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5.3.43 The graph atthe function ﬁx}: 1’4 —J(2 an the interval —25×52 is 1 a semicircle. The area under the graph is 51(212 a: 5.23319, to ﬁve decimal places. Use a Riemann sum with n =3 and midpeints tc estimate the area under the graph. Then cempute the errcr [the difference between the estimate and 5.23319}. The estimated area under the graph is E. {Type an integer er a decimal raunded te ﬁve decimal places as needed}

July 30, 2019/in Questions Uploads /by admin

5.3.43 The graph atthe function ﬁx}: 1’4 —J(2 an the interval —25×52 is
1
a semicircle. The area under the graph is 51(212 a: 5.23319, to ﬁve decimal places. Use a Riemann sum with n =3 and midpeints tc
estimate the area under the graph. Then cempute the errcr [the
difference between the estimate and 5.23319}. The estimated area under the graph is E.
{Type an integer er a decimal raunded te ﬁve decimal places as needed}