The explicit option maple may return the solution to

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Unformatted text preview: 404, 6.948819108, 6.923289158, 6.902789477, 6.888196449, 6.877830055, 6.870316621, 6.864739771, 6.860504866, 6.857222009, 6.854630209, 6.852550664 Use the option output = animation to create an animation for the left Riemann sum. 234 • Chapter 7: Solving Calculus Problems > ApproximateInt( f(x), x=0..4*Pi, method=left, partition=6, > output=animation, iterations=7); AnApproximationoftheIntegralof f(x)=1/2+sin(x) ontheInterval[0,4*Pi] UsingaLeft-endpointRiemannSum ApproximateValue:6.283185307 1.5 1.5 AnApproximationoftheIntegralof f(x)=1/2+sin(x) ontheInterval[0,4*Pi] UsingaLeft-endpointRiemannSum ApproximateValue:6.283185307 1.5 AnApproximationoftheIntegralof f(x)=1/2+sin(x) ontheInterval[0,4*Pi] UsingaLeft-endpointRiemannSum ApproximateValue:6.283185307 1.5 AnApproximationoftheIntegralof f(x)=1/2+sin(x) ontheInterval[0,4*Pi] UsingaLeft-endpointRiemannSum ApproximateValue:6.283185307 1 1 1 1 0.5 0.5 0.5 0.5 0 2 4 6 x 8 10 12 0 2 4 6 x 8 10 12 0 2 4 6 x 8 10 12 0 2 4 6 x 8 10 12 –0.5 –0.5 –0.5 –0.5 –1 Area:6.283185309 –1 Area:6.283185308 –1 Area:6.283185308 –1 Area:6.283185311 f(x) f(x) f(x) f(x) AnApproximationoftheIntegralof f(x)=1/2+sin(x) ontheInterval[0,4*Pi] UsingaLeft-endpointRiemannSum ApproximateValue:6.283185307 1.5 1.5 AnApproximationoftheIntegralof f(x)=1/2+sin(x) ontheInterval[0,4*Pi] UsingaLeft-endpointRiemannSum ApproximateValue:6.283185307 1.5 AnApproximationoftheIntegralof f(x)=1/2+sin(x) ontheInterval[0,4*Pi] UsingaLeft-endpointRiemannSum ApproximateValue:6.283185307 1 1 1 0.5 0.5 0.5 0 2 4 6 x 8 10 12 0 2 4 6 x 8 10 12 0 2 4 6 x 8 10 12 –0.5 –0.5 –0.5 –1 Area:6.283185308 –1 Area:6.283185309 –1 Area:6.283185309 f(x) f(x) f(x) In the limit, as the number of boxes tends to infinity, you obtain the definite integral. > Int( f(x), x=0..10 ); 10 0 1 + sin(x) dx 2 The value of the integral is > value( % ); −cos(10) + 6 and in floating-point numbers, this value is approximately > evalf( % ); 6.839071529 The indefinite integral of f is > Int( f(x), x ); 7.1 Introductory Calculus • 235 1 + sin(x) dx 2 > value( % ); 1 x − cos(x) 2 Define the function F to be the antiderivative. > F := unapply( %, x ); F := x → 1 x − cos(x) 2 Choose the constant of integration so that F (0) = 0. > F(x) - F(0); 1 x − cos(x) + 1 2 > F := unapply( %, x ); F := x → 1 x − cos(x) + 1 2 If you plot F and the left-boxes together, F increases faster when the corresponding box is larger, that is, when f is bigger. > with(plots): > display( [ plot( F(x), x=0..10, color=blue, legend="F(x)" ), > ApproximateInt( f(x), x=0..10, partition=14, > method=left, output=plot) ] ); 236 • Chapter 7: Solving Calculus Problems AnApproximationoftheIntegralof f(x)=1/2+sin(x) ontheInterval[0,10] UsingaLeft-endpointRiemannSum ApproximateValue:6.839071529 1.5 1 0.5 0 2 4 x 6 8 10 –0.5 –1 Area:6.954499888 F(x) f(x) By specifying method = right or method = midpoint when using the ApproximateInt command, yo...
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