The Integral of a Function. The Indefinite Integral Undoing a derivative: Antiderivative=Indefinite Integral Definition: A function is called an antiderivative of a function on same interval,, if differentiation undo 185
Note: Unlike derivatives, anti-derivatives are note unique But also for any constant because 1 3 1 3 0 186
Theorem: If is any antiderivative of on, Then so is Every antiderivative of on has the form for some Differentiation produces one derivative Antidifferentiation produces an infinite family of antiderivatives 187
differentiation 0 : A name for this family 188
The indefinite integral of the integral sign [elongated S ] - the integrand - indicates the independent variable - constant of integration - one of many antiderivative of The Indefinite Integral of represents the entire family of all antiderivatives of 189
So, Differentiation Antidifferentiation Note: Sometimes we write: [indefinite Integration] 1 1 190
Finding anti-derivatives: (1) Use derivatives we know to build a table Derivative 0 1 where 1 1 Corresponding anti-derivative 1 Add 1 to the power and divide by this new power 1 191
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 log 1 1 1 192
log 1 (2) Some Properties on Indefinite Integrals: a real number All applied earlier for limits + derivatives 193
Examples: Do not write 2 2 2 2 1 1 Note on constant of integration Do not forget constants of integrations Do not introduce them too soon Combine multiple constant into one What integration technique so far? (1) Use (create) a table (2) Rewrite an integrand (in order to use the table) 194
Examples: 2 2 2 3 3 3 3 3 195
The Indefinite Integration by Parts? Recall the product rule for derivatives, Integrate both sides Shorthand notation: The integration by part formula 196
Note: Identify something to call. The rest is. Compute by differentiation and by integration. Plug in the integration by parts formula and hope that the new integral is easier than the original one. Generally try to choose to be something that simplifies when you differentiate it. Examples: 1. 2 How to choose and? 2 and are easy to find: und But we cannot find the indefinite Integral of the product Then: 2 and 2 2 2 2 2 2 2 197
2. 2 2 2 2 2 2 198
3. 2 1 2 199
Idea: Suppose and exists Chain rule: The Indefinite Integration by Substitution So, Let, then, Substitution of for makes (when it works!) integration easier. 200
Straightforward Substitution Always consider Substitution first If on substitution fails, try another one! Always make a total change from to! Never mix variables! Substitution technique: Find something in the integrand to call to simplifies the appearance of the integral and whose is also present as a factor 201
Examples: 1 1 202
Exercises: Find the Indefinites Integral of the following functions by Substitution 1. 2. 1 3. 203
Solution: function substitution Integral 2 1 2 1 1 1 3 204
The Sigma shorthand for sums Area Defined as a Limit Greek S for sum Upper limit of summation term of the sum (common form) (integer): Index of summation:,,,, Lower limit of summation 205
Definition of Area under a Curve Continuous 0 Partition into equal subintervals Each width Choose any point in each interval to calculate rectangle heights 206
Definition: If is continuous on, 0 on, Then lim, 207
Net Area 0 Definition: net signed area Not! 0 0 a + - + b If is continuous on, 0 Then lim signed, 0 Approximating Area Numerically For large lim 208
The Definite Integral The Definite Integral Defined Extend our Net Area limit lim Continuous function Equal lenght subinterval To compute the area under the graph of and above the interval, we proceed as follows: 209
1. Subdivide the interval, into unequal subintervals with endpoints: For each 1,2, 1, let, Note: The largest of the will be denoted 2. Inside each, select a point, evaluate,,,, and compute,,,, 210
3. Form the Riemann Sum Approximation 4. Repeat Step 1-3 over and over with finer and finer subdivision of, (i.e. smaller and smaller and take a limit lim Partition in equal subinterval: means 0 guaranties each width shrinks Partition in unequal subinterval: 0 guaranties each width shrinks 211
