Accumulation & Riemann Sums Assessment Test

Reviewed by Editorial Team

The ProProfs editorial team is comprised of experienced subject matter experts. They've collectively created over 10,000 quizzes and lessons, serving over 100 million users. Our team includes in-house content moderators and subject matter experts, as well as a global network of rigorously trained contributors. All adhere to our comprehensive editorial guidelines, ensuring the delivery of high-quality content.

Learn about Our Editorial Process

| By Cripstwick

Cripstwick

Community Contributor

Quizzes Created: 636 | Total Attempts: 823,595

| Attempts: 133

Questions

Feedback

During the Quiz End of Quiz

Difficulty

Sequential Easy First Hard First

Share on Facebook Share on Twitter Share on Whatsapp Share on Pinterest Share on Email Copy to Clipboard Embed on your website

1/10 Questions

Integrate ∫ 3x+2 dx from x=1 to x=4.
- 28.5
- 28.6
- 28
- 28.7

About This Quiz

Accumulation and Riemann sums are used when the definite integral represents the area under the curve of a function between two vertical lines but we can't go find this area easily, so we can estimate it using simpler shapes like rectangles.

Accumulation & Riemann Sums Assessment Test - Quiz

2.

Approximate the area under the curve of ∫ cos(x) dx from x=1 to x=5.
- -3.14
- 3.14
- -1.8004
- 1.8004
Correct Answer

A. -1.8004

Explanation

The approximate area under the curve of the function cos(x) from x=1 to x=5 is -1.8004. This is calculated using numerical integration methods, such as the trapezoidal rule or Simpson's rule. These methods divide the interval into smaller sub-intervals, approximate the curve within each sub-interval as a straight line or a polynomial, and then sum up the areas of the trapezoids or the Simpson's rule approximations. The result is an approximation of the area under the curve, which in this case is -1.8004.

Rate this question:
3.

Integrate f(x)= x^3+4 from x=-2 to x=2 using left riemann sums with four equal subintervals.
- 6
- 7
- 8
- 9
Correct Answer

A. 8

Explanation

The left Riemann sum approximates the area under the curve by using the left endpoint of each subinterval to determine the height of the rectangle. In this case, we have four equal subintervals: [-2, -1], [-1, 0], [0, 1], and [1, 2]. Evaluating the function at the left endpoints of these intervals gives us the values: f(-2) = -4, f(-1) = 3, f(0) = 4, and f(1) = 5. The width of each rectangle is 1, since each subinterval has a width of 1. Calculating the area of each rectangle and summing them up gives us 8, which is the approximate value of the integral.

Rate this question:
4.

Integrate and solve ∫ cos(x) dx from x=-11 to x=4.
- 3.14
- -1.757
- -3.14
- 4.12
Correct Answer

A. -1.757

Explanation

The given question asks to integrate the function cos(x) with respect to x from x=-11 to x=4. The integral of cos(x) is sin(x), so we need to evaluate sin(x) from x=-11 to x=4. The value of sin(x) at x=-11 is -0.999 and at x=4 is -0.757. Subtracting the two values, we get -0.757 - (-0.999) = -0.757 + 0.999 = 0.242. Therefore, the correct answer is not available.

Rate this question:
5.

Integrate and solve ∫ 2x-cosx dx from x=1 to x=6.
- 36.121
- 24.75
- -36.121
- -24.75
Correct Answer

A. 36.121

Explanation

The given question asks us to find the definite integral of the function 2x - cosx with respect to x, from x=1 to x=6. By integrating the function and evaluating it at the upper and lower limits, we find that the result is 36.121.

Rate this question:
6.

Integrate and solve ∫ cos x dx from x=π to x=3π.
- 0
- 1
- 2
- 3
Correct Answer

A. 0

Explanation

The integral of cos x is sin x. Evaluating the integral from x=π to x=3π gives sin(3π) - sin(π). Since sin(3π) and sin(π) are both equal to 0, the result is 0.

Rate this question:
7.

Integrate and solve ∫1/x +2 dx from x=2 to x=5.
- 1n (5/2) +6
- -1n (5/2) +6
- 1n (3) +2
- -1n (3) +2
Correct Answer

A. 1n (5/2) +6

Explanation

The given integral is ∫(1/x + 2) dx from x=2 to x=5. Integrating 1/x with respect to x gives ln|x|, so integrating 1/x + 2 gives ln|x| + 2x. Evaluating this expression from x=2 to x=5 gives ln(5) + 2(5) - ln(2) - 2(2) = ln(5) + 10 - ln(2) - 4 = ln(5/2) + 6.

Rate this question:
8.

Integrate and solve ∫-sin(x)^2-cos(x) from x=12 to x=12.
- -2
- -1
- 0
- 1
Correct Answer

A. 0

Explanation

The integral of -sin(x)^2 is -1/2 * (x - sin(x)cos(x)), and the integral of -cos(x) is -sin(x). Evaluating the integral from x=12 to x=12, we get (-1/2 * (12 - sin(12)cos(12)) - sin(12)) - (-1/2 * (12 - sin(12)cos(12)) - sin(12)). Simplifying this expression, we get 0.

Rate this question:
9.

Integrate and solve ∫3x^5 dx from x=-5 to x=1.
- -42
- 42
- 7821
- -7821
Correct Answer

A. -7821

Explanation

The given question asks us to find the definite integral of the function 3x^5 from x=-5 to x=1. To solve this, we can use the power rule of integration, which states that the integral of x^n is (x^(n+1))/(n+1). Applying this rule, we get (3/6)(1^6 - (-5)^6) = (1/2)(1 - 15625) = (1/2)(-15624) = -7821. Therefore, the correct answer is -7821.

Rate this question:
10.

Integrate and solve ∫2x dx from intervals x=1 to x=3.
- -12
- 12
- -8
- 8
Correct Answer

A. 8

Explanation

The given question asks to integrate the function 2x with respect to x, and then evaluate the result within the interval from x=1 to x=3. The integral of 2x is x^2, so we can evaluate the definite integral by substituting the upper limit (x=3) and the lower limit (x=1) into the antiderivative. When we substitute x=3, we get 3^2=9, and when we substitute x=1, we get 1^2=1. Therefore, the result of the definite integral is 9-1=8.

Rate this question: