University:
Fresno Pacific University
Course:
MATH 210 | Calculus I
Academic year:

2019
Views:

141

Pages:

5
Author:

Lily Callahan

Honors Brief Calc. Name Period 15.6 Notes Approximating Definite Integrals How can we approximate the area under the curve? Steps for Approximating a Definite Integral f * f ( x )d x 1. Divide the interval [a,b] into n subintervals of equal length A x = 2. Pick a number u in each subinterval and evaluate f ( u) I- f ( un ) A x 3. The sum f i u f ) A x + f ( u2 ) A x + f { x )d x approximates — Ja g EX: Approximate the integral : f x 2 d x Right-hand Rectangular Method Left-hand Rectangular Method 1(0 se*y 1$ 9 •' - - TT -t v i'i A? 70 TP "’ -- \Q “ jb \ i* /M X o Ac - O 1r 5 A t j yuWn- vvAlS ^ - n 7 dy D > le 1 ^ ^ y ( Jr (x )) ' aCw 3 ' curve: - ' - C 3 0‘ O / 7. J " .A0 /' 'r| 74, + “72 1\> + 32 + \ j H,6 2n % = 340 4 EX: Approximate JQ ( 3 + 2 x ) d x Left-hand Rectangular Method Right-hand Rectangular Method y It 4- \v <4 -r * J 3 ^ 1.0 j, m 1 H ML > - jj *. M X f(x) Cl 3 $) b L 2yL - 3Y ^ - ^v 4 fix ) 3 n 5 \ [ vO ( • 5H + ± " W ' ) +- p a & ]? ^ - D = n* 'L AV KSVA 3 V>eUcr Riemann ^ Sums: If we divide each interval into different^lengths, then S ier 2> X jj ( fw f ( x ) d x = lim [/ (M1 )Ax1 + / ( w 2 ) Ax 2 + ... + f ( u„ )Ax „ ] where A -> 0 6 = max(Ax, Ax 2 ’ • This can be proven as a theorem and is called the Fundamental Theorem of Calculus , f f ( x ) d x = F ( b ) - F (a ) Ja + Right-hand Rectangular Method JbrVh YC^ ^ (artfrtf } 3y> . 3 \ j IS1 ** 3L A 3> '± \/v\ ALV\ ' + cs; / ' » ( *> ( 5 ) * 1( 0 * (? ) ' 0 - Y I Approximate area under the curve: Left-hand Rectangular Method ' # X a Exact area under the curve: 'I ’ O ^ 5 1 a - is Jr 3 i I £ - - * >. ' % - - where F is an antiderivative of f Ax„ ) - The Accumulation Function Classwork Now that we understand that the concept of a definite integral is nothing more than an area , let us consider a very special type of area problem. Suppose we are given a function f ( x ) = 2 . We know that to be a horizontal line at y = 2. First , realize that the equation of the graph of / ( -*) = 2 is the same as / ( f ) = 2 is the same as / ( ) 2 Whether we use *, /, or k , it does not matter . The graph is still a horizontal line. We are used to using x but we will see a good reason that we will occasionally be using another letter. *- -3.0 / (0 - 2 Consider the expression / /(<) <* - 1.0 0 2.0 1.0 Above, we have the graph f f ( 0 dt 4.0 5.0 / ( f ) = 2 . We now want to consider the expression function of x. As x changes, x 3.0 Jo / ( ) dt . What is this? It is a f -t Jo / ( f ) dt changes as well Complete the chart below. . 0 1 2 3 4 0 2 4 6 8 0 It should be apparent that as x gets bigger, Jo / ( ) dt increases as well. What we appear f to be doing is X “accumulating area ” and we call /(f ) = 2 Jo / ( ) dt ' f to describe our line rather than the accumulation function . Finally, it should be obvious why we use / ( T) = 2 . Jo / ( *) dx would be confusing. • X J ./( ) d t where the graph o f f ( x ) is below Remember * Example 1) Let = the same thing as . / o /( /) . (. ) Think of / r as the rate of snowfall over a period of time. For instance, at x = 1, snow is falling at 3 inches per hour, at x = 3, it is not snowing, and at x = 4, snow is melting at 4 inches per hour. y -M f a . Complete the chart . In the snow analogy, F [ x ) represents the accumulation of snow over time. 8 7 6.5 3 6 5 4 2 0 1 2 3.25 3 4.5 2.5 3 2 6.5 0 5 F(x) b. Now let’s consider F ' [ x ) = d dx x — J / ( ) d t . If f we take the derivative of an integral, what would you expect to 0 happen ? cancellation . So F '(.c ) = —J dx Knowing that, let’s complete the chart. 3 2 1 0 .*• 3 3 0 1 ' x F( ) f ( t ) d t is the same thing as _/(-*) " 0 c. On what subintervals of [0, 8] is F increasing? 4 -4 5 0 6 1 7 -1 6.5 0 ^ 8 -1 (0, 3) U (5, 6.5) Decreasing? (3, 5 U (6.5, 8) d . Where in the interval [0, 8] does F achieve its minimum and maximum value? What are those values? iMin = 0 at x = 0 IMax = 6.5 at * = 31 f . Find the concavity of F and any inflection points. Justify your answer Concave up ( 0, l ) U (4, 6) , Concave down (2, 4 ) U (6, 7) , Inflection pts: x = 4 , x = 6 h . Sketch a rough graph of ^ ( -*) .r Example 2) LetF ( x ) = Jo / ( ) dt where / is the function graphed below (consisting oflines and a semi-circle) / y - fM Find the following: a) F { 0) = 0 e) F ( - 1) — f ) F { -2) 1 g) f (-3) ~3 — -2 - h) F ( 4 ) = -2 3 1) F' { ~3 ) = -2 k) F' ( 6) = -2 i) ^'(2) = 2 i) F'(4) = 0 d) F(6) = 2 JZ c) F( 4 ) = 2n b) F { 2 ) = n [ j g where F > 0 (-2.5, 0 ) U ( 0, 4 ) m) On what subintervals of -4, 6 is F increasing and decreasing. Justify your answer . increasin n) Where in the interval o) Where in the interval ^ 4, -2, 5) U 4, 6 j decreasing where P < 0 [-4,6] does F achieve its minimum value? What is the minimum value? -3.5 atx = -2.31 [-4,6] does F achieve its maximum value? What is the minimum value? 127i at x = 4 p) Where on the interval [-4,6] is Fconcave up? Concave down ? Justify your answer. Concave up where F' is increasing: (-3, -2) U (0, 2) , Down where F ' is decreasing (-4,-3) U (-1,0) U (2, 6) q) Where does F have points of inflection ? X = -3, x = 0 , x = 2 r) Sketch the function F . A A A * Assume each tic mark on the y -axis is 2 units A

15.6 Notes Approximating Definite Integrals

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