Riemann sums (left, right, and midpoint evaluation points) - AP Calculus AB

Card 0 of 20

Question

Estimate the integral of from 0 to 3 using left-Riemann sum and 6 rectangles. Use $\Delta x=0.5$

Answer

Because our $\Delta x$ is constant, the left Riemann sum will be

$R_{L}=\Delta x(f(0)+f(0.5)+f(1)+f(1.5)+f(2)+f(2.5))$

$R_{L}=0.5\cdot (1+1.75+4+7.75+13+19.75)=23.625$

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Question

$\begin{align}&\text{The function }f(x)\text{ has the following values on the interval }x=[-2,13]:\&f(-2)=5\&f(1)=-19\&f(4)=5\&f(7)=-6\&f(10)=-18\&f(13)=0\&\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-2)=5\&f(1)=-19\&f(4)=5\&f(7)=-6\&f(10)=-18\&f(13)=0\&\&\text{Although the total interval has a length of }15\&\text{Notice the points are equidistant. This distance is our subinterval }3\&\text{and since we are using the right point of each interval, we disregard the first function value:}\&\int_{-2}^{13}f(x)\approx3[(-19)+(5)+(-6)+(-18)+(0)]\&\int_{-2}^{13}f(x)\approx-114\end{align}$

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Question

$\begin{align}&\text{Approximate }\int_{-8}^{1}f(x)\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\&f(-8)=-13\&f(-5)=-19\&f(-2)=6\&f(1)=-9\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-8)=-13\&f(-5)=-19\&f(-2)=6\&f(1)=-9\&\text{Although the total interval has a length of }9\&\text{Notice the points are equidistant. This distance is our subinterval: }3\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{-8}^{1}f(x)\approx3[(-19)+(6)+(-9)]\&\int_{-8}^{1}f(x)\approx-66\end{align}$

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Question

$\begin{align}&\text{Approximate }\int_{-2}^{23}f(x)\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\&f(-2)=-2\&f(3)=-15\&f(8)=14\&f(13)=15\&f(18)=-9\&f(23)=-12\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-2)=-2\&f(3)=-15\&f(8)=14\&f(13)=15\&f(18)=-9\&f(23)=-12\&\text{Although the total interval has a length of }25\&\text{Notice the points are equidistant, with }5\text{ intervals between. Each equidistant interval has length: }5\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{-2}^{23}f(x)\approx5[(-15)+(14)+(15)+(-9)+(-12)]\&\int_{-2}^{23}f(x)\approx-35\end{align}$

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Question

$\begin{align}&\text{Approximate }\int_{-7}^{13}f(x)\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\&f(-7)=18\&f(-2)=-17\&f(3)=-16\&f(8)=-15\&f(13)=-14\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-7)=18\&f(-2)=-17\&f(3)=-16\&f(8)=-15\&f(13)=-14\&\text{Although the total interval has a length of }20\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\&\text{Each equidistant interval has length: }5\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{-7}^{13}f(x)\approx5[(-17)+(-16)+(-15)+(-14)]\&\int_{-7}^{13}f(x)\approx-310\end{align}$

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Question

$\begin{align}&f(4)=9\&f(6)=10\&f(8)=-18\&f(10)=15\&f(12)=18\&\text{Approximate, using a right Riemann sum, }\int_{4}^{12}f(x)\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(4)=9\&f(6)=10\&f(8)=-18\&f(10)=15\&f(12)=18\&\text{Although the total interval has a length of }8\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\&\text{Each equidistant interval has length: }2\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{4}^{12}f(x)\approx2[(10)+(-18)+(15)+(18)]\&\int_{4}^{12}f(x)\approx50\end{align}$

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Question

$\begin{align}&\text{The function }f(x)\text{ has the following values on the interval }x=[-2,38]:\&f(-2)=-13\&f(6)=-4\&f(14)=-15\&f(22)=-19\&f(30)=18\&f(38)=-8\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-2)=-13\&f(6)=-4\&f(14)=-15\&f(22)=-19\&f(30)=18\&f(38)=-8\&\text{Although the total interval has a length of }40\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\&\text{Each equidistant interval has length: }8\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{-2}^{38}f(x)\approx8[(-4)+(-15)+(-19)+(18)+(-8)]\&\int_{-2}^{38}f(x)\approx-224\end{align}$

