Chapter 5 Integrals 机动 目录 上页 下页 返回 结束 5.2 Area 5.3 The Definite Integral 5.4 The Fundamental Theorem of Calculus 5.5 The Substitution Rule

4.2 Area Area Problem: Find the area of the region S that lies under the curve y=f(x) from a to b.(see Figure 1) y=f(x)y=f(x) S ab x y o Figure 1 机动 目录 上页 下页 返回 结束

Idea for problem solving : First approximate the region S by polygons, and then take the limit of the area of these polygons.(see the following example) Example 1 Find the area under the parabola y=x 2 from 0 to 1. y Solution: We start by dividing the interval [0, 1] into n-subintervals with equal length, and consider the rectangles whose bases are these subintervals and whose heights are the values of the function at the right-hand endpoints. ox (1,1) 1/n 1 Figure 2 机动 目录 上页 下页 返回 结束

Then the sum of the area of these rectangles is As n increases, S n becomes a better and better approximation to the area of the parabolic segment. Therefore we define the area A to be the limit of the sums of the areas of these rectangles, that is, Applying the idea of Example 1 to the more general region S of F.1, we introduce the definition of the area as following: Step 1: Partition--Divide the interval [a, b] into n smaller subintervals by choosing partition points x 0, x 1, x 2,…., x n so that a=x 0 < x 1 <x 2 <…< x n =b 机动 目录 上页 下页 返回 结束

Step 2: Approximation—By the partition above, the area of S can be approximated by the sum of areas of n rectangles. This subdivision is called a partition of [a,b] and we denote it by P. Let denote the length of ith subinterval [x i-1, x i ], and ||P|| (the norm of P) denotes the length of longest subinterval. Thus Using the partition P one can divide the region S into n strips (see F.3). Now, we choose a number in each subinterval, then each strip S i can be approximated by a rectangle R i (see F.4). The sum of areas of these rectangles as an approximation is 机动 目录 上页 下页 返回 结束

y=f(x)y=f(x) SiSi a b x y o Figure 3 XiXi X i-1 S1S1 S2S2 SnSn R2R2 y=f(x)y=f(x) RiRi a b x y o Figure 4 XiXi X i-1 R1R1 RnRn approximated by 机动 目录 上页 下页 返回 结束

Step 3: Taking limit—Notice that the approximation appears to become better and better as the strips become thinner and thinner. So we define the area of the region as the limit value (if it exists) of the sum of areas of the approximating rectangles, that is (1) Remark 2: In step 1, we have no need to divided the interval [a,b] into n subintervals with equal length. But for purposes of calculation, it is often convenient to take a partition that divides the interval into n subintervals with equal length.(This is called a regular partition) Remark 1: It can be shown that if f is continuous, then the limit (1) does exist. 机动 目录 上页 下页 返回 结束

Solution Since y=x 2 +1 is continuous, the limit (1) must exist for all possible partition P of the interval [a, b] as long as ||P|| 0. To simplify things let us take a regular partition. Then the partition points are x 0 =0, x 1 =2/n, x 2 =4/n, …, x i =2i/n, …, x n =2n/n=2 So the norm of P is ||P||=2/n Let us choose the point to be the right-hand endpoint: = x i =2i/n By definition, the area is Example 2: Find the area under the parabola y=x 2 +1 from 0 to 2. 机动 目录 上页 下页 返回 结束

Example 3 Find the area under the cosine curve from 0 to b, where Solution We choose a regular partition P so that ||P||=b/n and we choose to be the right-hand endpoint of the ith sub- interval: =x i =ib/n Since ||P|| 0 as n, the area under the cosine curve from 0 to b is Remark 3: If is chosen to be the left-hand endpoint, one will obtain the same answer. 机动 目录 上页 下页 返回 结束

