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

Card 0 of 297

Question

Use Left Riemann sums with 4 subintervals to approximate the area between the x-axis, , , and .

Answer

To use left Riemann sums, we need to use the following formula:

$Area \approx \sum_{i=1}^n f(x_i)\Delta x$ .

where is the number of subintervals, (4 in our problem),

is the "counter" that denotes which subinterval we are working with,(4 subintervals mean that will be 1, 2, 3, and then 4)

is the function value when you plug in the "i-th" x value, (i-th in this case will be 1-st, 2-nd, 3-rd, and 4-th)

, $\Delta x$ is the width of each subinterval, which we will determine shortly.

and $\sum_{i=1}^n$ means add all versions together (for us that means add up 4 versions).

This fancy equation approximates using boxes. We can rewrite this fancy equation by writing $f(x_i)\Delta x$ , 4 times; 1 time each for , , , and . This gives us

$f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x$

Think of $\Delta x$ as the base of each box, and as the height of the 1st box.

This is basically , 4 times, and then added together.

Now we need to determine what $\Delta x, x_1, x_2, x_3,$ and are.

To find $\Delta x$ we find the total length between the beginning and ending x values, which are given in the problem as and . We then split this total length into 4 pieces, since we are told to use 4 subintervals.

In short, $\Delta x = \frac{b - a}{n}= \frac{(2)- (-2)}{4} = 1$

Now we need to find the x values that are the left endpoints of each of the 4 subintervals. Left endpoints because we are doing Left Riemann sums.

The left most x value happens to be the smaller of the overall endpoints given in the question. In other words, since we only care about the area from to , we'll just use the smaller one, , for our first .

Now we know that .

To find the next endpoint, , just increase the first x by the length of the subinterval, which is $\Delta x = 1$ . In other words

$x_2 = x_1 + \Delta x$

Add the $\Delta x$ again to get

$x_3 = x_2 + \Delta x$

And repeat to find

$x_4 = x_3 + \Delta x$

Now that we have all the pieces, we can plug them in.

$f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x$

$f(-2)\cdot1 + f(-1) \cdot (1)+ f(0)\cdot (1) + f(1)\cdot (1)$

plug each value into and then simplify.

This is the final answer, which is an approximation of the area under the function.

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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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