A function g is decreasing and concave down on the interval [2,5]. An approximation for $\displaystyle\int_{2}^{5}$ g(x)dx is calculated using a midpoint Riemann sum with 6 equal subintervals. This approximation will:
Computing midpoint Riemann sums and midpoint-rule approximations.
24 questions
A function g is decreasing and concave down on the interval [2,5]. An approximation for $\displaystyle\int_{2}^{5}$ g(x)dx is calculated using a midpoint Riemann sum with 6 equal subintervals. This approximation will:
Underestimate the true value of the integral
Equal the true value of the integral
Be impossible to determine without knowing g explicitly
Overestimate the true value of the integral
The table provides data for approximating the integral of a function $$f$$ on the interval $$[2, 5]$$ using a midpoint Riemann sum with a single subinterval. Which statement correctly describes this approximation?
| Interval | [2,5] |
| Width | 3 |
| f(midpoint) | 7 |
The midpoint Riemann sum equals $$24$$ by using an incorrect subinterval width.
The midpoint Riemann sum equals $$21$$.
The midpoint Riemann sum equals $$18$$ due to a miscalculation of the midpoint value.
The midpoint Riemann sum equals $$20$$ reflecting a minor multiplication error.
Which of the following expressions represents the midpoint Riemann sum with four subintervals of equal width for the integral $$\int_{0}^{2} (3x^2+2\sqrt{x}) dx$$?
$$0.5*\Big[(3*(0.5)^2+2*\sqrt{0.5})+(3*(1)^2+2*\sqrt{1})+(3*(1.5)^2+2*\sqrt{1.5})+(3*(2)^2+2*\sqrt{2})\Big]$$
$$0.5*\Big[(3*(0)^2+2*\sqrt{0})+(3*(0.5)^2+2*\sqrt{0.5})+(3*(1)^2+2*\sqrt{1})+(3*(1.5)^2+2*\sqrt{1.5})\Big]$$
$$(3*(0.25)^2+2*\sqrt{0.25}+3*(0.75)^2+2*\sqrt{0.75}+3*(1.25)^2+2*\sqrt{1.25}+3*(1.75)^2+2*\sqrt{1.75})$$
$$0.5*\Big[(3*(0.25)^2+2*\sqrt{0.25})+(3*(0.75)^2+2*\sqrt{0.75})+(3*(1.25)^2+2*\sqrt{1.25})+(3*(1.75)^2+2*\sqrt{1.75})\Big]$$
Using the graph of $$f(x)=\frac{1}{x+1}$$ on the interval $$[0,4]$$, approximate the definite integral $$\int_{0}^{4} \frac{1}{x+1} dx$$ using a midpoint Riemann sum with 4 equal subintervals.
$$1.65$$
Approximately $$1.50$$ using a mistaken evaluation of midpoints.
$$1.75$$
$$1.57$$
Approximate the area under the curve of $$f(x)=3x+1$$ on the interval $$[0, 6]$$ using a midpoint Riemann sum with 3 subintervals of equal width.
$$54$$
$$56$$
$$60$$
$$64$$
Let g be the function defined by g(x) = ln(x²+1). Which of the following is the midpoint Riemann sum approximation of $\displaystyle\int_{1}^{3}$ g(x) dx with four subintervals of equal width?
0.5[ln(1.5²+1) + ln(2²+1) + ln(2.5²+1) + ln(3²+1)]
0.5[ln(1.25²+1) + ln(1.75²+1) + ln(2.25²+1) + ln(2.75²+1)]
0.5[ln(1²+1) + ln(1.5²+1) + ln(2²+1) + ln(2.5²+1)]
0.25[ln(1.25²+1) + ln(1.75²+1) + ln(2.25²+1) + ln(2.75²+1)]
What is the approximate value of $$\int_{0}^{4} \sqrt{x} dx$$ using a midpoint Riemann sum with four equal subintervals?
5.38
4.15
6.15
5.15
Let `g` be a function that is continuous on the interval [0, 4], with g(0)=3, g(1)=5, g(2)=4, g(3)=6, and g(4)=2. Which of the following is the midpoint Riemann sum approximation of $\displaystyle\int_{0}^{4}$ g(x) dx using four subintervals of equal length?
20
16
17
Cannot be determined from the given information
Using the graph of $$f(x)=\sin(x)$$ on the interval $$[0,\pi]$$, approximate the area under the curve using a midpoint Riemann sum with 4 equal subintervals.
$$1.90$$
Approximately $$2.10$$ when a rounding error occurs.
$$2.05$$
$$2.25$$
I.
$$M = 1*(1.5^2) + 1*(2.5^2)$$
II.
$$M = \frac{1}{2}*(2^2 + 3^2)$$
III.
$$M = 1*(1.25^2) + 1*(2.75^2)$$
When approximating the area under the curve of $$f(x)=x^2$$ on the interval [1,3] using two equal subintervals and the midpoint rule, which of the above expressions correctly computes the midpoint Riemann sum?
II only
I and III
III only
I only
In which of the following approximation methods are the heights of the rectangles determined by the function value at the midpoint of each subinterval?
Left Riemann Sum
Midpoint Riemann Sum
Right Riemann Sum
Simpson's Rule
In a Midpoint Riemann Sum, how is the height of each approximating rectangle determined?
By averaging the heights of the left and right sides of each rectangle.
By using the height for the right side of each base.
By using the height for the middle of each base.
By using the height for the left side of each base.
