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Calculus 2 : Riemann Sums

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Example Question #1 :Riemann Sums

Find the Left Riemann sum of the function

f(x)=x+7

on the interval [0,8] divided into four sub-intervals.

Possible Answers:

100

Correct answer:

Explanation:

The interval [0, 8] divided into four sub-intervals gives rectangles with vertices of the bases at

x=0,2,4,6,8

For the Left Riemann sum, we need to find the rectangle heights which values come from the left-most function value of each sub-interval, or f(0), f(2), f(4), and f(6).

f(0) = 0+7 = 7

f(2) = 2+7 = 9

f(4) = 4+7 = 11

f(6) = 6+7 = 13

Because each sub-interval has a width of 2, the Left Riemann sum is

A = bh = 2(7+9+11+13) = 80

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Example Question #1 :Riemann Sum: Left Evaluation

Given a function $y=x^{2}$ , find the Left Riemann Sum of the function on the interval [0,6] divided into three sub-intervals.

Possible Answers:

Correct answer:

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [0,6] divided intosub-intervals, we'll be using rectangles with vertices at x=0,2,4,6 .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is b=2 because the rectangles are spaced单位分开。因为我们正在寻找Riem左边ann Sum, we want to find the heightsof each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

$f(0)=(0)^{2}=0$

$f(2)=(2)^{2}=4$

$f(4)=(4)^{2}=16$

Putting it all together, the Left Riemann Sum is

A=bh=2(0+4+16)=2(20)=40 .

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Example Question #3 :Numerical Approximations To Definite Integrals

Given a function $y=\frac{1}{4}x+2$ , find the Left Riemann Sum of the function on the interval [0,4] divided into four sub-intervals.

Possible Answers:

$\frac{17}{2}$

$\frac{21}{2}$

$\frac{25}{2}$

$\frac{23}{2}$

$\frac{19}{2}$

Correct answer:

$\frac{19}{2}$

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [0,4] divided intosub-intervals, we'll be using rectangles with vertices at x=0,1,2,3,4 .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is b=1 because the rectangles are spacedunit apart. Since we're looking for the Left Riemann Sum, we want to find the heightsof each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

$f(0)=\frac{1}{4}(0)+2=0+2=2$

$f(1)=\frac{1}{4}(1)+2=\frac{1}{4}+\frac{8}{4}=\frac{9}{4}$

$f(2)=\frac{1}{4}(2)+2=\frac{1}{2}+\frac{4}{2}=\frac{5}{2}$

$f(3)=\frac{1}{4}(3)+2=\frac{3}{4}+\frac{8}{4}=\frac{11}{4}$

Putting it all together, the Left Riemann Sum is

$A=bh=1(2+\frac{9}{4}+\frac{5}{2}+\frac{11}{4})=\frac{8}{4}+\frac{9}{4}+\frac{10}{4}+\frac{11}{4}=\frac{38}{4}=\frac{19}{2}$

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Example Question #1 :Riemann Sums

Given a function y=3x , find the Left Riemann Sum of the function on the interval [0,3] divided into three sub-intervals.

Possible Answers:

Correct answer:

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [0,3] divided intosub-intervals, we'll be using rectangles with vertices at x=0,1,2,3 .

f(0)=3(0)=0

f(1)=3(1)=3

f(2)=3(2)=6

Putting it all together, the Left Riemann Sum is

A=bh=1(0+3+6)=9

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Example Question #2 :Riemann Sums

Given a function $y=\frac{1}{2}x^{2}$ , find the Left Riemann Sum of the function on the interval [0,8] divided into four sub-intervals.

Possible Answers:

Correct answer:

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [0,8] divided intosub-intervals, we'll be using rectangles with vertices at x=0,2,4,6,8 .

$f(0)=\frac{1}{2}(0)^{2}=0$

$f(2)=\frac{1}{2}(2)^{2}=2$

$f(4)=\frac{1}{2}(4)^{2}=8$

$f(6)=\frac{1}{2}(6)^{2}=18$

Putting it all together, the Left Riemann Sum is

A=bh=2(0+2+8+18)=2(28)=56 .

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Example Question #51 :Introduction To Integrals

Given a function y=3x+6 , find the Left Riemann Sum of the function on the interval [0,3] divided into three sub-intervals.

Possible Answers:

Correct answer:

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [0,3] divided intosub-intervals, we'll be using rectangles with vertices at x=0,1,2,3 .

f(0)=3(0)+6=0+6=6

f(1)=3(1)+6=3+6=9

f(2)=3(2)+6=6+6=12

Putting it all together, the Left Riemann Sum is

A=bh=1(6+9+12)=15+12=27 .

