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

Trapezoidal rule Quiz 2

mathematics

 Find the value of the function $f\left( x \right) = \int_0^2 {\left( {{x^2} - x - 1} \right)dx} $ by trapezoidal rule when the sub-interval is given $\left( {0,0.4,0.8,1.2,1.6,2} \right).$

 Apply the formula for trapezoidal rule of integral  $ \\ \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 $\left( {h = \dfrac{{b - a}}{n}} \right)$ 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2854ca84-9e49-4154-bee2-9213da14b8a33665940980997066732.png"/> $ \\ $ Apply the values in the formula $ = \dfrac{{\left( {\dfrac{{2 - 0}}{5}} \right)}}{2}\left[ {\left( { - 1 + (1)} \right) + 2\left( { - 1.24 - 1.16 - 0.76 - 0.04} \right)} \right] \\ = \dfrac{1}{5}\left[ {0 + 2\left( { - 3.2} \right)} \right]  = - 1.28$ 

 For a function $f\left( x \right) = \int_1^{1.6} {{{\left( {x + 2x} \right)}^2}dx,} $. Find the value by using trapezoidal rule of integration where $n = 3.$ 

 Apply the trapezoidal rule 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 $h = \left( {\dfrac{{b - a}}{n}} \right)$

  $h = \dfrac{{1.6 - 1}}{3} = \dfrac{{0.6}}{3} = 0.2 \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/eb66b66a-36b9-4c6a-9130-e4b66accafc86693631782122259603.png"/> $ \\ $ Apply the values in the formula $ = \dfrac{{0.2}}{2}\left[ {\left( {9 + 23.04} \right) + 2\left( {12.96 + 17.64} \right)} \right] \\ = 0.1\left( {93.24} \right) = 9.32$

 Find the value of integration function $f\left( x \right) = \int_1^3 {\left( {\sqrt {16 - {x^2}} } \right)dx} $ where $n = 2.$

 Apply the formula for trapezoidal rule $\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 $h = \left( {\dfrac{{b - a}}{n}} \right)$. 

$h = \left( {\dfrac{{3 - 1}}{2}} \right) = \dfrac{2}{2} = 1 \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5b51f59d-18e0-45c6-9c8d-cf2cb30dfa568776583544383231069.png"/> $ \\ $ Putting the value in the formula $ = \dfrac{1}{2}\left[ {\left( {\sqrt {15}  + \sqrt 7 } \right) + 2\sqrt {12} } \right] \\ = \dfrac{1}{2}\left[ {\left( {3.87 + 2.64} \right) + 2\left( {3.46} \right)} \right] \\ = \dfrac{1}{2}\left( {6.51 + 6.97} \right) = 6.71$ 

 Find the solution using trapezoidal rule. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/cdb46029-9c33-40da-8599-951f9f28c7363946322456003601914.png"/>

 Find the value of $h,$ where $h = \dfrac{{b - a}}{n}$ and apply the trapezoidal formula $ \\ \Rightarrow \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]$ 

 $h = \dfrac{{0.8 - 0}}{4} = 0.2 \\ $ Putting the value of $h$ and values in table in the 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] \\ = \dfrac{{0.2}}{2}\left[ {\left( {12 + 52} \right) + 2\left( {16 + 22 + 40} \right)} \right] \\ = 0.1\left( {220} \right) = 22$ 

 Find the solution of an equation $\left( {\dfrac{1}{{x + 2}}} \right)$ using trapezoidal rule with limits $\left( {1,2} \right)$ and with the value $\left( h \right) = 0.25.$ 

 Apply the formula for trapezoidal rule $\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 $h = \left( {\dfrac{{b - a}}{n}} \right)$ . 

 $h = \dfrac{{2 - 1}}{4} = 0.25 \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4ba105a6-bc97-4542-a469-9c4f6f8aa5e65170125555326085844.png"/> $ \\ \Rightarrow \dfrac{{0.25}}{2}\left[ {\left( {\dfrac{1}{3} + \dfrac{1}{4}} \right) + 2\left( {\dfrac{1}{{3.25}} + \dfrac{1}{{3.5}} + \dfrac{1}{{3.75}}} \right)} \right] \\ = \dfrac{{0.25}}{2}\left[ {\left( {0.33 + 0.25} \right) + 2\left( {0.30 + 0.28 + 0.266} \right)} \right] \\ = \dfrac{{0.25}}{2}\left( {2.26} \right) \\ = 0.125 \times 2.26 = 0.283$ 