Urgent! Need help with tricky calculus questions! 10pts
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Urgent! Need help with tricky calculus questions! 10pts

[From: ] [author: ] [Date: 12-04-28] [Hit: ]
Please help me with any of these.Minimum value of ∫₋₄³ f(x) dx is area of rectangle,Maximum value of ∫₋₄³ f(x) dx is area of rectangle,Since a = 1,Since this is Right Riemann sum,.......
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Please help me with any of these.~

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Interval is 7 units wide : from x = −3 to x = 4

Minimum value of ∫₋₄³ f(x) dx is area of rectangle, 1 unit high and 7 units wide
Maximum value of ∫₋₄³ f(x) dx is area of rectangle, 3 units high and 7 units wide

7 ≤ ∫₋₄³ f(x) dx ≤ 21

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You can factor 7/n from sum
This is width of subintervals (b-a)/n

Since a = 1, then b = 8

Values of x are
x0 = 1
x1 = 1 + 7/n
x2 = 1 + 14/n
xn = 1 + 7n/n = 8

Since this is Right Riemann sum, we use x1 to xn

Now we can clearly see that f(x) = 1/x

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∫₀⁶ f(x) dx = ∫₀² f(x) dx + ∫₂⁴ f(x) dx + ∫₄⁶ f(x) dx
∫₂⁴ f(x) dx = ∫₀⁶ f(x) dx − ∫₀² f(x) dx − ∫₄⁶ f(x) dx
∫₂⁴ f(x) dx = 5 − 3 − 5
∫₂⁴ f(x) dx = −3

∫₄² f(x) = −∫₂⁴ f(x) dx
∫₄² f(x) = −(−3) = 3

∫₄² (5 f(x) − 3) = 5 ∫₄² f(x) dx − 3 ∫₄² dx
. . . . . . . . . . . . = 5 (3) − 3 (−6)
. . . . . . . . . . . . = 33

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Good answer! (But beware the minor typo in limits of integration in the 1st part.)

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State the questions directly without any reference to any web-site.
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