Question

a) approximate the area under graph (a) of f(x) = 1/x² over the interval 2,10 by computing the area of each rectangle to four decimal places and then adding. b) approximate the area under graph (b) of f(x) = 1/x² over the interval 2,10 by computing the area of each rectangle to four decimal places and then adding. the area under graph (a) is approximately (round to four decimal places as needed.) the area under graph (b) is approximately (round to four decimal places as needed.)

Explanation:

Step1: Analyze Graph (a)

Graph (a) on the interval \([2,10]\) seems to have rectangles with widths. Let's assume the number of rectangles. From \(x = 2\) to \(x = 10\), let's check the intervals. Looking at the x - axis, the first rectangle is from \(2\) to \(4\), then \(4\) to \(6\), \(6\) to \(8\), \(8\) to \(10\). So number of rectangles \(n = 4\), width of each rectangle \(\Delta x=\frac{10 - 2}{4}=2\). The height of each rectangle is the function value at the left - hand endpoint (since it's a left - Riemann sum, looking at the graph).

For the first rectangle: \(x_0 = 2\), \(f(x_0)=\frac{1}{2^2}=\frac{1}{4}=0.25\), area \(A_1=\Delta x\times f(x_0)=2\times0.25 = 0.5\)

Second rectangle: \(x_1 = 4\), \(f(x_1)=\frac{1}{4^2}=\frac{1}{16}=0.0625\), area \(A_2 = 2\times0.0625=0.125\)

Third rectangle: \(x_2 = 6\), \(f(x_2)=\frac{1}{6^2}=\frac{1}{36}\approx0.027778\), area \(A_3=2\times0.027778\approx0.055556\)

Fourth rectangle: \(x_3 = 8\), \(f(x_3)=\frac{1}{8^2}=\frac{1}{64}\approx0.015625\), area \(A_4 = 2\times0.015625 = 0.03125\)

Now sum the areas: \(A_a=A_1 + A_2+A_3 + A_4=0.5+0.125 + 0.055556+0.03125\)
\(A_a=0.5 + 0.125=0.625\); \(0.625+0.055556 = 0.680556\); \(0.680556+0.03125=0.711806\approx0.7118\) (Wait, maybe I made a mistake in the number of rectangles. Wait, looking at the graph (a), maybe the intervals are different. Wait, the first rectangle is from \(2\) to \(3\)? No, the x - axis labels: 2,4,6,8,10. Wait, maybe the first rectangle is from \(2\) to \(3\)? No, the graph (a) has a yellow rectangle from \(2\) to \(4\), then \(4\) to \(6\), \(6\) to \(8\), \(8\) to \(10\). Wait, let's re - calculate.

Wait, \(f(x)=\frac{1}{x^2}\). For left - hand endpoints:

First rectangle: \(x = 2\), \(\Delta x = 2\) (from 2 to 4), \(f(2)=\frac{1}{4}=0.25\), area \(2\times0.25 = 0.5\)

Second rectangle: \(x = 4\), \(\Delta x = 2\) (from 4 to 6), \(f(4)=\frac{1}{16}=0.0625\), area \(2\times0.0625 = 0.125\)

Third rectangle: \(x = 6\), \(\Delta x = 2\) (from 6 to 8), \(f(6)=\frac{1}{36}\approx0.0277778\), area \(2\times0.0277778\approx0.0555556\)

Fourth rectangle: \(x = 8\), \(\Delta x = 2\) (from 8 to 10), \(f(8)=\frac{1}{64}\approx0.015625\), area \(2\times0.015625 = 0.03125\)

Sum: \(0.5+0.125 + 0.0555556+0.03125=0.7118056\approx0.7118\)

Step2: Analyze Graph (b)

Graph (b) has rectangles with different widths? Wait, looking at the x - axis: 2,3,4,5,6,7,8,9,10? No, the x - axis labels: 2,3,4,5,6,7,8,9,10? Wait, the first rectangle is from \(2\) to \(3\), then \(3\) to \(4\), \(4\) to \(5\), \(5\) to \(6\), \(6\) to \(7\), \(7\) to \(8\), \(8\) to \(9\), \(9\) to \(10\). Wait, no, the number of rectangles: from \(2\) to \(10\), if we have rectangles with width \(\Delta x = 1\) (since 2 - 3, 3 - 4, etc.). Let's check the left - hand endpoints.

