-1-(-(-2)^-1) = (-1/2)-(1/2) = -1
Is this true or false. I think it's false, but can someone explain to me why it's false.
The question is done correctly.
When you differentiate
-x^-1 you get x^-2
and all the arithmetic was done correctly.
If this an "area between curves" type, I have seen this before.
The function f(x) = 1/x^2 is discontinuous at x=0
so even though calculation-wise we come up with the answer above, it would be meaningless.
The answer is wrong because function x^(-2)>0 on [-2,2] except x=0 =>integral>0
This is Improper integral which is divergent.
x^(-1)-->infinity if x-->0
To determine if the statement is true or false, we need to evaluate the given integral.
The integral in question is ∫[from -2 to 2] x^(-2) dx.
To solve this integral, we use the power rule for integration, which states that the integral of x^n dx is equal to (x^(n+1))/(n+1), for any real number n (except -1).
Applying the power rule to the integral, we get:
∫[from -2 to 2] x^(-2) dx = [x^(-2+1)]/(-2+1) + C = -x^(-1).
However, it seems that there's an error in the provided solution steps. Let's go through the calculation again:
- Evaluating -x^(-1) from 2 to -2:
-Finding the value at x=2: -(2)^(-1) = -1/2.
-Finding the value at x=-2: -( (-2)^(-1) ) = -( -1/2 ) = 1/2.
Therefore, the correct evaluation of the integral is:
∫[from -2 to 2] x^(-2) dx = (-1/2) - (1/2) = -1/2 - 1/2 = -1.
So, the given statement is true, and the integral evaluates to -1.