Approximating Areas With Riemann Sums
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AP Calculus AB › Approximating Areas With Riemann Sums
Let $$g$$ be a function that is strictly decreasing on the interval $$a, b$$. Which of the following statements provides the best comparison between the right Riemann sum approximation ($$R_n$$) and the true value of the integral $$\int_a^b g(x) dx$$?
$$R_n$$ is an overestimate of the integral.
$$R_n$$ is equal to the value of the integral.
The relationship cannot be determined from the given information.
$$R_n$$ is an underestimate of the integral.
Explanation
For a strictly decreasing function on a given subinterval, the function's value at the right endpoint is the minimum value on that subinterval. Therefore, a right Riemann sum, which uses the function values at the right endpoints, will be an underestimate of the true value of the integral.
Let $$f$$ be a function such that $$f'(x) < 0$$ and $$f''(x) < 0$$ for all $$x$$ in the interval $$a, b$$. Let $$I = \int_a^b f(x) dx$$. For a given number of subintervals $$n$$, which of the following must be true about the left Riemann sum ($$L_n$$) and trapezoidal sum ($$T_n$$) approximations?
$$T_n < I < L_n$$
$$T_n < L_n < I$$
$$I < T_n < L_n$$
$$L_n < I < T_n$$
Explanation
Since $$f'(x) < 0$$, the function $$f$$ is decreasing. For a decreasing function, the left Riemann sum $$L_n$$ is an overestimate of the integral $$I$$. Since $$f''(x) < 0$$, the function is concave down. For a concave down function, the trapezoidal sum $$T_n$$ is an underestimate of the integral $$I$$. Combining these facts, we have $$T_n < I < L_n$$.
The approximation for $$\int_a^b f(x) dx$$ using a left Riemann sum is $$A_L$$ and using a right Riemann sum is $$A_R$$. If the trapezoidal approximation is $$A_T$$, which of the following gives an expression for $$A_T$$ in terms of $$A_L$$ and $$A_R$$, assuming equal subintervals?
$$\sqrt{A_L A_R}$$
$$A_R - A_L$$
$$A_L + A_R$$
$$\frac{A_L + A_R}{2}$$
Explanation
For equal subintervals of width $$\Delta x$$, the left Riemann sum is $$A_L = \Delta x \sum_{i=0}^{n-1} f(x_i)$$ and the right Riemann sum is $$A_R = \Delta x \sum_{i=1}^{n} f(x_i)$$. The trapezoidal sum is $$A_T = \frac{\Delta x}{2} \sum_{i=1}^{n} (f(x_{i-1}) + f(x_i)) = \frac{\Delta x}{2} [ (f(x_0)+f(x_1)) + ... + (f(x_{n-1})+f(x_n)) ] = \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + ... + 2f(x_{n-1}) + f(x_n)] = \frac{1}{2} [\Delta x(f(x_0)+...+f(x_{n-1})) + \Delta x(f(x_1)+...+f(x_n))] = \frac{A_L+A_R}{2}$$.
Using a left Riemann sum with three subintervals of equal width, what is the approximation of the total volume of water that flowed into the reservoir during the 6-hour period?
420 cubic meters
400 cubic meters
200 cubic meters
540 cubic meters
Explanation
The total volume is approximated by the integral of the rate, $$\int_0^6 R(t) dt$$. The interval is $$[0, 6]$$ with 3 subintervals, so the width is $$\Delta t = \frac{6-0}{3}=2$$ hours. A left Riemann sum uses the left endpoints $$t=0, 2, 4$$. The approximation is $$\Delta t [R(0) + R(2) + R(4)] = 2[50+70+80] = 2[200] = 400$$. The units are (hours) * (cubic meters/hour) = cubic meters.
The values of a continuous function $$f$$ for selected values of $$x$$ are given as follows: $$f(0)=5, f(1)=8, f(2)=13, f(3)=20$$. What is the value of a right Riemann sum approximation of $$\int_0^3 f(x) dx$$ with 3 equal subintervals?
26
36
41
46
Explanation
The interval of integration is $$[0, 3]$$ with $$n=3$$ subintervals, so the width of each subinterval is $$\Delta x = \frac{3-0}{3} = 1$$. The subintervals are $$[0,1], [1,2], [2,3]$$. A right Riemann sum uses the right endpoints $$x=1, 2, 3$$. The approximation is $$\Delta x [f(1) + f(2) + f(3)] = 1[8 + 13 + 20] = 41$$.
