DEFINITE INTEGRALS & NUMERIC INTEGRATION

Transcription

1 DEFINITE INTEGRALS & NUMERIC INTEGRATION Calculus answers two very important questions. The first, how to find the instantaneous rate of change, we answered with our study of the derivative. We are now ready to answer the second question: how to find the area of irregular regions. Let us start with simple way how to calculate area under the curve that is not well known shape from geometry Very often, we don t even have a function, ut only measurements taken from real life. Point to point. And we have to find area!!!!

2 Area under the curve. Left-hand Rectangular Approximation Method (LRAM / L n ) We could estimate the area under the curve y drawing rectangles touching at their left corners. Approximate area: y x 1 0 x 4 Right-hand Rectangular Approximation Method (RRAM / R n ) Approximate area:

3 Midpoint Rectangular Approximation Method (MRAM / M n ) Approximate area: M 4 = 6.625

4 Trapezoidal Approximation Method (T n) T Approximate area: T With 8 suintervals: Approximate area: M 8 = Is there an exact value for area under the curve? y x 1 0 x 4

5 SUM BECOMES INTEGRAL a x = n In limit as n x 0 n Area = lim f x i x = f x dx = F F(a) x 0 i=1 a

6 Area under the curve. y x 1 0 x 4 L 4 = 5.75 R 4 = 7.75 M 4 = T 4 = 6.75 M 8 =

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8 One can see the limiting process in action from efore. As n approaches infinity, the area approximations approach the actual area, each converging on the true value of the area. The process of finding the sum of the areas of rectangles to approximate area of a region is called Riemann Sums, after Bernhard Riemann, who pioneered the process. lim x 0 n i=1 f x i x a x = n x i = a + (n 1) x is called the definite integral of function f over interval a,. Leinitz introduced a simpler notation for the definite integral: lim x 0 n i=1 f x i x = f x dx = F F(a) a And as you are going to see definite integral is not automatically area; it is rather NET accumulation of the function over interval [a, ]

9 upper limit of integration Integration Symol a f x dx lower limit of integration integrand variale of integration (dummy variale) It is called a dummy variale ecause the answer does not depend on the variale chosen.

10 If y = f x is nonnegative and integrale over a closed interval a, then the area under the curve y = f x from a to is the definite integral of f from a to. A = a f x dx If y = f x is negative and integrale over a closed interval a,, then the area under the curve y = f x from a to is the OPPOSITE of the definite integral of f from a to. A = a f x dx If y = f x is oth positive and negative on closed interval a,, then a f x dx In general, will NOT give us the area. a f x dx AREA IS ALWAYS POSITIVE IN MATH does NOT give us the area ut rather the NET accumulation over the interval from x = a to x =. In velocity time graph, net accumulation of integral is displacement the area (asolute value) is distance!!!!!!

11 Example 1: The graph of y = f x is shown elow. If A 1 and A 2 are positive numers that represent the areas of the shaded regions, then shaded area A 1 and A 2 is: A 1 A 2 2( A 2 ) = A 1 + A 2 How else? Simple?

12 If we are integrating y hand, we must decide if and where the graph crosses the x-axis, then split up our interval, manually making negative regions positive. The following property will help accomplish this: a f x dx = f x dx + f x dx a c c if the calculator is permitted, then evaluate a f x dx If you use the asolute value function, you don t need to find the roots.

13 Difference etween the Value of a Definite Integral and Total Area Find the mean (average) value of the function f x = sinx over the interval 0, 2π Value of the Definite Integral 2π 0 sinx dx 2π = cosx = 0 cos 2π cos 0 = 1 1 = 0 Total Area 0 π 2π = sinx dx sinx dx 0 2π π = cosx sinx dx = π ( cosx) 0 = 1 1 = 4 2π π 1 1 =

14 AREA BETWEEN CURVES A = da = y 1 x y 2 (x) dx a a d d A = da = x 1 y x 2 (y) dy c c

15 Determine the area of the region ounded y y = 2x and y = 4x +16 etween x = -2 and x = 5 Determine the area of the region enclosed y y = sin x and y = cos x and the y -axis for 0 x π 2.

16 Determine the area of the enclosed area y x = 1 2 y2 3 and y = x 1 Intersection: (-1,-2) and (5,4).

17 Determine the area of the enclosed area y x = 1 2 y2 3 and y = x 1 Intersection: (-1,-2) and (5,4). This is the same that we got using the first formula and this was definitely easier than the first method. So, in this last example we ve seen a case where we could use either formula to find the area. However, the second was definitely easier.

18 Intersection points are: y = - 1 y = 3