UNDERSTAND THE DEFINITION OF RIEMANN SUM

EVALUATE A DEFINITE INTEGRAL USING LIMITS

EVALUATE A DEFINITE INTEGRAL USING PROPERTIES OF DEFINITE INTEGRALS

4.3a Riemann Sums and Definite Integrals

So far our rectangles have all been the same width. This was only done for computational convenience. The following example shows that it is not necessary to have subintervals of equal width.

http://youtu.be/W2309ZF5MYg Exact Area under a Curve

Georg Friedrich Bernhard Riemann (1826-1866) was a German mathematician attributed with the physics and mathematics that formed the structure on which Einstein’s Theory of General Relativity is based. Definite integrals had been used long before Riemann, but he generalized the concepts to cover a broader range of functions

Finding area under a curve is just one of many applications of integration. They are also used to find arc lengths, average values, centroids, volumes, work, accumulated distance, accumulated sales, and surface area.

The sums we found in 4.2 are some examples of Riemann sums, but there are more general Riemann sums than those we have already seen.∆ is what we are calling one of these partitions, and is ∆∆s is the norm which is the widest of all these partitions.If all the partitions are the same size, then we are back to f ∆∆f = ∆x = (b-a)/n As As∆∆s approaches zero, that forces n to approach infinity.

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://youtu

.be/fjJYR

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Notice that the definite integral looks a lot like the indefinite integral (antiderivative) that we saw in 4.1. We will see the connection with the Fundamental Theorem of Calculus, but for now, just know that the definite integral represents a NUMBER,

while the indefinite integral represents a family of functions!

For a function to be integrable on [a, b], it must be continuous on [a, b]. This is shown with the next theorem.

f x dxa

b

( )zf x dx( )z

22

1

x dx −zEvaluate the definite integral

22

1

01

x dx f c xii

n

i −

→=

z ∑= lim ( )∆

∆ =→∞

=∑lim ( )

ni

i

n

if c x1

∆

Let ci be chosen to come from right endpoint, c a i xi = + ∆Let Δxi =Δx = (b-a)/n = 3/n

ci = -2 + 3i/n

lim ( )n

i

n

fi

n n→∞=

− +∑ 23 3

1

f(x) = 2x

= − +FHG IKJ→∞=∑lim

ni

n i

n n2 2

3 3

1

= − +FHG IKJ→∞=∑lim

ni

n

n

i

n

62

3

1= − + +FHG IKJ→∞

lim( )

n nnn

n n62

3 1

2

= − + +LNM

OQP→∞

limn

n n

n12

18 18

2

2

2 = − + = −12 9 3

Since this is negative, it does not represent an area. For it to be area, the function must be continuous and nonnegative in the interval.

Ex 2 p. 274

We will learn in the next section some techniques to find definite integrals. For now you will find them by applying the limit definition or figuring them out by using geometry.

Ex 3 p. 275

a dx. 51

3z c dx. 9 - x 2

−z3

3

A bh= = ⋅ =2 5 10A b b h= +1

2 1 2( )

A = + =12 3 5 2 8( )

A r= 12

2π = =12 3 9

22π π

b dx. ( x + 2) 1

3z

a dx. 9 - x 2

3

3zEx 4 p. 276 Evaluating Definite Integrals

= 0

b dx. ( x + 2) 3

1z = − = −z x + 2) ( dx1

3

8

Ex 5 p276 Using the Additive Interval Property

x dx−z2

2

= +−z z-x x dx dx2

0

0

2

= ⋅ + ⋅12 2 2 1

2 2 2 = 4

Ex 6 p. 277 Find given: -x2 + −z 4 31

3

x dxc h x 2 dx1

3

263z = x dx

1

3

4z = dx1

3

2z =

= − + + −z z zx dx x dx dx2

1

3

1

3

1

3

4 3c h bg bg -x2 + −z 4 31

3

x dxc h= − + ⋅ + − ⋅ =26

3 4 4 3 2 43

= − + + −z z zx dx x dx dx2

1

3

1

3

1

3

4 3c h bg

4.3a p. 278/ 1-12, 15, 18, 21, 24, 27

Have a great weekend!