Chapter 13: Integral Calculus. SSMth2: Basic Calculus Science and Technology, Engineering and Mathematics (STEM) Strands Mr. Migo M.

Transcription

1 Chapter 13: Integral Calculus SSMth2: Basic Calculus Science and Technology, Engineering and Mathematics (STEM) Strands Mr. Migo M. Mendoza

2 Chapter 13: Integral Calculus Lecture 13.1: The Integrals Lecture 13.2: Basic Integral Formulas

3 Lecture 13.1: The Integrals SSMth2: Basic Calculus Science and Technology, Engineering and Mathematics (STEM) Strands Mr. Migo M. Mendoza

4 Did you know? Two fundamental problems that calculus can address are finding the area of a plane region bounded by various curves and finding the volume of a solid revolution. These are solved through the use of the process of integration which is very much related to differentiation. This relationship is made precise by the Fundamental Theorem of Calculus.

5 Did you know? Just like any other mathematical operation, the process of differentiation can be reversed.

6 How? For example, when we perform 3 2 differentiation in f ( x) x we get f '( x) 3x. 2 Now if you begin with F( x) 3x, reversing the process should yield the function y x 3, ory x 3 1, ory x 3 1.

7 Therefore, If differentiating x n yields nx n-1, the reverse operation from give you y x n C dy dx will, where C stands for an arbitrary constant. n nx 1

8 Did you know? What we have just discussed here referring to the reverse operation of differentiation is known as antidifferentiation or indefinite integration. If the derivative of x 3 is 3x 2, then we say that an antiderivative of 3x 2 is x 3 + C.

9 Antiderivative A function F is an antiderivative of f on an interval I if F (x) = f(x) for all x in the interval I.

10 Can you do this? Can you think of an antiderivative of the function f ( x) 5x 4?

11 Possible Answers: The following are some of the antiderivatives of f: F ( x) x 5, 1 F ( x) x 5 2 9, F ( x) x

12 Something to think about What have you observed from our previous example?

13 Our Observation: We have observed in the previous example that when the derivatives of two functions are equal in an interval, then they must be differ by at most a constant.

14 Property of Antiderivative If F is an antiderivative of f on an interval I, then G is an antiderivative of f on the interval I if and only if G is of the form G(x) = F(x) + C for all x in I where C is a constant.

15 Did you know? The property of antiderivatives also tells us that in order to find all antiderivatives of a given function f, we only need one F (so that F (x) = f(x)). From this one antiderivative, we obtain the general antiderivative of f and write it as: G( x) F( x) C.

16 Example : Find the general antiderivative of the function: f ( x) 6x 5 3

17 Final Answer: Therefore, the general antiderivative of f is: G( x) x 6 3x C.

18 Did you know? In order to indicate the process of antidifferentiation, we use the integral sign ʃ so that we write: f ( x) dx F( x) to mean the general antiderivative of f(x) is F(x) + C and say the integral of f(x)dx is F(x) +C. We also refer to the function f(x) being integrated as the integrand. C

19 Example : Evaluate: 1 4x 3 dx.

20 Final Answer: Therefore, 1 4x 3 dx x x 4 C

21 Example : Evaluate: 8x 3 2x dx.

22 Example : Our answer will be, 8x 3 2x dx 2x 4 x 2 C

23 Lecture 13.2: Basic Integration Formulas SSMth2: Basic Calculus Science and Technology, Engineering and Mathematics (STEM) Strands Mr. Migo M. Mendoza

24 Integration Rule 1: Zero Rule Differentiation Formula Integration Formula dy ( C) 0 0dx C dx

25 Integration Rule 2: Constant Rule Differentiation Formula Integration Formula dy dx ( kx) k kdx kx C

26 Integration Rule 3: Constant Multiple Rule Differentiation Formula Integration Formula dy dx kf( x) kf'( x) ( x) dx k kf f ( x) dx

27 Integration Rule 4: The Sum and Difference Rule Differentiation Formula Integration Formula dy dx ( f ( x) g( x) f '( x) g'( x) f ( x) g( x) f ( x) dx dx g( x) dx

28 Integration Rule 5: The Power Rule Differentiation Formula Integration Formula dy dx ( x n ) nx n1 n1 n x x dx n 1 wheren 1 C x n dx ln whenn 1 x C

29 Example Find: 2dx

30 Final Answer: Using the formula: 2dx 2x C

31 Example Find: 2 xdx

32 Final Answer: Simplifying the resulting expression, we have: 2 xdx 4 3 x 3 2 C

33 Example Find: x 2 3 x dx

34 Final Answer: Simplifying the resulting expression, we have: x 3 x 2 3 x dx x C

35 Example Find: x 3x 4 dx

36 Example Applying integration rules, we have: x x x x dx x C

37 Classroom Task 13.1: Please answer "Let's Practice (LP)" Numbers 49.