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Question

$\begin{align}&\text{Approximate }\int_{0}^{15}f(x)\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\&f(0)=-19\&f(5)=14\&f(10)=2\&f(15)=15\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(0)=-19\&f(5)=14\&f(10)=2\&f(15)=15\&\text{Although the total interval has a length of }15\&\text{, notice the points are equidistant, with }3\text{ intervals between.}\&\text{Each equidistant interval has length: }5\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{0}^{15}f(x)\approx5[(14)+(2)+(15)]\&\int_{0}^{15}f(x)\approx155\end{align}$

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Question

$\begin{align}&\text{Approximate }\int_{-8}^{0}f(x)\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\&f(-8)=7\&f(-6)=-7\&f(-4)=16\&f(-2)=-16\&f(0)=20\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-8)=7\&f(-6)=-7\&f(-4)=16\&f(-2)=-16\&f(0)=20\&\text{Although the total interval has a length of }8\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\&\text{Each equidistant interval has length: }2\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{-8}^{0}f(x)\approx2[(-7)+(16)+(-16)+(20)]\&\int_{-8}^{0}f(x)\approx26\end{align}$

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Question

$\begin{align}&\text{Approximate }\int_{-6}^{39}f(x)\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\&f(-6)=-4\&f(3)=-1\&f(12)=11\&f(21)=13\&f(30)=-16\&f(39)=-13\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-6)=-4\&f(3)=-1\&f(12)=11\&f(21)=13\&f(30)=-16\&f(39)=-13\&\text{Although the total interval has a length of }45\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\&\text{Each equidistant interval has length: }9\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{-6}^{39}f(x)\approx9[(-1)+(11)+(13)+(-16)+(-13)]\&\int_{-6}^{39}f(x)\approx-54\end{align}$

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Question

$\begin{align}&\text{Approximate }\int_{-5}^{13}f(x)\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\&f(-5)=-13\&f(1)=-12\&f(7)=17\&f(13)=7\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-5)=-13\&f(1)=-12\&f(7)=17\&f(13)=7\&\text{Although the total interval has a length of }18\&\text{, notice the points are equidistant, with }3\text{ intervals between.}\&\text{Each equidistant interval has length: }6\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{-5}^{13}f(x)\approx6[(-12)+(17)+(7)]\&\int_{-5}^{13}f(x)\approx72\end{align}$

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Question

$\begin{align}&\text{The function }f(x)\text{ has the following values on the interval }x=[1,11]:\&f(1)=10\&f(3)=3\&f(5)=-13\&f(7)=4\&f(9)=-8\&f(11)=-15\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(1)=10\&f(3)=3\&f(5)=-13\&f(7)=4\&f(9)=-8\&f(11)=-15\&\text{Although the total interval has a length of }10\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\&\text{Each equidistant interval has length: }2\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{1}^{11}f(x)\approx2[(3)+(-13)+(4)+(-8)+(-15)]\&\int_{1}^{11}f(x)\approx-58\end{align}$

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Question

$\begin{align}&\text{The function }f(x)\text{ has the following values on the interval }x=[-7,3]:\&f(-7)=-18\&f(-5)=-2\&f(-3)=-20\&f(-1)=16\&f(1)=-12\&f(3)=-17\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-7)=-18\&f(-5)=-2\&f(-3)=-20\&f(-1)=16\&f(1)=-12\&f(3)=-17\&\text{Although the total interval has a length of }10\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\&\text{Each equidistant interval has length: }2\&\text{Since we are using the right point of each interval, we disregard the first function value:}\&\int_{-7}^{3}f(x)\approx2[(-2)+(-20)+(16)+(-12)+(-17)]\&\int_{-7}^{3}f(x)\approx-70\end{align}$

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Question

$\begin{align}&\text{Approximate }\int_{5}^{13}f(x)\&\text{Using a left Riemann sum, if }f(x)\text{ has the values:}\&f(5)=-7\&f(7)=-8\&f(9)=-18\&f(11)=-8\&f(13)=-19\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(5)=-7\&f(7)=-8\&f(9)=-18\&f(11)=-8\&f(13)=-19\&\text{Although the total interval has a length of }8\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\&\text{Each equidistant interval has length: }2\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\&\int_{5}^{13}f(x)\approx2[(-7)+(-8)+(-18)+(-8)]\&\int_{5}^{13}f(x)\approx-82\end{align}$

