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Galina-37 [17]
3 years ago
14

The table gives the values of a function obtained from an experiment. Use them to estimate 9 f(x) dx 3 using three equal subinte

rvals with right endpoints, left endpoints, and midpoints. x 3 4 5 6 7 8 9 f(x) −3.4 −2.2 −0.6 0.2 0.8 1.5 1.7
Mathematics
2 answers:
alukav5142 [94]3 years ago
5 0
We shall use 3 equal subintervals, each with width = 2.

Integral with left end points:
A-left = -3.4*2 + (-0.6)*2 + 0.8*2 = -6.4

Integral with right end points:
A-right = -0.6*2 + 0.8*2 + 1.7*2 = 3.8

Integral with midpoints:
A-mid = -2.2*2 + 0.2*2 + 1.5*2 = -1.0

Notice that A-left < A-mid < A-right.
egoroff_w [7]3 years ago
3 0

Answer:

With right endpoints: -6.4

With left endpoints: 3.8

With midpoints: -1

Step-by-step explanation:

Given

x     3     4       5      6     7    8   9  

f(x) −3.4 −2.2 −0.6 0.2 0.8 1.5 1.7

We are asked to estimate the integral 3 to 9 f(x) dx.  

Using the intervals 3 to 5, 5 to 7 and 7 to 9, the length of the intervals is Δx = 2

With right endpoints the integral is:

f(3)*Δx + f(5)*Δx + f(7)*Δx  = (-3.4)*2 + (-0.6)*2 + 0.8*2 = -6.4

With left endpoints the integral is:  

f(5)*Δx + f(7)*Δx + f(9)*Δx  =  (-0.6)*2 + 0.8*2 + 1.7*2 = 3.8

With midpoint s the integral is:

f(4)*Δx + f(6)*Δx + f(8)*Δx = (-2.2)*2 + 0.2*2 + 1.5*2 = -1

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The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s
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Answer:

The maximum acceleration over that interval is A(6) = 28.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable t.

Notice that the interval of interest 0 \le t \le 6 is closed on both ends. In other words, this interval includes both endpoints: t = 0 and t= 6. Over this interval, the value of A(t) might be maximized when t is at the following:

  • One of the two endpoints of this interval, where t = 0 or t = 6.
  • A local maximum of A(t), where A^\prime(t) = 0 (first derivative of A(t)\! is zero) and A^{\prime\prime}(t) (second derivative of \! A(t) is smaller than zero.)

Start by calculating the value of A(t) at the two endpoints:

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  • A(6) = 28.

Apply the power rule to find the first and second derivatives of A(t):

\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}.

\displaystyle A^{\prime\prime}(t) = 6\, t - 15.

Notice that both t = 1 and t = 4 are first derivatives of A^{\prime}(t) over the interval 0 \le t \le 6.

However, among these two zeros, only t = 1\! ensures that the second derivative A^{\prime\prime}(t) is smaller than zero (that is: A^{\prime\prime}(1) < 0.) If the second derivative A^{\prime\prime}(t)\! is non-negative, that zero of A^{\prime}(t) would either be an inflection point (ifA^{\prime\prime}(t) = 0) or a local minimum (if A^{\prime\prime}(t) > 0.)

Therefore \! t = 1 would be the only local maximum over the interval 0 \le t \le 6\!.

Calculate the value of A(t) at this local maximum:

  • A(1) = 15.5.

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However, note that the maximum over this interval exists because t = 6\! is indeed part of the 0 \le t \le 6 interval. For example, the same A(t) would have no maximum over the interval 0 \le t < 6 (which does not include t = 6.)

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