Notice that if 0 on,, then the result of this procedure will be minus the area between the graph of and,. If takes both positive and negative value on,, then the procedure yield the net signed area between the graph of and the interval, 212
Definite Integral: Definition a. is integrable on, if exists and does not depend on lim the choice of partition or the choice of point b. If is integrable, then the limit lim is called the Definite Integral of over, [or from to ] and is denoted 213
: lower limit of integration : upper limit of integration Be careful not to confuse and. They are entirely different types of things. The first is a number, the second is a collection of functions. Notation: 214
The definite Integral of a continuous Function = Net Area under a curve Theorem: If is continuous on, Then is integrable on, And, Notation: 215
We will need methods for evaluating the number that defines them. other than computing the limit Some methods generally involve antidifferentiation, but some definite integrals can be evaluated by thinking of them as area, e.g. Definite Integrals using geometry 1 1 2 2 3 1 2 3 9 4 2 3 3 3 1 3 216
Finding Definite Integrals: A new definition and properties: a. If is in Domain of, define b. If is integrable on,, define 0 217
Theorem: If is integrable on any closed Interval containing,, Then No matter, how,, are ordered! Theorem: Suppose,, integrable on, a. If 0 for all in,, Then b. If for all in,, Then 0 218
The Fundamental Theorem of Calculus There are two parts to this. The Fundamental Theorem of Calculus, Part I Development: Suppose: is continuous on, and [ differentiable means continuous] Partition 1 219
on each interval: The Mean Value Theorem for derivatives applied to on each interval ; Taking a limit as 0 give us the definite Integral 220
FTC, Part I If is continuous on, and is any anti-derivative for on,. then Notice. If is any antiderivative of, So, we can always omit writing here. Thus 221
Idea: The Mean Value Theorem for Integrals to small? to large If is continuous on,, then there is at least one point in, such, that 222
Proof: By the Extreme Value Theorem assumes both a and min on,. So, Thus! By Intermediate Value Theorem for some in, 1 223
Thus 224
The Fundamental Theorem of Calculus, Part II FTC, I 225
The fundamental theorem of calculus says that is always true: If is continuous on the Interval, Then has an antiderivative on If is in Then is one such antiderivative for meaning 226
Differentiation and Integration are Inverse Processes: FTC, PartI Integral of derivative recovers original function FTC, PartII Derivative of integral recovers original function. 227
Definite and Indefinite Integrals Related: Is a function in, Is a number no involved! So, the variable of integration in a definite integral doesn t matter: The name of the variable is irrelevant. For this reason the variable in a definite integral is often referred to as dummy variable, place holder. 228
Some Examples: 1. 2. 3. 2 2 2 4 4 0 2 1 3 1 2 3 2 4 1 15 2 2 2 1 4 2 15 229
Definite Integration by Substitution. Extending the Substitution Method of Integration to definite Integrals to evaluate the number Substitution:,, Change - limits to -limits with the substitution: To get 230
Examples: 1. Find 1. substitution of : 2 2 2. limits substitution: lower limit: 1 2 upper limit: 1 2 1 2 1 2 231
2. Find: 2 1. substitution: 2 2. limits substitution: lower limit: 1 1 upper limit: 2 4 44 4 1 1 44 1 3 232
The Definite Integral Applied Total Area Although We can find that " " 233
Example. Compute the area between 0,5 0,5 2 2, the axis and the lines 2,5 and 2,5: f(x) = a x³ + b x² + c x + d 5 4 3 2 1 0-5 -3-1 -1 1 3 5-2 -3-4 -5 x 5 4 3 2 1 0-1 -2-3 -4-5 x 2, 1 2 234
Nullpoints 0,5 0,5 2 2 0,5 2 1 2 Area 0,5 1 4 0,5 1 3 2 1 2 2 1 8 1 6 2,, 2 2,5 1 2 2 1 2,5 2 4,67 3,76 0,96 4,66 0,67 0,96 1,03 0,67 0,90 5,625 0,29 0,36 7,175 235
Area between Two Curves [one floor, one ceiling] Example: Compute the area of the region between the graphs of and 6. To identify the top and the bottom and the interval, we need a sketch. 236
Intersections: 6 6 0 3 2 0 3, 2, 3,2 Area: 6 6 62 3 1 3 8 27 1 2 6 2 3 1 3 2 3 1 2 2 3 4 9 125 6 237
Compute the area of the region between two graphs 4 6 and 3 12 10 8 6 4 2 0 0 1 2 3 4 5 6 1 4 238
Intersections: 4 6 3 1, 4 Area: 3 4 6 2 1 3 2 6 4 1 16 32 24 2 2 6 5 239