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Both part of the fundamental Theorem establish connections between antiderivatives and definite integrals.
Part 1 says that if is continuous, then dt is an
Antiderivatives of f. part 2 says that dx can be
Found by evaluating F(b) – F(a), where F is an antiderivative of f. We need a convenient notation for antiderivatives that makes them easy to work with. Because of the relation given by the Fundamental Theorem between antiderivatives
and integrals, the natation dx is traditionally used for
Antiderivative of f and is called as indefinite integral.
Thus Means F ‘(x) = f(x)
For Example, we can write
So we can regard an indefinite integral as representing an entire family of function (one antiderivative for each value of the constant C). You should distinguish carefully between definite and
Indefinite integrals. A definite integrals is a
Number, whereas an indefinite integral is family
of functions. The connection between them is given by part 2 of the Fundamental Theorem. If f is continuous on [a, b] then
The effectiveness of the Fundamental Theorem depends on having a supply of antiderivatives of functions. We adopt the convention that when a formula for a general indefinite integral is given, it is valid only on an interval. Thus, we write
With the understanding that it is valid on the interval or on the interval. This true despite the fact that the general antiderivatives of the function is
THE AREA PROBLEM
Use rectangles to estimate the area under the parabola y = x2 from, 0 to 1
We first notice that the area of S must be somewhere between 0 and 1 because S is contained in a square with side length 1, but we can certainly do better than that. Suppose we divide S into four strips S1, S2, S3, and S4 by drawing the vertical lines x = ¼
X= and x= as in figure (a)
We can approximate each strip by a rectangle whose base is the same as the strip and whose height is the same as the right edge of the strip [see Figure (b)]. In other words, the heights of these rectangle are the values of the function
F(x) = x2 at the right end points of the subinterval
Each rectangle has width and the heights are ,
. If let R4 be the sum of the areas
of these approximating rectangle,
From the Figure (b) we see that the area A of S is less than
R4 , So
A < 0.46875
Instead of using the rectangles in Figure (b) we could use the smaller rectangles in Figure (c) whose heights are the values of f at the left endpoints of the subintervals. (The leftmost rectangle has collapsed because its heights is 0). The sum of the areas of these approximating rectangles is
We see that the area of S is larger than L4, so we have lower and upper estimates for A
0.21875 < A < 0.46875
We can repeat this procedure with a larger number of strips. Figure (d), (e) shows what happens when we divide the region S into eight strips of equal width.
(d) Using left endpoint
(e) Using right endpoint
By computing the sum of the areas of the smaller rectangles (L8) and the sum of the areas of the larger rectangles (R8), we obtain better lower and upper estimates for A : 0.2734375 < A < 0.3984375 So one possible answer the question is to say that the true area of S lies somewhere between 0.2734375 and 0.3984375. We could obtain better estimates by increasing the number of strips.
PROPERTIES OF DEFINITE INTEGRAL
P-1: CHANGE OF VARIABLE
The definite integral is a number, it does not
Depend on x. in fact, we could use any latter in place of x without changing the value of the integral;
P-2: CHANGE OF LIMIT
When we define the infinite integral we
Implicity assumed that a < b. But the definition as a limit of sum makes sense even if a > b. Notice that if we reverse a and b, then x changes from (b – a)/n to (a – b)/n.
P-3: ADDITIVITY WITH RESPECT TO THE INTERVAL OF INTEGRATION:
This not easy to prove to in general, but for the case where
and a < c < b Property 7 can be seen from the
Geometric interpretation in Figure: The area under y= f(x)
from a to c is equal to the total area from a to b.
Area Under the Curve
CALCULATING AREA BY USING
AREA BY HORIZONTAL ATRIPS
To determine the area bounded by the curve x=f(y), the y-
axis and abscissa at y=c & y=d is
CALCULATING AREA BY USING
AREA BY VERTICAL STRIPS
To determinant area bounded by curve y=f(x), the x-axis
and the ordinates at x= a & x= b is
If y=f(x) lies completely above the x-axis
If y=f(x) lies completely below the x-axis then A is negative.
The convention is to consider the magnitude only
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