Using the given values, approximate the definite integral $$\int_1^7 g(x) dx$$ with a right Riemann sum using 3 subintervals of equal width.
35
50
60
70
Explanation
The interval is $$[1, 7]$$ with $$n=3$$ subintervals, so the width of each subinterval is $$\Delta x = \frac{7-1}{3} = 2$$. The subintervals are $$[1,3], [3,5], [5,7]$$. For a right Riemann sum, we use the right endpoints $$3, 5, 7$$. The approximation is $$\Delta x [g(3) + g(5) + g(7)] = 2[8 + 12 + 15] = 2[35] = 70$$.
Use a midpoint Riemann sum with 2 equal subintervals to approximate the value of $$\int_0^2 x^3 dx$$.
1.0
3.5
5.0
9.0
Explanation
The interval is $$[0, 2]$$ with $$n=2$$ subintervals, so the width is $$\Delta x = \frac{2-0}{2} = 1$$. The subintervals are $$[0,1]$$ and $$[1,2]$$. The midpoints are $$0.5$$ and $$1.5$$. The midpoint Riemann sum is $$\Delta x [f(0.5) + f(1.5)] = 1[(0.5)^3 + (1.5)^3] = 1[0.125 + 3.375] = 3.5$$.
Let $$f$$ be a function that is strictly increasing on the interval $$a, b$$. Which of the following statements must be true about the approximations for $$\int_a^b f(x) dx$$ using a left Riemann sum ($$L_n$$) and a right Riemann sum ($$R_n$$) with $$n$$ subintervals?
$$L_n < \int_a^b f(x) dx < R_n$$
$$R_n < \int_a^b f(x) dx < L_n$$
$$\int_a^b f(x) dx < L_n < R_n$$
$$L_n < R_n < \int_a^b f(x) dx$$
Explanation
For a strictly increasing function on a given subinterval, the minimum value occurs at the left endpoint and the maximum value occurs at the right endpoint. Therefore, a left Riemann sum ($$L_n$$) will be an underestimate of the true integral, and a right Riemann sum ($$R_n$$) will be an overestimate. This leads to the inequality $$L_n < \int_a^b f(x) dx < R_n$$.
The function $$h(x)$$ is twice differentiable and $$h''(x) > 0$$ for all $$x$$ in the interval $$0, 4$$. Let $$T_4$$ be the trapezoidal sum approximation with 4 equal subintervals for $$\int_0^4 h(x) dx$$. Which statement about $$T_4$$ must be true?
The relationship cannot be determined without knowing if $$h(x)$$ is increasing or decreasing.
$$T_4$$ is an overestimate of $$\int_0^4 h(x) dx$$.
$$T_4$$ is equal to $$\int_0^4 h(x) dx$$.
$$T_4$$ is an underestimate of $$\int_0^4 h(x) dx$$.
Explanation
The condition $$h''(x) > 0$$ means that the graph of $$h(x)$$ is concave up. For a concave up function, the secant line segment connecting the endpoints of any subinterval lies above the curve. Since the trapezoidal rule uses these secant line segments to form the tops of the trapezoids, the area of each trapezoid is greater than the area under the curve on that subinterval. Thus, the trapezoidal sum $$T_4$$ is an overestimate.
The function $$f$$ is continuous on $$0, 6$$. Using three subintervals of equal width and right endpoints, the right Riemann sum approximation for $$\int_0^6 f(x) dx$$ is calculated. Which of the following expressions represents this approximation?
$$3(f(2)+f(4)+f(6))$$
$$2(f(0)+f(2)+f(4))$$
$$2(f(2)+f(4)+f(6))$$
$$3(f(0)+f(3)+f(6))$$
Explanation
The interval is $$[0, 6]$$ with $$n=3$$ subintervals of equal width. The width of each subinterval is $$\Delta x = \frac{6-0}{3} = 2$$. The subintervals are $$[0,2], [2,4], [4,6]$$. For a right Riemann sum, we use the function values at the right endpoints of these subintervals, which are $$x=2, x=4, x=6$$. The approximation is given by the sum of the areas of the three rectangles: $$f(2) \cdot \Delta x + f(4) \cdot \Delta x + f(6) \cdot \Delta x = \Delta x (f(2)+f(4)+f(6)) = 2(f(2)+f(4)+f(6))$$.