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Example Question #4 :Riemann Sums

Given a function $y=\frac{5}{x}$ , find the Right Riemann Sum of the function on the interval [0,4] divided into four sub-intervals.

Possible Answers:

$\frac{125}{12}$

$\frac{123}{12}$

$\frac{126}{12}$

$\frac{124}{12}$

$\frac{122}{12}$

Correct answer:

$\frac{125}{12}$

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [0,4] divided intosub-intervals, we'll be using rectangles with vertices at x=0,1,2,3,4 .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is b=1 because the rectangles are spacedunit apart. Since we're looking for the Right Riemann Sum of $y=\frac{5}{x}$ , we want to find the heightsof each rectangle by taking the values of each rightmost function value on each sub-interval, as follows:

$f(4)=\frac{5}{4}$

$f(3)=\frac{5}{3}$

$f(2)=\frac{5}{2}$

$f(1)=\frac{5}{1}=5$

Putting it all together, the Right Riemann Sum is

$A=bh=1(\frac{5}{4}+\frac{5}{3}+\frac{5}{2}+5)=\frac{15}{12}+\frac{20}{12}+\frac{30}{12}+\frac{60}{12}=\frac{125}{12}$ .

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Example Question #5 :Riemann Sums

Given a function $y=x^{3}$ , find the Left Riemann Sum of the function on the interval [0,4] divided into four sub-intervals.

Possible Answers:

Correct answer:

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [0,4] divided intosub-intervals, we'll be using rectangles with vertices at x=0,1,2,3,4 .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is b=1 because the rectangles are spaced单位分开。因为我们正在寻找Riem左边ann Sum, we want to find the heightsof each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

$f(0)=(0)^{3}=0$

$f(1)=(1)^{3}=1$

$f(2)=(2)^{3}=8$

$f(3)=(3)^{3}=27$

Putting it all together, the Left Riemann Sum is

A=bh=1(0+1+8+27)=36 .

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Example Question #6 :Riemann Sums

Given a function $y=\frac{x^{2}}{2}$ , find the Left Riemann Sum of the function on the interval [0,3] divided into three sub-intervals.

Possible Answers:

$\frac{9}{2}$

$\frac{11}{2}$

$\frac{7}{2}$

$\frac{3}{2}$

$\frac{5}{2}$

Correct answer:

$\frac{5}{2}$

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [0,3] divided intosub-intervals, we'll be using rectangles with vertices at x=0,1,2,3 .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is b=1 because the rectangles are spaced单位分开。因为我们正在寻找Riem左边ann Sum, we want to find the heightsof each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

$f(0)=\frac{(0)^{2}}{2}=0$

$f(1)=\frac{(1)^{2}}{2}=\frac{1}{2}$

$f(2)=\frac{(2)^{2}}{2}=\frac{4}{2}=2$

Putting it all together, the Left Riemann Sum is

$A=bh=1(0+\frac{1}{2}+2)=\frac{5}{2}$ .

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Example Question #7 :Riemann Sums

Given a function $y=\frac{2}{x}$ , find the Left Riemann Sum of the function on the interval [1,4] divided into three sub-intervals.

Possible Answers:

$\frac{17}{3}$

$\frac{11}{3}$

$\frac{13}{3}$

Correct answer:

$\frac{11}{3}$

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [1,4] divided intosub-intervals, we'll be using rectangles with vertices at x=1,2,3,4 .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is b=1 because the rectangles are spaced单位分开。因为我们正在寻找Riem左边ann Sum, we want to find the heightsof each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

$f(1)=\frac{2}{1}=2$

$f(2)=\frac{2}{2}=1$

$f(3)=\frac{2}{3}$

Putting it all together, the Left Riemann Sum is

$A=bh=1(2+1+\frac{2}{3})=3+\frac{2}{3}=\frac{9}{3}+\frac{2}{3}=\frac{11}{3}$ .

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Calculus 2 : Riemann Sums

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 :Riemann Sums

Example Question #1 :Riemann Sum: Left Evaluation

Example Question #3 :Numerical Approximations To Definite Integrals

Example Question #1 :Riemann Sums

Example Question #2 :Riemann Sums

Example Question #51 :Introduction To Integrals

Example Question #4 :Riemann Sums

Example Question #5 :Riemann Sums

Example Question #6 :Riemann Sums

Example Question #7 :Riemann Sums

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