Please calculate me 1, 2, 3, 4. Please write clearly for me. Thanks.
12 in 1h 13 next Sun Forum Grapher Calculus II: 14 Numerical Integration All lessons resize pop-out Live Notes EExpand All PD F'beta . The Fundamental Theorem of Calculus allows us to evaluate fab f($)d33 if we know an antiderivative F of f . In some cases nding F is hard and we use symbolic software. In many other cases, it is even impossible to nd an elementary exact expression for F. In such cases the Fundamental Theorem of Calculus is not usell to evaluate f: f (ac)dx. In these cases we resort to numerical approximations of f: f(z)dm by computing the corresponding Riemann sums. In order to estimate f: f (x)dac we divide the interval [L1, b] into n pieces each of the same length Ax = "'7\". In each of those intervals we pick a number 95;. We then sum the areas of the rectangles of width Ax and height f ($3 ). Rn: f(:c;)Ax + f(:c;)Aa: +...+f(:c;)A:c. area rectangle 1 area rectangle 2 Then recall f:f($)d$ = limnaoo Rn The limit is the same regardless of how we choose 9:; e [:L'i, art-+1]. y=f($) What is the height of this rectangle? 1 Heights of the approximating rectangles In order to approximate the area under the graph, we use only a nite number of rectangles and the approximations may differ according to heights of the approximating rectangle based on the interval [as.-,x,+1] is f (m2) where x;- S x: S mm. The rectangles extend till then hit the graph of f(:c). There is some exiblity how high the rectangles should be as long as the graph interects the top side of the rectangle. For example, the upper left vertex of the rectangle may be on the graph, or the right upper vertex of the rectangle be on the graph, or the graph may interest the top side in the middle. 1 2 Left and right endpoints Left endpoints. If we always choose 22;? = 33, then we obtain rectangles such that the upper left vertex is on the graph y = f (:13). The gure below illustrates the approximation of the area under 11 = x2 between :6 = 0 and :c = 2 using four rectangles and the left endpoints. Here, the width of each rectangle A1 = #. Thus R4 = f($i)AI + - - - + f(IZ)A$ :
2%:12-%:e)2-l1-75- 2 0 ' 2 2 2 roll- The choice of as: only matters when we approximate areas. The limit A = limn_,co Rn is the same regardless of the choices of 2:}. Right endpoints. If we always choose always x: = 1m, then we obtain rectangles such that the upper left vertex is on the graph 9 = f (z). rectangles are those for which the upper right vertex is on the graph 9 =f(I)- O V .4 \fTheorem (Error Bounds for Midpoint Rule). Suppose |f\"(:c)| s M for m e [a, b]. Let E be the error in the Midpoint Rule with n rectangles, then M(b a)3
