MATH241 Lecture Notes - Lecture 29: Antiderivative
25 May 2018
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MATH241 - Lecture 29 - Antidifferentiation, Areas and Reimann Sums, and The Definite
Integral
4.9: Antidifferentiation (conclusion)
Rectilinear Motion (continued)
Example:
A ball is thrown upward with a speed of 48 ft/sec from the edge of a building 160 ft above the
ground
A) If the gravitational acceleration of the object is −32 𝑓𝑡/𝑠𝑒𝑐2, find its height 𝑡 seconds
later
B) When does the ball hit the ground?
Solution:
A) 𝑎(𝑡)= −32
𝑉(𝑡) = −32𝑡+6
𝑉(0)=48
→ 𝑐 = 48 thus 𝑉(𝑡)= −32𝑡+48
𝑠(𝑡)= −16𝑡2+48𝑡+𝐷
𝑠(0)=160
→ 𝐷 = 160 thus 𝑠(𝑡)= −16𝑡2+48𝑡+160
B) 𝑠(𝑡)=0
Solve for 𝑡
−16𝑡2+48𝑡+160 =0
−16(𝑡2−3𝑡−10)=0
−16(𝑡−𝑠)(𝑡+2)=0 thus 𝑡 = −2 or 𝑡 = 𝑠
Thus 𝑡 = 5 seconds
Example:
Find the position of a particle if its acceleration is 𝑎(𝑡)=6𝑡+4
Initial velocity is 𝑉(0)= −6 and initial position is 𝑠(0)=8
Solution:
𝑠′′(𝑡)=6𝑡+4
𝑠′(𝑡)=3𝑡2+4𝑡+𝑐
𝑠′(0)= −6 → 𝑐 = −6
Thus 𝑠′(𝑡)=3𝑡2+4𝑡 −6
𝑠(𝑡)= 𝑡2+2𝑡2−6𝑡 +𝐷
𝑠(0)=8 → 𝐷 = 8
Thus 𝑠(𝑡)= 𝑡3+2𝑡2−6𝑡 +8
5.1: Areas and Reimann Sums
Find the area of the region that lies under the curve from 𝑥 = 𝑎 to 𝑥 = 𝑏
We can estimate the area under the curve by adding the area of 𝑛 rectangles
The height of each rectangle is 𝑓(𝑥𝑖), the width of each rectangle is 𝛥𝑥 = 𝑏 − 𝑎
𝑛
Thus the area is ≃∑𝑛
𝑖 = 1𝑓(𝑥𝑖)𝛥𝑥
Example:
A) Estimate the area under the curve 𝑦 = 10 −2𝑥 between 𝑥 = 0 and 𝑥 = 𝑠 using 5
rectangles
B) Estimate the same area using 10 rectangles
C) Find the exact area under the curve
Solution:
A) 𝛥𝑥 = 5 − 0
5=1
Let 𝑥𝑖 = 𝑖𝛥𝑥
Then 𝑥1=1 𝑓(𝑥1)=10 −2=8
𝑥2=2 𝑓(𝑥2)=10 −4=6
𝑥3=3 𝑓(𝑥3)=4
