# Consider the following. f(x) = 49 - x^{2} from x = 1 to x = 7, 4 subintervals (a) Approximate the area under the curve over the specified interval by using the indicated number of subintervals (or rectangles) and evaluating the function at the right-hand endpoints of the subintervals. (b) Approximate the area under the curve by evaluating the function at the left-hand endpoints of the subintervals.

Question
Confidence intervals
Consider the following.
$$f(x) = 49 - x^{2}$$
from $$x = 1 to x = 7, 4$$ subintervals
(a) Approximate the area under the curve over the specified interval by using the indicated number of subintervals (or rectangles) and evaluating the function at the right-hand endpoints of the subintervals.
(b) Approximate the area under the curve by evaluating the function at the left-hand endpoints of the subintervals.

2020-11-15

(a) Given that, the function is $$f(x) = 49 - x^{2}$$ on the interval $$[1, 7], n = 4.$$
It is known that, the formula for the Riemann-sum is $$\int_{a}^{b}f(x)dx=\Delta x \sum_{i=0}^{\infty} f(xi),$$ where $$\Delta_{x} = \frac{b-a}{n}.$$
Obtain the value of $$\Delta_{x} = \frac{b-a}{n}$$
$$(a = 1, b = 7, n = 4)$$
$$\Delta_{x} = \frac{7-1}{4}$$
$$= \frac{6}{4}$$
$$= \frac{3}{2}$$
Divide the interval $$[1, 7]$$ into $$n = 4$$ sub-intervals with length $$\Delta_{x} = \frac{3}{2}$$ as $$[1, \frac{5}{2}], [\frac{5}{2}, 4] [4, \frac{11}{2}], [\frac{11}{2}, 7], [\frac{11}{2}, 7]$$ and the right end-points are $$x_{1} = \frac{5}{2}, x_{2} = 4, x_{3} = \frac{11}{2}, x_{4} = 7$$. Find the area under the curve over $$[1,7]$$ by using the right-end points as follows,
$$\int_{1}^{7}(49-x^{2})dx \approx \Delta x(f(x_{1})+f(x_{2})+f(x_{3})+f(x_{4}))$$
$$= \frac{3}{2} ((49 - \frac{5}{2}^2) + (49 - (4)^2) + (49 - \frac{11}{2}^2) + (49 - (7)^2))$$
$$= \frac{3}{2} (42.75 + 33 + 18.75 + 0)$$
$$= \frac{3}{2} (94.5)$$
$$= 141.75$$
Therefore, the area under the given curve over $$[1,7]$$ by the right-end points approximation is 141.75.
(b) The left end-points are $$x_{0} = 1, x_{1} = \frac{5}{2}, x_{2} = 4, x_{3} = \frac{11}{2}.$$
Find the area under the curve over $$[1, 7]$$ by using the left-end points as follows,
$$\int_{1}^{7}(49-x^{2})dx \approx \Delta x(f(x_{1})+f(x_{2})+f(x_{3})+f(x_{4}))$$
$$= \frac{3}{2}(((49 - (1)^{2}) + 49 - (\frac{5}{2})^{2}) + (49 - (4)^2) + (49 - (\frac{11}{2})^{2}))$$
$$= \frac{3}{2} (48 + 42.75 + 33 + 18.75)$$
$$= \frac{3}{2} (142.5)$$
$$= 213.75$$
Therefore, the area under the given curve over $$[1,7]$$ by the left-end points approximation is 213.75.

### Relevant Questions

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​(Round to one decimal place as​ needed.)
$$A. 20602060xf(x)$$
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$$B. 20602060xf(x)$$
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$$C. 20602060xf(x)$$
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$$D.20602060xf(x)$$
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$$\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$
The mileage for the tire pressure $$\displaystyle{29}\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$ is
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Consider a vehicle moving with constant velocity $$\displaystyle\vec{{{v}}}$$. Find the power dissipated by form drag.
Express your answer in terms of $$\displaystyle{C}_{{d}},{A},$$ and speed v.
Part B:
A certain car has an engine that provides a maximum power $$\displaystyle{P}_{{0}}$$. Suppose that the maximum speed of thee car, $$\displaystyle{v}_{{0}}$$, is limited by a drag force proportional to the square of the speed (as in the previous part). The car engine is now modified, so that the new power $$\displaystyle{P}_{{1}}$$ is 10 percent greater than the original power ($$\displaystyle{P}_{{1}}={110}\%{P}_{{0}}$$).
Assume the following:
The top speed is limited by air drag.
The magnitude of the force of air drag at these speeds is proportional to the square of the speed.
By what percentage, $$\displaystyle{\frac{{{v}_{{1}}-{v}_{{0}}}}{{{v}_{{0}}}}}$$, is the top speed of the car increased?
Express the percent increase in top speed numerically to two significant figures.
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