the first fundamental theorem of calculus. first fundamental theorem take the antiderivative....
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The First Fundamental Theorem of Calculus
First Fundamental Theorem
• Take the Antiderivative.• Evaluate the Antiderivative at the Upper Bound.• Evaluate the Antiderivative at the Lower Bound.• Subtract the Lower Bound Value from the Upper Bound Value.
)()()( aFbFdxxfb
a
The Definite IntegralIf f is a continuous function, the definite integral of f from a to b is defined to be
1
0
( ) limb n
kn
ka
f x dx f x x
The function f is called the integrand, the numbers a and b are called the limits of integration, and the variable x is called the variable of integration.
The Definite Integral
is read “the integral, from a to b of f(x)dx.”
( )b
a
f x dx
Also note that the variable x is a “dummy variable.”
( ) ( )b b
a a
f x dx f t dt
The Definite Integral As a Total
If r(x) is the rate of change of a quantity Q (in units of Q per unit of x), then the total or accumulated change of the quantity as x changes from a to b is given by
Total change in quantity ( )b
a
Q r x dx
Net Change
f F F .b
a
x dx b a
F F F .b
a
x dx b a
This can be rewritten as follows
The quantity F(b) – F(a) is the net change of the function F over the interval [a,b]. The derivative F’(x) is the rate of change of the function F.
By the Fundamental Theorem of Calculus we have, for an indefinite integral function F of f:
Definition
So now we can do all kinds of summing problems before we even mention an antiderivative.
Historically, that’s what scientists had to do before calculus.
Here’s why it mattered to them:
mi/hr
hr1 4
40
20
60v(t)= 40
d = 120 mi
The Definite Integral As a Total
Ex. If at time t minutes you are traveling at a rate of v(t) feet per minute, then the total distance traveled in feet from minute 2 to minute 10 is given by
10
2
Total change in distance ( )v t dt
mi/hr
hr1 4
40
20
60
v(t)
d = 120 mi
The calculus pioneers knew that the area would still yield distance, but what was the connection to tangent lines?
And was there an easy way to find these irregularly-shaped areas? The Definite Integral
Area Under a Graph
a bIdea: To find the exact area under the graph of a function.
( )y f x
Method: Use an infinite number of rectangles of equal width and compute their area with a limit.
Width:b a
xn
(n rect.)
Geometric Interpretation (All Functions)
( )b
af x dx Area of R1 – Area of R2 + Area of R3
a b
( )y f xR1
R2
R3
Area Using GeometryEx. Use geometry to compute the integral
5
1
1x dx
Area = 2
5
1
1 4 2 2x dx
Area =4
Computing Area Ex. Find the area enclosed by the x-axis, the vertical lines x = 0, x = 2 and the graph of
23
02x dx
Gives the area since 2x3 is nonnegative on [0, 2].
22
3 4
00
12
2x dx x 4 41 1
2 02 2
8
Antiderivative Fund. Thm. of Calculus
22 .y x
Evaluate:
Evaluate:
Evaluate Definite Integralswith your calculator.
• Enter the function in y =.• GRAPH & CALC (2nd TRACE) Option 7 or MATH Option 9 fnInt(y1,x, __, __)• Enter Lower Bound• Enter Upper Bound
AREA under f (x) down to the x-axis
from x = a to x = b
Use your graphing calculator to graph the integrand and determine whether the integral is… positive, negative, or zero.
•
•
b
a
dxxf )(
2
2
3dxx
2
1
24 )723( dxxx