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# fxy d12f fxy x y 2 x2 y 2 x2 x2 x2 y

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Unformatted text preview: , − 1 (−sin(4) esin(4) 720 + 16 cos(4)2 esin(4) − 15 sin(4)2 esin(4) + 75 sin(4) cos(4)2 esin(4) − 20 cos(4)4 esin(4) − 15 sin(4)3 esin(4) + 45 sin(4)2 cos(4)2 esin(4) − 15 sin(4) cos(4)4 esin(4) 1 + cos(4)6 esin(4) )(2 − π )6 , − (−sin(2) esin(2) 720 + 16 cos(2)2 esin(2) − 15 sin(2)2 esin(2) + 75 sin(2) cos(2)2 esin(2) − 20 cos(2)4 esin(2) − 15 sin(2)3 esin(2) + 45 sin(2)2 cos(2)2 esin(2) − 15 sin(2) cos(2)4 esin(2) + cos(2)6 esin(2) )(4 − π )6 , 0.01542298119 (4 − π )6 , − −sin(2) esin(2) + 16 cos(2)2 esin(2) − 15 sin(2)2 esin(2) + 75 sin(2) cos(2)2 esin(2) − 20 cos(2)4 esin(2) − 15 sin(2)3 esin(2) + 45 sin(2)2 cos(2)2 esin(2) − 15 sin(2) cos(2)4 esin(2) 1 + cos(2)6 esin(2) )(2 − π )6 , − (−sin(4) esin(4) 720 + 16 cos(4)2 esin(4) − 15 sin(4)2 esin(4) + 75 sin(4) cos(4)2 esin(4) − 20 cos(4)4 esin(4) − 15 sin(4)3 esin(4) + 45 sin(4)2 cos(4)2 esin(4) − 15 sin(4) cos(4)4 esin(4) + cos(4)6 esin(4) )(4 − π )6 , 0.04005698601 (2.180293062 − π )6 } Find the maximal element. The max command expects a sequence of numbers, so you must use the op command to convert the set of values into a sequence. &gt; max_error := max( op(%) ); 1 ( 720 max _error := 0.07333000221 (2 − π )6 Approximately, this number is &gt; evalf( max_error ); 0.1623112756 232 • Chapter 7: Solving Calculus Problems You can now plot f , its Taylor approximation, and a pair of curves indicating the error band. &gt; plot( [ f(x), poly, f(x)+max_error, f(x)-max_error ], &gt; x=2..4, &gt; color=[ red, blue, brown, brown ] ); 2.5 2 1.5 1 0.5 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 x The plot shows that the actual error stays well inside the error estimate. The Integral The integral of a function can be considered as a measure of the area between the x-axis and the graph of the function. The deﬁnition of the Riemann integral relies on this graphical interpretation of the integral. &gt; f := x -&gt; 1/2 + sin(x); f := x → 1 + sin(x) 2 For example, the ApproximateInt command with method = left, partition = 6, and output = plot speciﬁed, from the Student[Calculus1] package draws the graph of f along with 6 boxes. The height of each box is the value of f evaluated at the left-hand side of the box. &gt; with(Student[Calculus1]): &gt; ApproximateInt( f(x), x=0..10, method=left, partition=6, &gt; output=plot); 7.1 Introductory Calculus AnApproximationoftheIntegralof f(x)=1/2+sin(x) ontheInterval[0,10] UsingaLeft-endpointRiemannSum ApproximateValue:6.839071529 1.5 • 233 1 0.5 2 –0.5 4 x 6 8 10 –1 Area:6.845601763 f(x) Change output = plot to output = sum to calculate the area of the boxes. &gt; ApproximateInt( f(x), x=0..10, method=left, partition=6, &gt; output=sum); 5 3 5 i=0 1 5 ( + sin( i)) 2 3 Approximately, this number is &gt; evalf( % ); 6.845601766 The approximation of the area improves as you increase the number of boxes. Increase the number of boxes to 12 and calculate the value of ApproximateInt for each of these boxes. &gt; seq( evalf(ApproximateInt( f(x), x=0..10, method=left, &gt; partition=n^2)), n=3..14); 6.948089...
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