Improper Integrals - Integrating a Discontinuous Integrand. In this problem our goal is to determine whether the improper integral below converges or diverges. If it converges, we will determine the value of the imprope integral. 1 da 7+3z – 40 Part 1. 1 Begin by finding the area under the curve y 1² + 3x - from z =t to a = 9, t > 5. This area can be written as the definite integral 40 A(t) Upon evaluating the definite integral you found above, we have A(t) – t>5 Part 2. Now we will find the limit of A(t) as t → 5+. lim A(t) Part 3. Based on your answer for Part 2. above, the improper integral either converges to a finite value or diverges. If the integral converges, state that it converges ar what value it converges. Otherwise, state that it diverges to infinity. The integral to

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Improper Integrals - Integrating a Discontinuous Integrand.
In this problem our goal is to determine whether the improper integral below converges or diverges. If it converges, we will determine the value of the improper
integral.
1
dz
40
Part 1.
1
Begin by finding the area under the curve y =
12 + 3x
from z =t to r = 9, t> 5. This area can be written as the definite integral
40
A(t) =
Upon evaluating the definite integral you found above, we have
A(t) =t>:
Part 2.
Now we will find the limit of A(t) as t → 5+.
lim A(t)
t-5+
Part 3.
Based on your answer for Part 2. above, the improper integral either converges to a finite value or diverges. If the integral converges, state that it converges and to
what value it converges. Otherwise, state that it diverges to infinity.
The integral
to
Transcribed Image Text:Improper Integrals - Integrating a Discontinuous Integrand. In this problem our goal is to determine whether the improper integral below converges or diverges. If it converges, we will determine the value of the improper integral. 1 dz 40 Part 1. 1 Begin by finding the area under the curve y = 12 + 3x from z =t to r = 9, t> 5. This area can be written as the definite integral 40 A(t) = Upon evaluating the definite integral you found above, we have A(t) =t>: Part 2. Now we will find the limit of A(t) as t → 5+. lim A(t) t-5+ Part 3. Based on your answer for Part 2. above, the improper integral either converges to a finite value or diverges. If the integral converges, state that it converges and to what value it converges. Otherwise, state that it diverges to infinity. The integral to
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