/section 5.2 end 机动 目录 上页 下页 返回 结束

5.3 The Definite Integral In Chapters 6 and 8 we will see that limit of form occurs in a wide variety of situations not only in mathematics but also in physics, Chemistry, Biology and Economics. So it is necessary to give this type of limit a special name and notation. 1. Definition of a Definite Integral If f is a function defined on a closed interval [a, b], let P be a partition of [a, b] with partition points x 0, x 1, x 2,…., x n, where a=x 0 < x 1 <x 2 <…< x n =b 机动 目录 上页 下页 返回 结束

Choose points in [x i-1, x i ] and let and. Then the definite integral of f from a to b is if this limit exists. If the limit does exist, then f is called integrable on the interval [a, b]. upper limit integrand Note 1: lower limit integral sign 机动 目录 上页 下页 返回 结束

Note 2: is a number; it does not depend on x. In fact, we have Note 3: The sum is usually called a Riemann sum. Note 4: Geometric interpretations For the special case where f(x)>0, = the area under the graph of f from a to b. In general, a definite integral can be interpreted as a difference of areas: x y o a b ++ -

Note 5: In the case of a>b and a=b, we extend the definition of as follows: If a>b, then If a=b, then Example 1 Express as an integral on the interval [0, ]. Example 2 Evaluate the integral by interpreting in terms of areas. 机动 目录 上页 下页 返回 结束

Solution We compute the integral as the difference of the areas of the two triangles: 1 3 o x y y=x-1 A1A1 A2A2 2. Existence Theorem Theorem If f is either continuous or monotonic on [a, b], then f is integrable on [a, b]; that is, the definite integral exists. Remark 1: If f is discontinuous at some points, then might exist or it might not exist. But if f is piecewise continuous, then f is integrable. 机动 目录 上页 下页 返回 结束

Remark 2: It can be shown that if f is integrable on [a, b], then f must be a bounded function on [a, b]. 3. Integral Formulas under Regular Partition If f is integrable on [a, b], it is often convenient to take a regular partition. Then and If we choose to be the right endpoint in each subinterval, then Since ||P||=(b-a)/n, we have ||P|| 0 as n, so the definition gives 机动 目录 上页 下页 返回 结束

Theorem If f is integrable on [a, b], then Example 3 Express as an integral on the interval [1, 2]. Answer: If the purpose is to find an approximation to an integral, it is usually better to choose to be midpoint of the subinterval, which we denote by. Any Riemann sum is an approximation to an integral, but if we use midpoints and a regular partition we get the following approximation: 机动 目录 上页 下页 返回 结束

Midpoint Rule where and Using the Midpoint Rule with n=5 we can get an approximation of integral (see page 277). 机动 目录 上页 下页 返回 结束

4. Properties of the Integral Suppose all of the following integrals exist. Then Example 4 Using the properties above and the results 机动 目录 上页 下页 返回 结束 to evaluate

Order properties of the integral Suppose the following integrals exist and a<b. 5. If f(x)>0 and a<x<b, then 6. If f(x)>g(x) for a<x<b, then 7. If m<f(x)<M for a<x<b, then 8. Proof of Property 7 Since m<f(x)<M, Property 6 gives Using Property 1, we obtain 机动 目录 上页 下页 返回 结束

Example 5 Show that Solution Notice that for 1<x<4. Example 6 Proof that /section 4.3 end 机动 目录 上页 下页 返回 结束 Since |x|=x for x>0, we have Thus, by Property 7,

4.4 The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus in this section gives the precise inverse relationship between the derivative and the integral. It enables us to compute areas and integrals very easily without having to compute them as limits of sums as we did in sections 4.2 and Fundamental Theorem For a continuous function f on [a, b], we define a new function g by Computing the derivative of g(x) we obtain 机动 目录 上页 下页 返回 结束

The Fundamental Theorem of Calculus, Part 1 If f is continuous on [a, b], then the function defined by is continuous on [a, b] and differentiate on (a, b), and Proof If x and x+h are in [a, b], then and so, for h, (2) 机动 目录 上页 下页 返回 结束

