Another look at the Fundamental Theorem 72. Without evaluating integrals , prove that ∫ 0 2 d d x ( 12 sin π x 2 ) d x = ∫ 0 2 d d x ( x 10 ( 2 − x ) 3 ) d x .
Another look at the Fundamental Theorem 72. Without evaluating integrals , prove that ∫ 0 2 d d x ( 12 sin π x 2 ) d x = ∫ 0 2 d d x ( x 10 ( 2 − x ) 3 ) d x .
Solution Summary: The author explains how f and g have continuous derivatives on an interval, without evaluating integrals.
∫
0
2
d
d
x
(
12
sin
π
x
2
)
d
x
=
∫
0
2
d
d
x
(
x
10
(
2
−
x
)
3
)
d
x
.
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
I need help in construct a matlab code to find the voltage, the currents, and the watts based on that circuit.
Objective
Implement Bottom-Up Iterative MergeSort and analyze its efficiency compared to
recursive MergeSort. Unlike the recursive approach, which involves multiple function
calls and stack overhead, the bottom-up version sorts iteratively by merging small
subarrays first, reducing recursion depth and improving performance.
Task
1. Implement Bottom-Up Iterative MergeSort
о Start with single-element subarrays and iteratively merge them into larger
sorted sections.
Use a loop-based merging process instead of recursion.
○ Implement an efficient in-place merging strategy if possible.
2. Performance Analysis
Compare execution time with recursive MergeSort on random, nearly
sorted, and reversed datasets.
○ Measure and plot time complexity vs. input size.
O
Submission
Explain why the iterative version reduces function call overhead and when
it performs better.
•
Code implementation with comments.
•
A short report (1-2 pages) comparing performance.
•
Graph of execution time vs. input size for…
Given a shared data set, we allow multiple readers to read at the same time, and only one single writer can access the shared data at the same time. In the lecture slides, a solution is given. However, the problem is that the write cannot write forever, if there are always at least one reader.
How to ensure that the writer can eventually write? Propose your solution by using semaphores and implemented in Python
from threading import Thread, Semaphore
from time import sleep
from sys import stdout
class Reader(Thread):
def__init__(self, name):
self.n=name; Thread.__init__(self)
defrun(self):
globalnr, nw, dr, dw
whileTrue:
# ⟨await nw == 0 then nr += 1⟩
e.acquire()
ifnw>0:
#if nw > 0 or dw > 0 :
dr+=1; e.release(); r.acquire()
nr+=1
ifdr>0: dr-=1; r.release()
else: e.release()
# read data
stdout.write(self.n+' reading\n')
sleep(1)
# ⟨nr -= 1⟩
e.acquire()
nr-=1
ifnr==0anddw>0:
dw-=1 ; w.release()
else: e.release()
class Writer(Thread):
def__init__(self, name):…
Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)
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