\(x = 2\), \(f(2)=\frac{1}{4}=0.25\), area \(1\times0.25 = 0.25\)

\(x = 3\), \(f(3)=\frac{1}{9}\approx0.111111\), area \(1\times0.111111\approx0.1111\)

\(x = 4\), \(f(4)=\frac{1}{16}=0.0625\), area \(1\times0.0625 = 0.0625\)

\(x = 5\), \(f(5)=\frac{1}{25}=0.04\), area \(1\times0.04 = 0.04\)

\(x = 6\), \(f(6)=\frac{1}{36}\approx0.027778\), area \(1\times0.027778\approx0.0278\)

\(x = 7\), \(f(7)=\frac{1}{49}\approx0.020408\), area \(1\times0.020408\approx0.0204\)

\(x = 8\), \(f(8)=\frac{1}{64}=0.015625\), area \(1\times0.015625 = 0.0156\)

\(x = 9\), \(f(9)=\frac{1}{81}\approx0.012346\), area \(1\times0.012346\approx0.0123\)

Now sum these areas:

\(0.25+0.1111+0.0625 + 0.04+0.0278+0.0204+0.0156+0.0123\)

First, \(0.25+0.1111 = 0.3611\)

\(0.3611+0.0625 = 0.4236\)

\(0.4236+0.04 = 0.4636\)

\(0.4636+0.0278 = 0.4914\)

\(0.4914+0.0204 = 0.5118\)

\(0.5118+0.0156 = 0.5274\)

\(0.5274+0.0123 = 0.5397\)

Wait, but maybe I misread the graph. Wait, the graph (b) has a yellow area from 2 - 3, 3 - 4, etc. Wait, let's recalculate the sum properly:

\(0.25+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\frac{1}{49}+\frac{1}{64}+\frac{1}{81}\)

Calculate each term:

\(\frac{1}{4}=0.25\)

\(\frac{1}{9}\approx0.1111111\)

\(\frac{1}{16}=0.0625\)

\(\frac{1}{25}=0.04\)

\(\frac{1}{36}\approx0.0277778\)

\(\frac{1}{49}\approx0.0204082\)

\(\frac{1}{64}=0.015625\)

\(\frac{1}{81}\approx0.0123457\)

Now sum them:

\(0.25+0.1111111 = 0.3611111\)

\(0.3611111+0.0625 = 0.4236111\)

\(0.4236111+0.04 = 0.4636111\)

\(0.4636111+0.0277778 = 0.4913889\)

\(0.4913889+0.0204082 = 0.5117971\)

\(0.5117971+0.015625 = 0.5274221\)

\(0.5274221+0.0123457 = 0.5397678\approx0.5398\)

Wait, but maybe the graph (a) has 4 rectangles with \(\Delta x = 2\) (2 - 4, 4 - 6, 6 - 8, 8 - 10) and graph (b) has 8 rectangles with \(\Delta x = 1\) (2 - 3, 3 - 4, ..., 9 - 10).

Re - calculating graph (a):

\(A_a=2\times f(2)+2\times f(4)+2\times f(6)+2\times f(8)\)

\(f(2)=\frac{1}{4}=0.25\), \(2\times0.25 = 0.5\)

\(f(4)=\frac{1}{16}=0.0625\), \(2\times0.0625 = 0.125\)

\(f(6)=\frac{1}{36}\approx0.0277778\), \(2\times0.0277778\approx0.0555556\)

\(f(8)=\frac{1}{64}=0.015625\), \(2\times0.015625 = 0.03125\)

Sum: \(0.5 + 0.125+0.0555556 + 0.03125=0.7118056\approx0.7118\)

Graph (b):

\(A_b=1\times f(2)+1\times f(3)+1\times f(4)+1\times f(5)+1\times f(6)+1\times f(7)+1\times f(8)+1\times f(9)\)

\(f(2)=0.25\), \(f(3)=\frac{1}{9}\approx0.111111\), \(f(4)=0.0625\), \(f(5)=0.04\), \(f(6)\approx0.027778\), \(f(7)\approx0.020408\), \(f(8)=0.015625\), \(f(9)=\frac{1}{81}\approx0.012346\)

Sum: \(0.25+0.111111 + 0.0625+0.04+0.027778+0.020408+0.015625+0.012346\)

Calculate step - by - step:

\(0.25+0.111111 = 0.361111\)

\(0.361111+0.0625 = 0.423611\)

\(0.423611+0.04 = 0.463611\)

\(0.463611+0.027778 = 0.491389\)

\(0.491389+0.020408 = 0.511797\)

\(0.511797+0.015625 = 0.527422\)

\(0.527422+0.012346 = 0.539768\approx0.5398\)

Answer:

The area under graph (a) is approximately \(\boldsymbol{0.7118}\) (rounded to four decimal places).

The area under graph (b) is approximately \(\boldsymbol{0.5398}\) (rounded to four decimal places).