How can I evaluate the improper integral of these problems? 1. y = 1/2sqrt(3-x), x = 3, coordinate axes 2. y = 1/(3-x)2, x = 3, coordinate axes
How can I evaluate the improper
1. y = 1/2sqrt(3-x), x = 3, coordinate axes
2. y = 1/(3-x)2, x = 3, coordinate axes
To evaluate the improper integrals for these problems, you need to calculate definite integrals over certain intervals where the functions may have singularities or other issues. Let's evaluate each of these integrals
(1). ∫(1 / (2√(3 - x))) dx from x = 3 to the coordinate axes:
The integral has a singularity at x = 3, so we need to split it into two parts and evaluate each separately.
First, consider the integral from x = 0 to x = 3:
∫(1 / (2√(3 - x))) dx from 0 to 3
To solve this integral, perform a substitution:
Let u = 3 - x, then du = -dx
When x = 0, u = 3, and when x = 3, u = 0. The integral becomes:
-∫(1 / (2√u)) du from 3 to 0
Now integrate:
-∫(1 / (2√u)) du = -√u from 3 to 0 = -√0 - (-√3) = √3
Now, let's consider the integral from x = 3 to x = 0:
∫(1 / (2√(3 - x))) dx from 3 to 0
You can see that this integral is the negative of the previous one because we are integrating in the opposite direction. So, this integral is also -√3.
Therefore, the improper integral ∫(1 / (2√(3 - x))) dx from x = 3 to the coordinate axes equals 2√3.
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