Let E(n) represent the error in the approximation of $\displaystyle\int_{0}^{1}$ eˣ dx by a midpoint Riemann sum with n subintervals. Which of the following statements best describes the relationship between E(n) and n?
E(n) is proportional to 1/n
E(n) is proportional to 1/√n
E(n) is proportional to 1/n³
E(n) is proportional to 1/n²
Which of the following expressions represents the midpoint Riemann sum for $\displaystyle\int_{1}^{3}$ sin(πx) dx with n subintervals of equal width?
Σᵢ₌₀ⁿ⁻¹ sin(π(1+i·2/n))·(2/n)
Σᵢ₌₁ⁿ sin(π(1+i·2/n-2/n))·(2/n)
Σᵢ₌₁ⁿ sin(π(1+(2i-1)·1/(2n)))·(2/n)
Σᵢ₌₁ⁿ sin(π(1+i·2/n))·(2/n)
Using the data in the table for a function $$f$$, what is the midpoint Riemann sum approximation for the area under the graph of $$f$$ on the interval $$[0, 7]$$?
| Interval | [0,3] | [3,7] |
| Width | 3 | 4 |
| f(midpoint) | 2 | 6 |
The midpoint Riemann sum equals $$26$$ due to an underestimated width calculation.
The midpoint Riemann sum equals $$30$$.
The midpoint Riemann sum equals $$34$$ by overestimating the function values.
The midpoint Riemann sum equals $$28$$ as a result of an arithmetic error.
The table provides values for a function on the interval [0, 4]. Using a midpoint Riemann sum with two subintervals of equal length, approximate the area under the curve of the function.
| Interval | [0,2] | [2,4] |
|---|---|---|
| Endpoint values | 2, 5 | 5, 10 |
| Midpoint values | 3 | 7 |
$$20$$
$$18$$
$$24$$
$$22$$
The table provides values for a function `f` at the midpoints of three subintervals of equal width. Which of the following represents the midpoint Riemann sum approximation for the integral of `f`?
| Midpoint (x) | 1.5 | 4.5 | 7.5 |
|---|---|---|---|
| f(x) | 2 | 5 | 10 |
Midpoint Sum = $$3*(2+5+10)=51$$
Midpoint Sum = $$3*2+5+10=21$$
Midpoint Sum = $$\frac{2+5+10}{3}=5.67$$
Midpoint Sum = $$\frac{3*(2)+3*(5)+3*(10)}{3}=17$$
The table below shows the velocity of a vehicle, in meters per second, at the midpoints of the time intervals $[0, 2]$, $[2, 4]$, and $[4, 6]$. Using a midpoint Riemann sum, what is the approximate distance, in meters, traveled by the vehicle from $t=0$ to $t=6$ seconds?
| t (s) | Velocity (m/s) at Midpoint |
|---|---|
| 1 | 10 |
| 3 | 14 |
| 5 | 18 |
90
84
96
80
The table provides values for a continuous function $$f$$ on the interval $$[0, 6]$$.
| $$x$$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| $$f(x)$$ | 0 | 5 | 2 | -1 | -2 | 0 | 3 |
What is the midpoint Riemann sum approximation of $$\displaystyle\int_{0}^{6} f(x) dx$$ using the three subintervals $$[0,2]$$, $$[2,4]$$, and $$[4,6]$$?
8
0
3
6
Which of the following expressions represents $\displaystyle\int_{2}^{5}$ √x dx as the limit of a midpoint Riemann sum with n equal subintervals?
lim(n→∞) Σ(i=0 to n-1) √(2 + i·3/n)·(3/n)
lim(n→∞) Σ(i=1 to n) (2 + i·3/n)·(3/n)
lim(n→∞) Σ(i=1 to n) √(2 + i·3/n)·(3/n)
lim(n→∞) Σ(i=1 to n) √(2 + (i-1/2)·3/n)·(3/n)
Let $$f$$ be a continuous function. If the midpoint Riemann sum with $$n = 4$$ subintervals for $$\int_{1}^{3} f(x) dx$$ is 12, which of the following statements must be true?
The average value of $$f$$ on $$[1,3]$$ is 6
The integral is approximately 12, but the actual value may differ depending on $$f(x)$$
$$f(2) = 6$$
$$\int_{1}^{3} f(x) dx = 12$$
All of the following statements about midpoint Riemann sums are true EXCEPT:
In a midpoint Riemann sum, the function is evaluated at the left endpoint of each subinterval.
The midpoint of a subinterval is calculated as $$\frac{a_i + a_{i+1}}{2}$$.
Using the function value at the midpoint can reduce error when the function behaves approximately linearly over the subinterval.
Midpoint Riemann sums often yield a more accurate approximation than left or right Riemann sums.
Which of the following statements about midpoint Riemann sums is false?
The midpoint Riemann sum always provides a less accurate approximation than either the left or right Riemann sums.
Increasing the number of subintervals in the midpoint approximation generally enhances its accuracy.
For many functions, the midpoint rule yields an approximation closer to the actual integral than the left or right rules.
The midpoint Riemann sum uses the value of the function at the center of each subinterval.
Consider the approximation of $$\int_{1}^{3} \frac{1}{x^2} dx$$ with $$n$$ uniform subintervals. Let $$E_T$$ be the error resulting from a trapezoidal sum and $$E_M$$ be the error resulting from a midpoint Riemann sum. Which of the following equations correctly relates $$E_T$$ and $$E_M$$?
E_T = 2E_M
E_T = -2E_M
E_T = -E_M/2
E_T = -E_M