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This is one of our many quizzes on Riemann sums. Click on this to access more such competitive quizzes on Calculus.

Riemann sums Quiz 2

mathematics

 Find the area under the given curve $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7d0329c7-fc19-4b10-82e7-5acae802a17c5931102145538272168.png"/>

 Substitute the function directly in the left endpoint approximation to form an integral value, use the limits $a=2$ and $b=10$ to integrate.

  $\text{The table according to the graph will be formed as} \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1c7814de-5242-4e84-929e-e98946b3fc1d1961674300769284905.png"/> $ \\ \text{The equation formed will be} \int\limits_{2}^{10}{f(x)dx} \\ \text{Since} a=2 , b=10 \\ \text{According to Riemann Sums the equation formed will be} \\  \int\limits_{a}^{b}{f(x)dx\approx \Delta x(f({{x}_{0}})+f({{x}_{1}})...+f({{x}_{n-2}})+f({{x}_{n-1}}))} \\  \Delta x=\dfrac{b-a}{n} \\ \text{Since integral over right end points is asked} \\ \int\limits_{2}^{10}{f(x)dx\approx (4-(2))\times (3)+(6-(4))\times (5)+(8-(6))\times (6)+(10-(8))\times (0)=28}$ 

 An NGO decided to clean a seashore to ensure that tortoises could come and lay eggs there. The area of seashore was very huge and curved so they decided to form 3 teams. Sameeksha, a student of maths helped the NGO by telling that the fence was in shape of a curve with equation $\int\limits_{0}^{8}{2\sqrt[3]{{{x}^{7}}+1}dx}$ with limit of maximum length of the shore as 6units to scale. Since Sameeksha had to leave early, all the teams needed to cover using her information.

 n takes a value of 3$($since the curve is divided into 3 equal parts$)$, curve to find the area under is $\int\limits_{0}^{8}{2\sqrt[3]{{{x}^{7}}+1}dx}$. Use the limits $[0,8]$ to evaluate.

  Since the curve was divided into $3$ equal parts that means $n$ takes a value of $3$. Similarly the curve to find the area under is $ \\ \int\limits_{0}^{8}{2\sqrt[3]{{{x}^{7}}+1}dx} \\ $ And limits are $[0,8]$  $ \\ $ We know that according to Riemann Sums the equation formed will be $ \\   \int\limits_{a}^{b}{f(x)dx\approx \Delta x(f({{x}_{0}})+f({{x}_{1}})...+f({{x}_{n-2}})+f({{x}_{n-1}}))} \\  \Delta x=\dfrac{b-a}{n} \\ $ Since integral over right end points is asked. $ \\ $ Placing values $ \\\Delta x=2 \\ $ Dividing the interval into $3$ equal parts $ \\ a=\left[ 0,2 \right],\left[ 2,4 \right],\left[ 4,6 \right]=b \\ $ Now finding the values of function at each left endpoint $ \\ f({{x}_{0}})=f(a)=f(0)=2.00 \\  f({{x}_{1}})=f(2)=\dfrac{2\sqrt[3]{129}}{1}=10.10 \\  f({{x}_{2}})=f(4)=\dfrac{2\sqrt[3]{16385}}{1}=50.79 \\ $  Finding the sum of all values and multiplying it by $\Delta x$  $ \\ =2\times (2+10.10+50.79)=125.80$ 

 The locus of a particle is observed uniformly at the interval of 2 units starting from 3. The values obtained by the movement of the object are uniform. Starting with 3 the results increase till 12 before the lifetime of the object terminates. Find the area covered by the particle

 Use the results provided to make a table.  The lower limit of integral is 3 and upper limit of integral is 9.

Since the observations are made in every 2 units starting from 3 the value of $x$ will be $ \\ 3,3+2,3+2+2,... \\ $ The results increase from 3 to 12 so $ \\ \Rightarrow \dfrac{12}{3}=4 \\ $ Therefore, the table of the movement by the particle would look like follows $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/11723da3-111a-4155-89f4-40cc42558bf03310478440111559851.png"/>$ \\ $ Since the value of x varies from 3 to 9 $ \\ $ The lower limit of integral is 3 and upper limit of integral is 9 $ \\ $ We know that according to Riemann Sums $ \\  \int\limits_{a}^{b}{f(x)dx\approx \Delta x(f({{x}_{0}})+f({{x}_{1}})...+f({{x}_{n-2}})+f({{x}_{n-1}}))} \\  \Delta x=\dfrac{b-a}{n} \\ $ Since integral over right end points is asked $ \\ \int\limits_{3}^{9}{f(x)dx\approx (5-(3))\times (3)+(7-(5))\times (6)+(9-(7))\times (9)=36}$ 

