The figure below shows rectangles approximating the area under the function f(z)=√ over an interval of the x-axis. 3 4 5

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answerhappygod
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The figure below shows rectangles approximating the area under the function f(z)=√ over an interval of the x-axis. 3 4 5

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The Figure Below Shows Rectangles Approximating The Area Under The Function F Z Over An Interval Of The X Axis 3 4 5 1
The Figure Below Shows Rectangles Approximating The Area Under The Function F Z Over An Interval Of The X Axis 3 4 5 1 (37.91 KiB) Viewed 9 times
The Figure Below Shows Rectangles Approximating The Area Under The Function F Z Over An Interval Of The X Axis 3 4 5 2
The Figure Below Shows Rectangles Approximating The Area Under The Function F Z Over An Interval Of The X Axis 3 4 5 2 (50.88 KiB) Viewed 9 times
The figure below shows rectangles approximating the area under the function f(z)=√ over an interval of the x-axis. 3 4 5 6 7 8 1 Write the sum of the areas of the rectangles in the form f(zo)Az+ f(z) Az + f(2₂) Az + f(z) Az: (Use square roots in your answer.) Now, write the same Riemann sum using sigma notation: B ¡M- 4-0
Area = 1 ✓ Let's see how the ideas that help us find areas of more complicated shapes can work here. We can approximate the area by dividing the interval [0, 1] into three intervals and using right-endpoint approximation: 2y => 2 0= Give the following values as fractions: f(a). 0 What is the width of each interval? (Write your answer as a fraction.) Az X ХО X₁ X2 X3 3 Part 2 of 6 1
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