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Translating words to math Quiz 2

mathematics

 Solve this function by trapezoidal rule of integration $f\left( x \right) = \int_0^5 {\left( {2x - {x^2}} \right)dx} $ when the value of $n = 10.$ 

 Apply the trapezoidal rule’s formula $ \\ \int_a^b {f\left( x \right)} dx = \dfrac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + .........{y_{n - 1}}} \right)} \right]$ where $\left( {h = \dfrac{{b - a}}{n}} \right)$ 

  $h = \dfrac{{b - a}}{n} = \dfrac{{5 - 0}}{{10}} = 0.5\\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/60ab64b0-dd33-4539-bf82-990ae0ae7ada2308228135365824355.png"/>$ \\ $ Substitute the values in formula  $ \\  = \dfrac{{0.5}}{2}\left[ {\left( {0 - 15} \right) + 2\times\left[ {0.75 + 1 + 0.75 + 0 - 1.25 - 3 - 5.25 - 8 - 11.25} \right]} \right] \\ = \dfrac{1}{4}\left[ { - 15 + 2 \times ( - 26.25)} \right] \\ = \dfrac{1}{4}\left[- 15 - 52.5\right] \\  = - \dfrac{1}{4} \times 67.5 \\ =  - \dfrac{{33.75}}{2}$ 

 Solve this by trapezoidal rule of integration $f\left( x \right) = \int_{ - 2}^2 {\left( {4{x^2} - 3x} \right)dx} $ and $n = 8.$ 

 Apply the formula of trapezoidal rule for integration  $ \\ \int_a^b {f(x)dx = \dfrac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + ......{y_{n - 1}}} \right)} \right]} $ , where $\left( {h = \dfrac{{b - a}}{n}} \right)$.

  $h = \dfrac{{b - a}}{n} = \dfrac{{2 + 2}}{8} = \dfrac{4}{8} = 0.5 \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/8c6445db-c759-4a6c-84ba-8cafd31fed637273091484307582436.png"/> $ \\ $ Putting the values in formula  $ \\  \Rightarrow \int_{ - 2}^2 {\left( {4{x^2} - 3x} \right)dx = \dfrac{{0.5}}{2}\left[ {\left( {22 + 10} \right) + 2\left( {13.5 + 7 + 2.5 + 0.5 + 1 + 4.5} \right)} \right]} \\ \dfrac{{0.5}}{2}\left[ {32 + 2.8} \right] \\  = 31 \times 0.5 = 15.5$ 

 Use the trapezoidal rule of integration and solve this $f\left( x \right) = \int_1^7 {\left( {{x^2} + 4x} \right)dx,n = 3.} $ 

 Apply the formula $ \Rightarrow \int_a^b {f(x)dx = \dfrac{h}{2}\left[ {({y_0} + {y_n}) + 2\left( {{y_1} + {y_2} + .......{y_{n - 1}}} \right)} \right]} $ where $\left( {h = \dfrac{{b - a}}{n}} \right)$ 

$h = \dfrac{{b - a}}{n} = \dfrac{{7 - 1}}{3} = \dfrac{6}{3} = 2 \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9f3d1412-18a2-4983-9498-6b5340ac47615658544700790148948.png"/>$ \\ $ Putting the values in the trapezoidal formula. $ \\ = \dfrac{2}{2}\left[ {\left( {5 + 77} \right) + 2 \times \left( {21 + 45} \right)} \right]\\ = 82 + 66 \times 2 \\ = 82 + 132 = 214$ 

 By trapezoidal rule only, solve this function and get the value $f\left( x \right) = \int_0^1 {\left( {\dfrac{{{x^2} - 2}}{{x - \sqrt 2 }}} \right)dx} $  where $n = 5.$ 

 Apply the formula  $\int_a^b {f\left( x \right)dx = \dfrac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + \left( {{y_1} + {y_2}........{y_{n - 1}}} \right)} \right]} $ where $h = \left( {\dfrac{{b - a}}{n}} \right)$ 

 $h = \dfrac{{1 - 0}}{5} = \dfrac{1}{5} = 0.2 \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9be8b5bb-010a-4428-8e87-3af572fb57424419388806366581946.png"/> $ \\ $ Putting the values in the formula $ \\ = \dfrac{{0.2}}{2}\left[ {\left( {1.41 + 2.43} \right) + 2\left( {1.61 + 1.82 + 2.02 + 2.22} \right)} \right] \\ = 0.1\left[ {3.84 + 15.34} \right] \\  = 0.1\left[ {19.18} \right] \\ = 1.918$ 

 By trapezoidal rule, find the value of the integration function $f\left( x \right) = \int_0^4 {\left( {4{x^3} - 9{x^2} + 7x + 3} \right)dx} $ where the value of $n = 4.$

 $h = \dfrac{{4 - 0}}{4} = \dfrac{4}{4} = 1 \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9588b33f-e316-4e60-abe0-d513440b51cd4670448951694629833.png"/> $ \\ $ Putting the values in the formula of trapezoidal rule  $ \\ \int_0^4 {\left( {4{x^3} - 9{x^2} + 7x + 3} \right)dx = \dfrac{1}{2}\left[ {\left( {143 + 3} \right) + 2\left( {51 + 13 + 5} \right)} \right]} \\ = \dfrac{1}{2}\left[ {146 + 138} \right] = 142$ 