 A jeweller wishes to find the area of a gold sheet. You find that the sheet is shaped such that three sides are a rectangle and the remaining side is a curve with equation $\cos \theta \sin \theta $. Three trapezoids to be used find the area with 0 to 6 units. What will you get as an answer? It is known that $ \\   \sin (2)\cos 2=-0.37 \\  \sin (4)\cos (4)=0.49 \\ $ 

 Divide the interval $[0,6]$ into 3 subintervals such since 3 trapezoids are used. The endpoints are $a=0,2,4,6=b$ 

 Since trapezoids are used to calculate the area, Trapezoidal rule of Riemann series is used which states that $ \\ \int\limits_{a}^{b}{f(\theta )d\theta \approx \Delta \theta \left( f\left( {{\theta }_{0}} \right)+2f\left( {{\theta }_{1}} \right)+2f\left( {{\theta }_{2}} \right)+...+2f\left( {{\theta }_{n-1}} \right)+2f\left( {{\theta }_{n}} \right) \right)} \\  \Delta \theta =\dfrac{b-a}{a} \\ $ We have $a=0,b=6,n=3$  $ \\ \Delta \theta =\dfrac{6-0}{3}=2 \\ $ Divide the interval $[0,6]$ into 3 subintervals since 3 trapezoids are used. $ \\ $  The endpoints are $a=0,2,4,6=b$ $ \\ $ Evaluating the function at these endpoints: $ \\ f\left( {{\theta }_{0}} \right)=f\left( a \right)=f\left( 0 \right)=0 \\ f\left( {{\theta }_{1}} \right)=f\left( 2 \right)=\sin (2)\cos (2)=-0.37 \\ f\left( {{\theta }_{2}} \right)=f\left( 4 \right)=\sin (4)cos(4)=0.49 \\ $ Add up the above values and multiply by $\Delta \theta $  $ \\ =2(0-0.37+0.49)=0.232$ 

 Anish wants to find the area of the base of the building as one side of the plot is in shape of a curve. He finds out the equation of the curve to be $\sqrt[7]{{{x}^{5}}+\dfrac{1}{2}}$ .He divides the base of the building into 4 equal parts and finds area for 2 unit length using midpoint of each section. What is the area of the base of the building?

 Since no lower limit is given take it as $0$. The upper limit will be $2$ and intervals will be divided in 4 parts.

 We know that according to the midpoint rule of approximation of Riemann sums $ \\ \int\limits_{a}^{b}{f(x)dx\approx \Delta x\left( f\left( \dfrac{{{x}_{0}}+{{x}_{1}}}{2} \right)+f\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)+...f\left( \dfrac{{{x}_{n-1}}+{{x}_{n}}}{2} \right) \right)} \\  \Delta x=\dfrac{b-a}{a} \\ $ From question we see that the Curve or function $ \\ \Rightarrow \sqrt[7]{{{x}^{5}}+\dfrac{1}{2}} \\ $ Value of $a$ is $0$ since lower limit is not given we take x-axis as lower limit $ \\ $ Value of $b$ is $2$ $ \\ $ Since land is divided into 4 parts, the value of $n$ is $4$ $  \\ $ Dividing the interval $[0,2]$ into 3 subintervals of length $\Delta x=\dfrac{1}{2}$  $ \\ a=\left[ 0,\dfrac{1}{2} \right],\left[ \dfrac{1}{2},1 \right],\left[ 1,\dfrac{3}{2} \right],\left[ \dfrac{3}{2},2 \right]=b \\ f\left( \dfrac{{{x}_{0}}+{{x}_{1}}}{2} \right)=f\left( \dfrac{(0)+\dfrac{1}{2}}{2} \right)=f\left( \dfrac{1}{4} \right)=\dfrac{\sqrt[\frac{4}{7}]{2}\times \sqrt[7]{513}}{4}=0.905 \\ f\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)=f\left( \dfrac{\dfrac{1}{2}+1}{2} \right)=f\left( \dfrac{3}{4} \right)=\dfrac{\sqrt[\frac{4}{7}]{2}\times \sqrt[7]{755}}{4}=0.957$  $f\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2} \right)=f\left( \dfrac{\dfrac{3}{2}+1}{2} \right)=f\left( \dfrac{5}{4} \right)=\dfrac{\sqrt[\frac{4}{7}]{2}\times \sqrt[7]{3637}}{4}=1.198 \\ f\left( \dfrac{{{x}_{3}}+{{x}_{4}}}{2} \right)=f\left( \dfrac{\dfrac{3}{2}+2}{2} \right)=f\left( \dfrac{7}{4} \right)=\dfrac{\sqrt[\frac{4}{7}]{2}\times \sqrt[7]{17319}}{4}=1.497 \\ $ Finding the sum of calculated values and multiplying them by $\Delta x=\dfrac{1}{2}$  $ \\ \dfrac{1}{2}\times (0.905+0.957+1.198+1.149)=2.279$