Unit 5. Applications of Derivatives Tangent and Normal lines Rectilinear Motion Related Rates Curve Sketching Optimization Differentials
We say f say f ’ is the derivative of f of f . We can also say that f that f is is an antiderivative antiderivative of of f f ’ .
Note that g is also an antiderivative of f of f ’, since the derivative of g is equal to f to f ’ .
If
, then there is a constant C such such that
If F If F is is an antiderivative of a function function f f , then all the antiderivatives of f of f will will be of the form
where C is is an arbitrary a rbitrary constant.
Antidifferentiation (or integration Antidifferentiation (or integration)) is the process of finding a function from its derivative. Notation:
The expression on the left-hand side is also called an indefinite integral.. integral The equation above is equivalent to
or
Verify that Simply show that the derivative of the function on the right is the function in the integrand. If then Therefore,
, .
Verify that
Verify that
Evaluate the following integrals. 1. 2.
3.
4.
5.
Evaluate the following indefinite integrals.
1 x 3
x dx
3 x 4
2
dx
2
x 5 x 4 x
2
dx
1 x x
2
dx
Let g be a differentiable function, and let the range of g be an interval I. Suppose that f is a function defined on I and that F is an antiderivative of f on I. Then
f g x g ' x dx F g x c
This chain rule simply involves a change in variable in the integral. Suppose u Then du
g x .
g ' x dx.
and the integral becomes
f g x g ' x dx F g x c
f u du F u c
Evaluate the following indefinite integrals.
x 3 x
2
3
4
2 dx
Evaluate the following indefinite integrals.
x
2
3
x 2 dx
Evaluate the following indefinite integrals.
8 x 3
2
dx
x 2
3
Evaluate the following indefinite integrals.
2
x dx 4
3
x 2
Evaluate the following indefinite integrals.
x 3 dx 3
2
x 6 x
Wednesday: Exercise on Optimization Optimization and and Differentials
Evaluate the following indefinite integrals.
6 x
3
dx
2
x 1 2
3
Let u x 1 du 2 xdx
x 2
2x
dx
3
2
x 1 u
3
1/ 2
u
u 1 du
1/ 2
u
du
u 3/ 2 u1/ 2 3 c 3 / 2 1/ 2
Evaluate the following indefinite integrals.
6 x
3
2
dx
x 1
u 3/ 2 u1/ 2 3 c 3 / 2 1/ 2 2u
3/ 2
2
1/ 2
6u
3/ 2
2 x 1
c
2
1/ 2
6 x 1
c
Evaluate the following indefinite integrals.
1
dx 3
cos 3 x
Let u 3x
Then du 3dx
1 3
1 3 1 3
cos 3 x 3dx
cos udu
sin u c
c
sin 3 x
Evaluate the following indefinite integrals.
x sin dx 2
Let u Then du
x 2
1 2
dx
2
2
x 1 sin dx 22 sin udu
2 cos u c x 2cos c 2
Evaluate the following indefinite integrals.
sec
x tan x x
Let u Then du
x dx 2 x
dx 2
2
sec
x tan x 2
x
sec u tan udu
2sec u c 2sec x c
dx
VOLUNTEER?
sin x cos
2
dx
x
1 tan x
2
dx
x
2
dx
cos 4 x
3
Evaluate the following indefinite integrals.
x
2
sin
5
x x dx
Let u sin 1 12
2
3
4
cos 4
4x
12 x sin
5
3
3
Then du 12 x 2 cos
x x dx 4
3
1
cos 4
u
sin
6
12 6
3
c
6
1 12
dx 4x
5
u du
x c 4
72
3
3
We want to determine the area of a plane region region that lies under the curve , above the the x x -axis -axis and between the lines
First, let’s subdivide the region into strips of equal width. Since there are n strips, each strip has width
Thus, we divide the interval
into subintervals
From each interval, we will choose any number. Choose
in the i -th -th subinterval
.
Using these ‘sample’ points, we form rectangles whose heights are .
Note the following
Note that the sum of the areas of the rectangles is ‘near’ the area of the plane region.
Now, let us get the area of the rectangles. The width of the i -th -th rectangle is
, while its height is
Then the sum of the areas of the rectangles is
This sum can also be expressed as We call this the Riemann sum.
.
.
Note what happens when we increase the number of the rectangles.
The area of the plane region that lies under the curve above the x the x -axis -axis and between the lines and
, is
Determine the area of the region in the figure. The interval is [0,1], while the function is
Subdivide the interval into n equal parts. The width of each interval is For the sample point, let us choose the right endpoint of each subinterval. For the subinterval
, we choose
Thus, the area of the region is given by
The area of the plane region is 1/3.
Determine the area of the region below the curve above the x the x -axis -axis from -1 to 2.
,
Divide the interval [-1,2] into n subintervals of width
This time, let us choose the left endpoint as the sample point for each subinterval. For the subinterval
, choose
.
Then, the area of the region is given by
The area of the plane region is 21/2.
If f If f is is a function defined on the closed interval [a,b [a,b], ], then the definite integral of f of f from from a to b is given by
if this limit exists. If the limit exists, then we say f say f is is integrable integrable on on [a,b [a,b]. ]. a is the lower limit (or bound), bound), while b is the upper limit (or bound) of bound) of the integral.
If the function is continuous on the closed interval [a,b [ a,b], ], then it is integrable on the interval [a,b [a,b]. ]. If
, then
if it exists. If
exists, then
Friday: more on Riemann Integrals
