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# 5.2 Definite Integrals - PowerPoint PPT Presentation

5.2 Definite Integrals. When we find the area under a curve by adding rectangles, the answer is called a Riemann sum. The width of a rectangle is called a subinterval. The entire interval is called the partition. subinterval. partition. Subintervals do not all have to be the same size.

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When we find the area under a curve by adding rectangles, the answer is called a Riemann sum.

The width of a rectangle is called a subinterval.

The entire interval is called the partition.

subinterval

partition

Subintervals do not all have to be the same size.

If the partition is denoted by the answer is called a P, then the length of the longest subinterval is called the norm of P and is denoted by .

As gets smaller, the approximation for the area gets better.

subinterval

partition

if P is a partition

of the interval

is called the the answer is called a definite integral of

over .

If we use subintervals of equal length, then the length of a subinterval is:

The definite integral is then given by:

Note that the very small change in x becomes dx.

It is called a dummy variable because the answer does not depend on the variable chosen.

upper limit of integration

Integration

Symbol

integrand

variable of integration

(dummy variable)

lower limit of integration

Function is continuous on [a,b] then its definite integral over

[a,b] exists.

We have the notation for integration, but we still need to learn how to evaluate the integral.

time

In section 5.1, we considered an object moving at a constant rate of 3 ft/sec.

Since rate . time = distance:

If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

After 4 seconds, the object has gone 12 feet.

Distance:

(C=0 since s=0 at t=0)

After 4 seconds:

The distance is still equal to the area under the curve!

Notice that the area is a trapezoid

That can be solved graphically.

We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example. We would use the sum expressions to solve.

It seems reasonable that the distance travelled will again equal the area under the curve.

We can use anti-derivatives to find the area under a curve!

Note – for some problems, using algebraic integration is either very difficult or even impossible!

– drawing a diagram and solving pictorially is still an excellent method!

Use the graph of the integrand and areas to evaluate the integral:

Since this is really asking

for the area of a quarter

circle of radius 5, we can

use the area formula!

Area

Let area under the curve from either very difficult or even impossible!a to x.

(“a” is a constant)

Let’s look at it another way:

Then:

min either very difficult or even impossible!f

max f

h

The area of a rectangle drawn under the curve would be less than the actual area under the curve.

The area of a rectangle drawn above the curve would be more than the actual area under the curve.

As either very difficult or even impossible!h gets smaller, min f and max f get closer together.

This is the definition of derivative!

initial value

Take the anti-derivative of both sides to find an explicit formula for area.

As either very difficult or even impossible!h gets smaller, min f and max f get closer together.

Area under curve from a to x = antiderivative at x minus antiderivative at a.

Area either very difficult or even impossible!

= F(b) – F(a)

Example: either very difficult or even impossible!

Find the area under the curve from x=1 to x=2.

=

=

=

Area under the curve from x=1 to x=2.

Area from x=0

to x=2

Area from x=0

to x=1

ENTER either very difficult or even impossible!

Example:

Find the area under the curve from x=1 to x=2.

To do the same problem on the TI-83:

fnInt(x^2,x,1,2)

Math 9

Example: either very difficult or even impossible!

Find the area between the

x-axis and the curve

from to .

pos.

neg.

On the TI-83:

p

fnInt(abs(cos(x)),x,0,3 /2)

If you use the absolute value function, you don’t need to find the roots.

p

values, then we can interpret the integral as the difference of

areas or the net area of the region.

Where A1 is the region

above the x-axis and

below the graph of f

Where A2 is the region

below the x-axis and

above the graph of f

A1

A2