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Question

$\begin{align}&\text{Approximate }\int_{-9}^{26}f(x)\&\text{Using a left Riemann sum, if }f(x)\text{ has the values:}\&f(-9)=-17\&f(-2)=11\&f(5)=17\&f(12)=1\&f(19)=-16\&f(26)=13\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-9)=-17\&f(-2)=11\&f(5)=17\&f(12)=1\&f(19)=-16\&f(26)=13\&\text{Although the total interval has a length of }35\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\&\text{Each equidistant interval has length: }7\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\&\int_{-9}^{26}f(x)\approx7[(-17)+(11)+(17)+(1)+(-16)]\&\int_{-9}^{26}f(x)\approx-28\end{align}$

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Question

$\begin{align}&f(-10)=-19\&f(-3)=7\&f(4)=4\&f(11)=1\&\text{Approximate, using a left Riemann sum, }\int_{-10}^{11}f(x)\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-10)=-19\&f(-3)=7\&f(4)=4\&f(11)=1\&\text{Although the total interval has a length of }21\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\&\text{Each equidistant interval has length: }7\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\&\int_{-10}^{11}f(x)\approx7[(-19)+(7)+(4)]\&\int_{-10}^{11}f(x)\approx-56\end{align}$

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Question

$\begin{align}&\text{Approximate }\int_{-6}^{34}f(x)\&\text{Using a left Riemann sum, if }f(x)\text{ has the values:}\&f(-6)=8\&f(2)=2\&f(10)=-4\&f(18)=-18\&f(26)=11\&f(34)=-7\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-6)=8\&f(2)=2\&f(10)=-4\&f(18)=-18\&f(26)=11\&f(34)=-7\&\text{Although the total interval has a length of }40\&\text{Notice the points are equidistant, with }5\text{ intervals between.}\&\text{Each equidistant interval has length: }8\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\&\int_{-6}^{34}f(x)\approx8[(8)+(2)+(-4)+(-18)+(11)]\&\int_{-6}^{34}f(x)\approx-8\end{align}$

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Question

$\begin{align}&\text{The function }f(x)\text{ has the following values on the interval }x=[-8,2]:\&f(-8)=2\&f(-6)=-1\&f(-4)=16\&f(-2)=12\&f(0)=10\&f(2)=-18\&\text{Approximate the integral of }f(x)\text{ on this interval using left Riemann sums.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(-8)=2\&f(-6)=-1\&f(-4)=16\&f(-2)=12\&f(0)=10\&f(2)=-18\&\text{Although the total interval has a length of }10\&\text{Notice the points are equidistant, with }5\text{ intervals between.}\&\text{Each equidistant interval has length: }2\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\&\int_{-8}^{2}f(x)\approx2[(2)+(-1)+(16)+(12)+(10)]\&\int_{-8}^{2}f(x)\approx78\end{align}$

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Question

$\begin{align}&\text{The function }f(x)\text{ has the following values on the interval }x=[5,29]:\&f(5)=8\&f(13)=-15\&f(21)=9\&f(29)=-16\&\text{Approximate the integral of }f(x)\text{ on this interval using left Riemann sums.}\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(5)=8\&f(13)=-15\&f(21)=9\&f(29)=-16\&\text{Although the total interval has a length of }24\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\&\text{Each equidistant interval has length: }8\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\&\int_{5}^{29}f(x)\approx8[(8)+(-15)+(9)]\&\int_{5}^{29}f(x)\approx16\end{align}$

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Question

$\begin{align}&f(0)=10\&f(4)=3\&f(8)=10\&f(12)=-11\&f(16)=10\&\text{Approximate, using a left Riemann sum, }\int_{0}^{16}f(x)\end{align}$

Answer

$\begin{align}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{In our case, we have function values over an interval:}\&f(0)=10\&f(4)=3\&f(8)=10\&f(12)=-11\&f(16)=10\&\text{Although the total interval has a length of }16\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\&\text{Each equidistant interval has length: }4\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\&\int_{0}^{16}f(x)\approx4[(10)+(3)+(10)+(-11)]\&\int_{0}^{16}f(x)\approx48\end{align}$

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