For now let us assume that h>0. Since f is continuous on [x, x+h], the Extreme Value Theorem says that there are number u and v in [x, x+h] such that f(u)=m and f(v)=M, where m and M are the absolute minimum and maximum values of f on [x, x+h]. 机动 目录 上页 下页 返回 结束 Inequality (3) can be proved in a similar manner for the case where h<0. (3) Combining it with (2) gives Since h>0, we can divide this inequality by h: By Property 7, we have

Since f is continuous at x, and u, v lie between x and x+h, we have 机动 目录 上页 下页 返回 结束 If x=a and b, then Equation (4) can be interpreted as one- side limit. Then Theorem shows that g is continuous on [a, b]. (4) We conclude, from (3) and the Squeeze Theorem, that

Note: This Theorem can be written in Leibniz notation as 机动 目录 上页 下页 返回 结束 Roughly speaking, equation (5) says that we first integrate f and then differentiate the result, we get back to the original function f. (5)

Example 1 Find the derivative of the function g(x)= Example 2 Find Example 3 Find Example 4 Find The second part of the Fundamental Theorem of Calculus provides us with a much simpler method for the evaluation of integrals. The Fundamental Theorem of Calculus, Part 2 If f is continuous on [a, b], then 机动 目录 上页 下页 返回 结束 where F is any antiderivative of f, that is

Proof We know from part 1 that g is an antidervitive of f. If F is any other antidervitive of f on [a, b], then f and g differ only by a constant: (6) F(x)=g(x)+C for a<x<b. But both f and g are continuous on [a, b] and so, by taking limits of Equation (6)(as x a - and x b + ), we see that it also holds when x=a and x=b. Noticing g(a)=0 and using Eq.(8) with x=a and x=b, we have F(b)-F(a)=[g(b)+C]-[g(a)+C]=g(b)-g(a) = Notation Set F(x)| =F(b)-F(a). So we have 机动 目录 上页 下页 返回 结束

Example 5 Evaluate the integral Example 6 Find the area under the cosine curve from 0 to b, where 2. The Indefinite Integral Because of the relation given by the Fundamental Theorem between antidervitive and integral, the notation is traditionally used for an antidervative of f and called an indefinite integral. Thus (7) Attention A definite integral is a number, whereas an indefinite integral is a function. Then the connection between them is given by Part 2 of the Fundamental Theorem, i.e.,

(7) 3. Table of Indefinite Integrals 机动 目录 上页 下页 返回 结束

Example 7 Find the general indefinite integral Example 8 Evaluate Example 9 Evaluate Example 10 What is wrong with the following calculation? End /section 4.4 机动 目录 上页 下页 返回 结束

4.5 The Substitution Rule Because of the Fundamental Theorem of Calculus, we can integrate a function if we know an antiderivative. But the anti-differentiation formulas in Section 4.4 do not suffice to evaluate integrals of more complicated functions. It is necessary for us to develop some techniques to transform a given integral into one of the forms in the table. The substitution rule drives from the Chain Rule 机动 目录 上页 下页 返回 结束 Integrating it, we get (1)

If we make the substitution u=g(x), then Eq.(1) becomes Thus we have proved the following rule: The Substitution Rule for Indefinite Integral If u=g(x) is a differentiable function whose range is an interval I and f is continuous on I, then 机动 目录 上页 下页 返回 结束 or, writing, we get

Example 1 Find Example 2 Evaluate Example 3 Evaluate Example 4 Evaluate Example 5 Calculate 机动 目录 上页 下页 返回 结束

The Substitution Rule for Definite Integral If g'(x) is continuous on [a, b] and f is continuous on the range of g, then Example 6 Evaluate Example 7 Evaluate Example 8 Evaluate 机动 目录 上页 下页 返回 结束

The next theorem uses the Substitution Rule for Definite Integral to simplify the calculation of integral of functions that possess symmetry properties. Integrals of Symmetric Functions Suppose f is continuous on [-a, a]. (a)If f is even [f(-x)=f(x)], then (b)If f is odd [f(-x)=-f(x)], then Example 9 Evaluate Example 10 Show that 机动 目录 上页 下页 返回 结束

END 机动 目录 上页 下页 返回 结束