Accumulation & Riemann Sums Assessment Test

1) Integrate and solve ∫ cos(x) dx from x=-11 to x=4.

3.14

-1.757

-3.14

4.12

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.

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.

2) Integrate and solve ∫ 2x-cosx dx from x=1 to x=6.

36.121

24.75

-36.121

-24.75

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.

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.

3) Integrate ∫ 3x+2 dx from x=1 to x=4.

28.5

28.6

28

28.7

The given question asks to find the definite integral of the function 3x+2 from x=1 to x=4. To find the definite integral, we need to find the antiderivative of the function and evaluate it at the upper limit (4) and subtract the value at the lower limit (1). The antiderivative of 3x+2 is (3/2)x^2+2x. Evaluating this at x=4 gives (3/2)(4)^2+2(4) = 24+8 = 32. Evaluating at x=1 gives (3/2)(1)^2+2(1) = 3/2 + 2 = 7/2. Subtracting the value at the lower limit from the value at the upper limit gives 32 - 7/2 = 57/2 = 28.5. Therefore, the correct answer is 28.5.

Explanation

The given question asks to find the definite integral of the function 3x+2 from x=1 to x=4. To find the definite integral, we need to find the antiderivative of the function and evaluate it at the upper limit (4) and subtract the value at the lower limit (1). The antiderivative of 3x+2 is (3/2)x^2+2x. Evaluating this at x=4 gives (3/2)(4)^2+2(4) = 24+8 = 32. Evaluating at x=1 gives (3/2)(1)^2+2(1) = 3/2 + 2 = 7/2. Subtracting the value at the lower limit from the value at the upper limit gives 32 - 7/2 = 57/2 = 28.5. Therefore, the correct answer is 28.5.

4) Integrate and solve ∫2x dx from intervals x=1 to x=3.

-12

12

-8

8

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.

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.

5) Integrate f(x)= x^3+4 from x=-2 to x=2 using left riemann sums with four equal subintervals.

6

7

8

9

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.

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.

6) Integrate and solve ∫ cos x dx from x=π to x=3π.

0

1

2

3

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.

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.

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

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.

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.

8) Integrate and solve ∫-sin(x)^2-cos(x) from x=12 to x=12.

-2

-1

0

1

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.

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.

9) Integrate and solve ∫3x^5 dx from x=-5 to x=1.

-42

42

7821

-7821

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.

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.

10) Approximate the area under the curve of ∫ cos(x) dx from x=1 to x=5.

-3.14

3.14

-1.8004

1.8004

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.

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.