{"id":377,"date":"2025-02-13T19:44:26","date_gmt":"2025-02-13T19:44:26","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/approximating-areas-learn-it-2\/"},"modified":"2025-02-13T19:44:26","modified_gmt":"2025-02-13T19:44:26","slug":"approximating-areas-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/approximating-areas-learn-it-2\/","title":{"raw":"Approximating Areas: Learn It 2","rendered":"Approximating Areas: Learn It 2"},"content":{"raw":"\n

Approximating Area Cont.<\/h2>\n

Right-Endpoint Approximation<\/h3>\n

The second method for approximating area under a curve is the right-endpoint approximation. It is almost the same as the left-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of each subinterval.<\/p>\n

\n

Right-Endpoint Approximation<\/h3>\n

In the right-endpoint approximation, we estimate the area under a curve by constructing rectangles whose heights are determined by the function values at the right endpoints of subintervals.<\/p>\n

 <\/p>\n

The approximation of the area [latex]A[\/latex] using [latex]n[\/latex] subintervals is given by the formula:<\/p>\n

[latex]A \\approx R_n = \\displaystyle\\sum_{i=1}^{n} f(x_i)\\Delta x [\/latex]<\/div>\n

 <\/p>\n

where [latex]\\Delta x =\\frac{b-a}{n}[\/latex] is the width of each subinterval, and [latex]x_{i}[\/latex] are the right endpoints of the subintervals.<\/p>\n<\/section>\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 3. In the right-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the right of each subinterval. Note that the right-endpoint approximation differs from the left-endpoint approximation in (Figure).[\/caption]\n\n

\n

Since we have already seen how to solve using left-endpoint approximation, the right-endpoint approximation follows a similar process. The key difference is that the heights of the rectangles are determined by the function values at the right endpoints of the subintervals, rather than the left endpoints. This means:<\/p>\n

    \n\t
  • Left-Endpoint Approximation:<\/strong> Uses [latex]f(x_{i -1})[\/latex] for each subinterval.<\/li>\n\t
  • Right-Endpoint Approximation:<\/strong> Uses [latex]f(x_i)[\/latex] for each subinterval.<\/li>\n<\/ul>\n

    By adjusting the endpoint used, we slightly alter the position and height of the rectangles, which can affect the accuracy of the approximation depending on the behavior of the function.<\/p>\n<\/section>\n

    \n

    The graphs in Figure 4 represent the curve [latex]f(x)=\\frac{x^2}{2}[\/latex].<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"708\"]\"Diagrams Figure 4. Methods of approximating the area under a curve by using (a) the left endpoints and (b) the right endpoints.[\/caption]\n\n

    In graph (a) we divide the region represented by the interval [latex][0,3][\/latex] into six subintervals, each of width [latex]0.5[\/latex]. Thus, [latex]\\Delta x=0.5[\/latex].<\/p>\n

    We then form six rectangles by drawing vertical lines perpendicular to [latex]x_{i-1}[\/latex], the left endpoint of each subinterval.<\/p>\n

    We determine the height of each rectangle by calculating [latex]f(x_{i-1})[\/latex] for [latex]i=1,2,3,4,5,6[\/latex].<\/p>\n

    The intervals are [latex][0,0.5], \\, [0.5,1], \\, [1,1.5], \\, [1.5,2], \\, [2,2.5], \\, [2.5,3][\/latex].<\/p>\n

    We find the area of each rectangle by multiplying the height by the width.<\/p>\n

    Then, the sum of the rectangular areas approximates the area between [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis.<\/p>\n

    When the left endpoints are used to calculate height, we have a left-endpoint approximation. Thus,<\/p>\n

    [latex]\\begin{array}{ll} A \\approx L_6 & =\\displaystyle\\sum_{i=1}^{6} f(x_{i-1})\\Delta x=f(x_0)\\Delta x+f(x_1)\\Delta x+f(x_2)\\Delta x+f(x_3)\\Delta x+f(x_4)\\Delta x+f(x_5)\\Delta x \\\\ & =f(0)0.5+f(0.5)0.5+f(1)0.5+f(1.5)0.5+f(2)0.5+f(2.5)0.5 \\\\ & =(0)0.5+(0.125)0.5+(0.5)0.5+(1.125)0.5+(2)0.5+(3.125)0.5 \\\\ & =0+0.0625+0.25+0.5625+1+1.5625 \\\\ & =3.4375 \\end{array}[\/latex] <\/div>\n

