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Using left and right Riemann sums based on the diagrams above, we definitively conclude that + 2x dx < 4 x² + 2x dx < 4 -x2 + 2x dx < 4 Hint: For the last integral, you should consistently choose either to underestimate or overestimate the area. This may require that you use the left Riemann sum for some x-intervals and the right Riemann sum for other x-intervals.

Using left and right Riemann sums based on the diagrams above, we definitively conclude that + 2x dx < 4 x² + 2x dx < 4 -x2 + 2x dx < 4 Hint: For the last integral, you should consistently choose either to underestimate or overestimate the area. This may require that you use the left Riemann sum for some x-intervals and the right Riemann sum for other x-intervals.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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The rectangles in the graph below illustrate a left and right endpoint Riemann sum

Using left and right Riemann sums based on the diagrams above, we definitively conclude that 4 -x² + 2x dx < 4 6 x² + 2x dx < 4 6 + 2x dx < 4 Hint: For the last integral, you should consistently choose either to underestimate or overestimate the area. This may require that you use the left Riemann sum for some x-intervals and the right Riemann sum for other x-intervals.
Transcribed Image Text:Using left and right Riemann sums based on the diagrams above, we definitively conclude that 4 -x² + 2x dx < 4 6 x² + 2x dx < 4 6 + 2x dx < 4 Hint: For the last integral, you should consistently choose either to underestimate or overestimate the area. This may require that you use the left Riemann sum for some x-intervals and the right Riemann sum for other x-intervals.
-x² + 2x on the interval 2, 6|. 4 The rectangles in the graph below illustrate a left endpoint Riemann sum for f(x) The value of this left endpoint Riemann sum is and it is an underestimate of v the area of the region enclosed by y = f(x), the x-axis, and the vertical lines x = 2 and x = 6. F1 Left endpoint Riemann sum for y + 2x on [2, 6] – x² + 2x on the interval (2, 6). 4 The rectangles in the graph below illustrate a right endpoint Riemann sum for f(x) The value of this right endpoint Riemann sum is and it is an an underestimate of v the area of the region enclosed by y = f(x), the x-axis, and the vertical lines x = 2 and x = 6. X 9 Right endpoint Riemann sum for y = 4 + 2x on [2,6]
Transcribed Image Text:-x² + 2x on the interval 2, 6|. 4 The rectangles in the graph below illustrate a left endpoint Riemann sum for f(x) The value of this left endpoint Riemann sum is and it is an underestimate of v the area of the region enclosed by y = f(x), the x-axis, and the vertical lines x = 2 and x = 6. F1 Left endpoint Riemann sum for y + 2x on [2, 6] – x² + 2x on the interval (2, 6). 4 The rectangles in the graph below illustrate a right endpoint Riemann sum for f(x) The value of this right endpoint Riemann sum is and it is an an underestimate of v the area of the region enclosed by y = f(x), the x-axis, and the vertical lines x = 2 and x = 6. X 9 Right endpoint Riemann sum for y = 4 + 2x on [2,6]
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Using left and right Riemann sums based on the diagrams above, we definitively conclude that 6 [.. -x² + 2x dx < 8 O≤ f * = 22²2 + 6 6 + 2x dx 8 -X 0≤ √² = 2² ²³ + 6 4 + 2x dx < Hint: For the last integral, you should consistently choose either to underestimate or overestimate the area. This may require that you use the left Riemann sum for some x-intervals and the right Riemann sum for other x-intervals.
Transcribed Image Text:Using left and right Riemann sums based on the diagrams above, we definitively conclude that 6 [.. -x² + 2x dx < 8 O≤ f * = 22²2 + 6 6 + 2x dx 8 -X 0≤ √² = 2² ²³ + 6 4 + 2x dx < Hint: For the last integral, you should consistently choose either to underestimate or overestimate the area. This may require that you use the left Riemann sum for some x-intervals and the right Riemann sum for other x-intervals.
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