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Final

MA 141 Lecture 22: Definite IntegralExam


Department
Mathematics
Course Code
MA 141
Professor
Kevin Flores
Study Guide
Final

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MA 141 Chapter 4
Section 4.2: Definite Integral
Now we will consider functions that may be above or below the x-axis. We will consider
rectangles above the x-axis to have positive signed area and rectangles below the x-axis to
have negative signed area. The Riemann sum will then calculate the net area.
Def: Let f(x)be defined and continuous on the interval [a, b]. For each partition a=x0<
x1< x2< . . . < xn1< xn=bof the interval [a, b]into nequal parts, the length of each
subinterval is x=ba
nand each xi=a+ix,i= 1,2, . . . , n. The definite integral of
fon [a, b]is denoted and defined by
Zb
a
f(x)dx = lim
n→∞
n
X
i=1
f(xi)x
provided this limit exists. If the limit exists, we call fintegrable on the interval [a, b].
Here, aand bare called the limits of integration. The term f(x)dx is a differential as in
Section 3.6.
Theorem: If f(x)is a continuous function on [a, b], then Zb
a
f(x)dx exists.
Ex: Evaluate the definite integral Z2
3
(2x+ 6)dx using geometry.
Let’s draw this function over the interval [3,2]:
= =
This region is a triangle with base b= 5
and height h=f(2) = 2(2) + 6 = 10.
Then the area of the shaded region is
A=1
2(5)(10) = 25
Since the definite integral Z2
3
(2x+ 6)dx =the area between the "curve" and the x-axis, we
know
Z2
3
(2x+ 6)dx = 25
1
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Ex: Evaluate the definite integral Z3
0
9x2dx using geometry.
Let’s graph again - what shape does 9x2have? This is the upper half of a circle of
radius 3, centered at the origin.
The full circle would have area π·32= 9π.
Since we are only concerned with the
quarter of the circle on the right, the area
is
Z3
0
9x2dx =1
4·9π=9π
4.
Ex: Evaluate the Z2
3
(2x+ 6)dx using the limit definition of the definite integral.
In the previous example, we showed this definite integral to be equal to 25 using geometry.
Let’s verify using the limit definition of definite integral.
Z2
3
(2x+ 6)dx = lim
n→∞
n
X
i=1
f(xi)x= lim
n→∞
n
X
i=1
(2xi+ 6)x
We need to find xiand x. Since the interval is [3,2],x=2(3)
n=5
n. Recall that
xi=a+ix=3 + 5i
n.
lim
n→∞
n
X
i=1
(2xi+ 6)x= lim
n→∞
n
X
i=1 23 + 5i
n+ 6·5
n
= lim
n→∞
n
X
i=1 6 + 10i
n+ 6·5
n
= lim
n→∞
n
X
i=1 50i
n2
= lim
n→∞
50
n2
n
X
i=1
i
= lim
n→∞
50
n2·n(n+ 1)
2
= lim
n→∞
50n+ 50
2n
= 25
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