5.1 ... 2 40For the region under f(x) = 5x² on [0, 2], show that the sum of the areas of the upper approximating rectangle approaches that is,lim R =00-5SolutionR is the sum of the areas of the n rectangles in the figure below.yII4012Each rectangle has widthR, = ² - 5(²)² + Z² - 5( ² )² + 2² · 5(0)² + ... + ² · s(²^)²-10- (1² +2²+3²+...+ n²)lim7-8= limThus we have the following.lim R= lim²+2²+ 3² + ... + n²).Here we need the formula for the sum of the squares of the first n positive integers.1²++3²+...+ n² = n(n + 1)(2n + 1)6(2, 20)2Perhaps you have seen this formula before. Putting this formula into our expression for R, we get the following.n(n + 1)(2n + 1)6and the heights are the values of the function f(x) = 5x² at the points.-1.2=246n nn) (^+ ¹) (²0+¹)]) (¹ + ²)(²+¹)⠀that is, the heights are 5(²)²³, 5(=)², 5(-)-²… …….. (²ª) ². THThus,

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40 For the region under f(x) = 5x² on [0, 2], show that the sum of the areas of the upper approximating rectangle approaches that is, lim R = 00-5 Solution R is the sum of the areas of the n rectangles in the figure below. y II 40 12 Each rectangle has width R, = ² - 5(²)² + ² - 5( ² )² + 2² · 5(0)² + ... + ² - 5(²^)² 10 - (1² +2²+3²+...+n²) lim 7-8 = lim = lim ²+2²+ 3² + ... + n²). Here we need the formula for the sum of the squares of the first n positive integers. 1²+ n(n+ 1) (2n + 1) 6 Thus we have the following. lim R (2, 20) +3²+...+ n² = 2 and the heights are the values of the function f(x) = 5x² at the points. Perhaps you have seen this formula before. Putting this formula into our expression for R, we get the following. n(n + 1)(2n + 1) 6 -1.2= 246 n nn ) (^+¹)(²^+¹) ]) (¹ + ²)(²+¹) ⠀ that is, the heights are 5(²)², 5(=)²³, 5(-) ³²... (²7) ²³. - Thus,