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Integral Calculus

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Integral Calculus PDF Notes, Important Questions and Formulas

Indefinite Integration

Introduction:

Both part of the fundamental Theorem establish connections between antiderivatives and definite integrals.

Part 1 says that if is continuous, then begin mathsize 12px style integral subscript straight a superscript straight x text   end text straight f left parenthesis straight t right parenthesis end style dt is an

Antiderivatives of f. part 2 says that begin mathsize 12px style integral subscript straight alpha superscript straight b straight F left parenthesis straight x right parenthesis end style 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 begin mathsize 12px style integral straight f left parenthesis straight x right parenthesis end style dx is traditionally used for

Antiderivative of f and is called as indefinite integral.

Thus begin mathsize 12px style integral straight f left parenthesis straight x right parenthesis dx equals straight F left parenthesis straight x right parenthesis plus straight C end style  Means F ‘(x) = f(x)

For Example, we can write

begin mathsize 12px style integral straight x to the power of blank squared end exponent dx equals straight x cubed over 3 plus straight C end style  because begin mathsize 12px style straight d over dx left parenthesis straight x cubed over 3 plus straight C right parenthesis equals straight x squared end style

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 begin mathsize 12px style integral subscript straight alpha superscript straight b straight f left parenthesis straight x right parenthesis dx end style is a

Number, whereas an indefinite integral begin mathsize 12px style integral straight f left parenthesis straight x right parenthesis dx end style 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

 begin mathsize 12px style integral subscript straight alpha superscript straight b straight f left parenthesis straight x right parenthesis dx equals integral straight f left parenthesis straight x right parenthesis dx vertical line subscript straight alpha superscript straight b end style

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

 begin mathsize 12px style integral 1 over straight x squared dx equals negative 1 over straight x plus straight C end style

With the understanding that it is valid on the interval begin mathsize 12px style left parenthesis 0 comma straight infinity right parenthesis end style or on the intervalbegin mathsize 12px style left parenthesis negative straight infinity comma 0 right parenthesis end style. This true despite the fact that the general antiderivatives of the function begin mathsize 12px style straight f left parenthesis straight x right parenthesis equals 1 divided by straight x squared comma straight x not equal to 0 comma end style is

 begin mathsize 12px style straight f left parenthesis straight x right parenthesis left curly bracket table attributes columnalign left end attributes row cell negative 1 over straight x plus straight C subscript 1 text  if x < 0 end text end cell row cell negative 1 over straight x plus straight C subscript 2 text  if x > 0 end text end cell end table end style

 

 

 

Definite Integration

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= begin mathsize 12px style 1 half end style and x= begin mathsize 12px style 3 over 4 end style 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 begin mathsize 12px style open square brackets 0 comma 1 fourth close square brackets comma end style

 begin mathsize 12px style left square bracket 1 fourth comma 1 half right square bracket comma left square bracket 1 half comma 3 over 4 right square bracket comma and left square bracket 3 over 4 comma 1 right square bracket end style

 

 

Each rectangle has width begin mathsize 12px style 1 fourth end style and the heights are begin mathsize 12px style open parentheses fraction numerator begin display style 1 end style over denominator 4 end fraction close parentheses squared end style ,



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. If let R4 be the sum of the areas

of these approximating rectangle,

we get,

 

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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

 

 

 begin mathsize 12px style straight L subscript 4 equals 1 fourth.0 squared plus 1 fourth. open parentheses fraction numerator begin display style 1 end style over denominator begin display style 4 end style end fraction close parentheses squared plus 1 fourth. open parentheses fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction close parentheses squared 1 fourth plus 1 fourth. open parentheses fraction numerator begin display style 3 end style over denominator begin display style 4 end style end fraction close parentheses squared end style

 

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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 begin mathsize 12px style integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx end style 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;

 begin mathsize 12px style integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis straight space dx equals integral subscript straight a superscript straight b straight f left parenthesis straight t right parenthesis space dt  = integral subscript straight a superscript straight b straight f left parenthesis straight r right parenthesis space dr end style

 

P-2: CHANGE OF LIMIT

When we define the infinite integral begin mathsize 12px style integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx end style 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.

 

Therefore, begin mathsize 12px style integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis dx equals negative integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx end style

 

P-3: ADDITIVITY WITH RESPECT TO THE INTERVAL OF INTEGRATION:

 begin mathsize 12px style integral subscript straight a superscript straight c straight f left parenthesis straight x right parenthesis dx plus integral subscript straight c superscript straight b straight f left parenthesis straight x right parenthesis dx equals integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx end style

This not easy to prove to in general, but for the case where

begin mathsize 12px style straight f left parenthesis straight x right parenthesis greater or equal than 0 straight space end style 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

HORIZONTAL STRIP

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

 

 

i.e. begin mathsize 12px style straight A equals integral integral from straight c to straight d of straight f left parenthesis straight y right parenthesis dy equals integral from straight c to straight d of xdy end style

 

CALCULATING AREA BY USING

VERTICAL STRIP

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

Case-I

If y=f(x) lies completely above the x-axis

i.e.  A= begin mathsize 12px style integral from straight a to straight b of straight f left parenthesis straight x right parenthesis dx equals integral integral from straight a to straight b of ydx end style

 

Case –II

If y=f(x) lies completely below the x-axis then A is negative.

The convention is to consider the magnitude only

 

i.e. begin mathsize 12px style straight A equals integral from straight a to straight b of straight f left parenthesis straight x right parenthesis dx equals integral from straight a to straight b of ydx end style

 

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