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Calculus 2 : Integrals

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Example Questions

Example Question #1 : Riemann Sums

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 into sub-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 units apart. Since we're looking for the Left Riemann Sum, we want to find the heights of 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 #1 : Riemann Sums

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:

$\frac{17}{2}$

$\frac{23}{2}$

$\frac{21}{2}$

$\frac{19}{2}$

$\frac{25}{2}$

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 into sub-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 unit apart. Since we're looking for the Left Riemann Sum, we want to find the heights of 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 #53 : Integrals

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 into sub-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 #54 : Integrals

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 into sub-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 #55 : 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 into sub-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 #2 : 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:

$\frac{122}{12}$

$\frac{123}{12}$

$\frac{125}{12}$

$\frac{124}{12}$

$\frac{126}{12}$

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 into sub-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 unit apart. Since we're looking for the Right Riemann Sum of $y=\frac{5}{x}$ , we want to find the heights of 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 #57 : Integrals

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 into sub-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 units apart. Since we're looking for the Left Riemann Sum, we want to find the heights of 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 #58 : Integrals

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:

$\frac{11}{2}$

$\frac{5}{2}$

$\frac{3}{2}$

$\frac{7}{2}$

$\frac{9}{2}$

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 into sub-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 units apart. Since we're looking for the Left Riemann Sum, we want to find the heights of 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 #1 : 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:

$\frac{17}{3}$

$\frac{13}{3}$

$\frac{11}{3}$

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 into sub-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 units apart. Since we're looking for the Left Riemann Sum, we want to find the heights of 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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Example Question #60 : Integrals

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

Possible Answers:

120

240

180

Correct answer:

Explanation:

$f(0)=-6(0)^{2}=0$

$f(2)=-6(2)^{2}=-24$

$f(4)=-6(4)^{2}=-96$

Putting it all together, the Left Riemann Sum is

A=bh=2(0+(-24)+(-96))=2(-120)=-240 .

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Riemann Sums

Example Question #1 : Riemann Sums

Example Question #53 : Integrals

Example Question #54 : Integrals

Example Question #55 : Integrals

Example Question #2 : Riemann Sums

Example Question #57 : Integrals

Example Question #58 : Integrals

Example Question #1 : Riemann Sums

Example Question #60 : Integrals

All Calculus 2 Resources

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