The Meaning of Integration
Author
Thomas Rike
Title
The Meaning of Integration
Description
This lesson helps students understand the meaning of integration, including: Integration can be used to find area under a curve. The value of the integral is an area if the integrand is always nonnegative on an interval. Integrals can also have a value which is negative or zero. Integrals can be evaluated by using the geometric interpretation of the integral to simplify the calculations in some cases. The integral can be used to find the average value of a function over an integral. Symmetry can be used to simplify integrals. Integration can be used to accumulate a rate of change. An integral can be used to find the value of a function at a point. An Integral can be used to define a new function. The integral is an infinite sum of products. (Riemann Sum)
Category
Educational Materials
Keywords
URL
http://www.notebookarchive.org/2018-10-10pzec6/
DOI
https://notebookarchive.org/2018-10-10pzec6
Date Added
2018-10-02
Date Last Modified
2018-10-02
File Size
47.93 kilobytes
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Redistribution rights reserved
This notebook has not been updated since 2011. 
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The Meaning of Integration
The Meaning of Integration
Integration can be used to find area under a curve. The value of the integral is an area if the integrand is always nonnegative on an interval.
Integration can be used to find area under a curve. The value of the integral is an area if the integrand is always nonnegative on an interval.
If f(x) ≥ 0 on [ a , b ] then b∫af(x)x ≥ 0.
If f(x) ≥ 0 on [ a , b ] then f(x)x ≥ 0.
b
∫
a
Find the area of the region bounded by y = 3x and the lines x = 2 and x = 3
Find the area of the region bounded by y = and the lines x = 2 and x = 3
3
x
Integrate[x^3,{x,2,3}]
Integrals can also have a value which is negative or zero.
Integrals can also have a value which is negative or zero.
Two examples of integrals which evaluate to zero. 2∫23xx and π∫0cosxx.
Two examples of integrals which evaluate to zero. x and cosxx.
2
∫
2
3
x
π
∫
0
Integrate[x^3,{x,2,2}]
Two examples of integrals which have negative values. 1∫-23xx and 1∫22xx.
Two examples of integrals which have negative values. x and x.
1
∫
-2
3
x
1
∫
2
2
x
Integrate[x^3,{x,-2,1}]
Integrals can be evaluated by using the geometric interpretation of the integral to simplify the calculations in some cases.
Integrals can be evaluated by using the geometric interpretation of the integral to simplify the calculations in some cases.
For example, to evaluate 3∫-2|x|x, one would look at the graph of |x|.
For example, to evaluate |x|x, one would look at the graph of |x|.
3
∫
-2
Plot[{Abs[x],y=0},{x,-2,3},PlotRangeAutomatic,FillingAxis]
Now add the areas of the two isosceles right triangles to evaluate the integral.3∫-2|x|x= 12(2)(2) + 12(3)(3) =132.
Now add the areas of the two isosceles right triangles to evaluate the integral.|x|x= (2) + (3)(3) =
3
∫
-2
1
2
(2)
1
2
13
2
.
The integral can be used to find the average value of a function over an integral.
The integral can be used to find the average value of a function over an integral.
Find the average value of f(x) = xe on the interval [ 1, 3 ].
Find the average value of f(x) = on the interval [ 1, 3 ].
x
e
Integrate[E^x,{x,1,3}]/(3-1)
Symmetry can be used to simplify integrals.
Symmetry can be used to simplify integrals.
If f is an even function then a∫-af(x)x=2a∫0f(x)x.If f is an odd function then a∫-af(x)x=0.Find 1∫-12x+3xx. First note that 2xis an even function and 3x is an odd function, so that1∫-12x+3xx= 21∫02xx= 2 (1/3) = 2/3. Check by using Mathematica.
If f is an even function then f(x)x=2f(x)x.If f is an odd function then f(x)x=0.Find +x. First note that is an even function and is an odd function, so that+x= 2x= 2 (1/3) = 2/3. Check by using Mathematica.
a
∫
-a
a
∫
0
a
∫
-a
1
∫
-1
2
x
3
x
2
x
3
x
1
∫
-1
2
x
3
x
1
∫
0
2
x
Integrate[x^2+x^3,{x,-1,1}]
Integration can be used to accumulate a rate of change.
Integration can be used to accumulate a rate of change.
If water is leaking from a pool at a rate of r(t) = 100-2te gallons per hour, where t is the time in hours. How many gallons will have leaked out after 24 hours?
If water is leaking from a pool at a rate of r(t) = 100 gallons per hour, where t is the time in hours. How many gallons will have leaked out after 24 hours?
-2t
e
Integrate[100E^(-2t),{t,0,24}]
An integral can be used to find the value of a function at a point.
