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**preview**shows page 1. to view the full**5 pages of the document.**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< . . . < xn−1< xn=bof the interval [a, b]into nequal parts, the length of each

subinterval is △x=b−a

nand each xi=a+i△x,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

√9−x2dx using geometry.

Let’s graph again - what shape does √9−x2have? 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

√9−x2dx =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+i△x=−3 + 5i

n.

lim

n→∞

n

X

i=1

(2xi+ 6)△x= lim

n→∞

n

X

i=1 2−3 + 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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