    In Figure 4(b), we draw vertical lines perpendicular to [latex]x_i[\/latex] such that [latex]x_i[\/latex] is the right endpoint of each subinterval, and calculate [latex]f(x_i)[\/latex] for [latex]i=1,2,3,4,5,6[\/latex].<\/p>\n

    We multiply each [latex]f(x_i)[\/latex] by [latex]\\Delta x[\/latex] to find the rectangular areas, and then add them. This is a right-endpoint approximation of the area under [latex]f(x)[\/latex]. Thus,<\/p>\n

    [latex]\\begin{array}{ll} A \\approx R_6 & =\\displaystyle\\sum_{i=1}^{6} f(x_i)\\Delta x=f(x_1)\\Delta x+f(x_2)\\Delta x+f(x_3)\\Delta x+f(x_4)\\Delta x+f(x_5)\\Delta x+f(x_6)\\Delta x \\\\ & =f(0.5)0.5+f(1)0.5+f(1.5)0.5+f(2)0.5+f(2.5)0.5+f(3)0.5 \\\\ & =(0.125)0.5+(0.5)0.5+(1.125)0.5+(2)0.5+(3.125)0.5+(4.5)0.5 \\\\ & =0.0625+0.25+0.5625+1+1.5625+2.25 \\\\ & =5.6875 \\end{array}[\/latex]<\/div>\n<\/section>\n
    \n

    Use both left-endpoint and right-endpoint approximations to approximate the area under the curve of [latex]f(x)=x^2[\/latex] on the interval [latex][0,2][\/latex]; use [latex]n=4[\/latex].<\/p>\n

    [reveal-answer q=\"fs-id1170572368402\"]Show Solution[\/reveal-answer]
    \n[hidden-answer a=\"fs-id1170572368402\"]\n\n

    First, divide the interval [latex][0,2][\/latex] into [latex]n[\/latex] equal subintervals.<\/p>\n

    Using [latex]n=4, \\, \\Delta x=\\frac{(2-0)}{4}=0.5[\/latex]. This is the width of each rectangle.<\/p>\n

    The intervals [latex][0,0.5], \\, [0.5,1], \\, [1,1.5], \\, [1.5,2][\/latex] are shown in Figure 5.<\/p>\n

    \n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 5. The graph shows the left-endpoint approximation of the area under [latex]f(x)=x^2[\/latex] from 0 to 2.[\/caption]\n<\/div>\n

    Using a left-endpoint approximation, the heights are [latex]f(0)=0, \\, f(0.5)=0.25, \\, f(1)=1, \\, f(1.5)=2.25[\/latex]. Then,<\/p>\n

    [latex]\\begin{array}{ll} L_4 & =f(x_0)\\Delta x+f(x_1)\\Delta x+f(x_2)\\Delta x+f(x_3)\\Delta x \\\\ & =0(0.5)+0.25(0.5)+1(0.5)+2.25(0.5) \\\\ & =1.75 \\end{array}[\/latex]<\/div>\n

    The right-endpoint approximation is shown in Figure 6. The intervals are the same, [latex]\\Delta x=0.5[\/latex], but now use the right endpoint to calculate the height of the rectangles.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 6. The graph shows the right-endpoint approximation of the area under [latex]f(x)=x^2[\/latex] from 0 to 2.[\/caption]\n\n

    We have<\/p>\n

    [latex]\\begin{array}{ll} R_4 & =f(x_1)\\Delta x+f(x_2)\\Delta x+f(x_3)\\Delta x+f(x_4)\\Delta x \\\\ & =0.25(0.5)+1(0.5)+2.25(0.5)+4(0.5) \\\\ & =3.75 \\end{array}[\/latex]<\/div>\n
    \n
     <\/div>\n<\/div>\n

    The left-endpoint approximation is [latex]1.75[\/latex]; the right-endpoint approximation is [latex]3.75[\/latex].<\/p>\n

    Watch the following video to see the worked solution to this example.<\/p>\n