An integral can be used to find the value of a function at a point.
If the derivative is given as f'(x) and f(a) is known, then the Fundamental Theorem of Calculus says that f(b) - f(a) = b∫af'(x)x, so that f(b) = f(a) +b∫af'(x)x.Find the approximate value of y at x = 3 , if dydx = (Sin[x])^3 and y= 2 when x = 1.
If the derivative is given as f'(x) and f(a) is known, then the Fundamental Theorem of Calculus says that f(b) - f(a) = f'(x)x, so that f(b) = f(a) +f'(x)x.Find the approximate value of y at x = 3 , if = ([x])^3 and y= 2 when x = 1.
b
∫
a
b
∫
a
dy
dx
Sin
2+NIntegrate[(Sin[x])^3,{x,1,3},WorkingPrecision20]
An Integral can be used to define a new function.
An Integral can be used to define a new function.
One of the most notable examples is the new function, ln(x) for logarithms with base e. The definition of ln(x) is x∫11tt. Then ln(2) = 2∫11ttand ddxln(x) = ddxx∫11tt = 1x.Let g(x) =x∫0cos3tt. What is the maximum value of g on the interval [ 0, 2 ] ?Note g(0) = 0 and g'(x) = Cos[x^3]. Since Cos (x) is zero at π2, to find the zero of g'(x) take the cube root of π2
One of the most notable examples is the new function, ln(x) for logarithms with base e. The definition of ln(x) is t. Then ln(2) = tand ln(x) = t = .Let g(x) =cost. What is the maximum value of g on the interval [ 0, 2 ] ?Note g(0) = 0 and g'(x) = Cos[x^3]. Since Cos (x) is zero at , to find the zero of g'(x) take the cube root of
x
∫
1
1
t
2
∫
1
1
t
d
dx
d
dx
x
∫
1
1
t
1
x
x
∫
0
3
t
π
2
π
2
N[(Pi/2)^(1/3)]
The integral is an infinite sum of products. (Riemann Sum)
The integral is an infinite sum of products. (Riemann Sum)
Find limn∞n2n+n2(n+1)+ ⋯ + n2(2n)
Find ++ ⋯ +
lim
n∞
n
2
n
n
2
(n+1)
n
2
(2n)
■
Note that the sum can be rewritten as a product by multiplying by = nn.This gives the equivalent problem: limn∞[2nn+1+2nn+2+ ⋯ + 2n2n]1n= limn∞n∑k=1211+kn1n= 2∫112xx
Note that the sum can be rewritten as a product by multiplying by = .This gives the equivalent problem: [++ ⋯ + ]= = x
n
n
lim
n∞
2
n
n+1
2
n
n+2
2
n
2n
1
n
lim
n∞
n
∑
k=1
2
1
1+
k
n
1
n
2
∫
1
1
2
x
Integrate[1/x^2,{x,1,2}]
Exercises
▼
1. Find the area of the region bounded by x = 1, x = 2, f(x) = and the x-axis.
1
3
x
▼
2. Evaluate the following: a) (x-2)x b) (x-2)x c) cosxx d) sinxx
3
∫
1
0
∫
3
2π
∫
0
0
∫
π
▼
3. Use the geometric interpretation of the integral to evaluate |x-2|x.
3
∫
-1
▼
4. Use the geometric interpretation of the integral to evaluate
2
∫
0
4-
x.2
x
▼
5. What is the average value of f(x) = x + sin x on the interval [ 0, π]?
▼
6. Use symmetry to evaluate (sinx+x+-cosx+3)x.
π
∫
-π
3
x
▼
7. If the velocity of an object is v(t) = -16t + 56 and the object is 5 feet off the ground at t = 0, then how high will the object be seven seconds from now?
▼
8. Wine is leaking from a vat at the rate of R(t) = 1000 gallons per hour, where t is measured in hours. How much wine has leaked out of the vat after 12 hours?
-0.1t
e
▼
9. If f(0) = 1 and f'(x) = tan(), then what is f(1)?
3
x
▼
10. If g(x) = sin()t on the closed interval [ 0, 3 ], then for what value of x does g have a local minimum and for what value of x does g have a local maximum? Plot the graph to verify your results.
x
∫
0
2
t
▼
11. Find [++⋯+]. Hint: Divide by and factor out the to form a Riemann sum.
lim
n∞
2
n
3
(n+1)
2
n
3
(n+2)
2
n
3
(2n)
3
n
1
n
Cite this as: Thomas Rike, "The Meaning of Integration" from the Notebook Archive (2011), https://notebookarchive.org/2018-10-10pzec6
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