Let f be the function given by f(x)=x2+1x√+x+5. It is known that f is increasing on the interval [1,7]. Let R3 be the value of the right Riemann sum approximation for ∫71f(x)ⅆx using 3 intervals of equal length. Which of the following statements is true?
R3=13.133 and is an overestimate for ∫71f(x)ⅆx∫17f(x)ⅆx.
We can find the length of each interval by dividing the total length of the interval [1,7] by the number of intervals, which is 3 in this case.
Length of each interval = (7-1)/3 = 2
So, the intervals are [1,3], [3,5], and [5,7].
The right Riemann sum for n intervals is given by:
Rn = f(x1)(x2-x1) + f(x2)(x3-x2) + … + f(xn)(xn+1-xn)
where xi is the right endpoint of the ith interval.
Plugging in the values:
R3 = f(3)(3-1) + f(5)(5-3) + f(7)(7-5)
To find f(3), f(5), and f(7), we simply plug in those values into the given formula for f:
f(3) = 32+1(3)√+3+5 = 29/√6
f(5) = 52+1(5)√+5+5 = 61/√10
f(7) = 72+1(7)√+7+5 = 101/√12
Plugging these values into the equation for R3, we get:
R3 = (29/√6)(2) + (61/√10)(2) + (101/√12)(2)
R3 = (58/√6) + (122/√10) + (202/√12)
To simplify this expression, we can rationalize the denominators:
R3 = (58/√6)(√6/√6) + (122/√10)(√10/√10) + (202/√12)(√12/√12)
R3 = (58√6/6) + (122√10/10) + (202√12/12)
R3 = 29√6 + 61√10 + 101√3
Therefore, the value of the right Riemann sum approximation for ∫71f(x)ⅆx using 3 intervals of equal length is R3 = 29√6 + 61√10 + 101√3. None of the given statements seems to be true, so we cannot select any one of them.
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