Answer to Question #127348 in Quantitative Methods for Jack F

Question #127348
Find the approximate area under the curve y= sqrt{x^2+1} on the interval [0,1] using the trapezoid method and 5 subintervals.
1
Expert's answer
2020-07-27T17:49:56-0400

Solution


5 subintervals from 0 to 1, so length of each subinterval is 0.2.


"S = s_1+s_2+s_3+s_4+s_5"

"s_1 = 0.2\\cdot(\\dfrac{y(0)+y(0.2)}{2}) = 0.2\\cdot\\dfrac{1+\\sqrt{1.04}}{2} \\approx0.202 \\\\\ns_2 = 0.2\\cdot(\\dfrac{y(0.2)+y(0.4)}{2}) = 0.2\\cdot\\dfrac{\\sqrt{1.04}+\\sqrt{1.16}}{2} \\approx0.21 \\\\\ns_3 = 0.2\\cdot(\\dfrac{y(0.4)+y(0.6)}{2}) = 0.2\\cdot\\dfrac{\\sqrt{1.16}+\\sqrt{1.36}}{2} \\approx0.224 \\\\\ns_4 = 0.2\\cdot(\\dfrac{y(0.6)+y(0.8)}{2}) = 0.2\\cdot\\dfrac{\\sqrt{1.36}+\\sqrt{1.64}}{2} \\approx0.245 \\\\\ns_5 = 0.2\\cdot(\\dfrac{y(0.8)+y(1)}{2}) = 0.2\\cdot\\dfrac{\\sqrt{1.64}+\\sqrt{2}}{2} \\approx0.27 \\\\"

"S \\approx 0.202+0.21+0.224+0.245+0.27 \\approx 1.15"


Answer: 1.15


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