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This is one of our many quizzes on Mean value theorem for integrals. Click on this to access more such competitive quizzes on Calculus.

Mean value theorem for integrals Quiz 2

mathematics

 Find the value of $c$ of the following function $f\left( x \right) = \dfrac{8}{{{x^3}}}$ over the interval $\left( {1,3} \right).$

 Use the mean value theorem:$f\left( c \right) = \dfrac{1}{{b - a}}\int_a^b {f\left( x \right)dx} $to find the value of $c$.

$  f\left( c \right) = \dfrac{1}{{b - a}}\int_a^b {f\left( x \right)dx}  \\   \Rightarrow \dfrac{8}{{{c^3}}} = \dfrac{1}{{3 - 1}}\int_1^3 {\dfrac{8}{{{x^3}}}dx}  \\   \Rightarrow \dfrac{8}{{{c^3}}} = \dfrac{1}{2} \times 8 \times \left[ {\dfrac{{{x^{ - 2}}}}{{ - 2}}} \right]_1^3 \\   \Rightarrow \dfrac{8}{{{c^3}}} =  - 2\left[ {{x^{ - 2}}} \right]_1^3 \\   \Rightarrow \dfrac{8}{{{c^3}}} =  - 2\left[ {\dfrac{1}{9} - 1} \right] \\   \Rightarrow \dfrac{8}{{{c^3}}} =  - 2 \times \left( { - \dfrac{8}{9}} \right) \\   \Rightarrow \dfrac{8}{{{c^3}}} = \dfrac{{16}}{9} \\   \Rightarrow c = \sqrt[3]{{\dfrac{9}{2}}} \\  $

 Find $c$ guaranteed by Mean Value Theorem for the function $f\left( x \right) = {x^3}$ over the interval $\left( {0.3} \right).$

 Apply formula $f\left( c \right) = \left( {\dfrac{1}{{b - a}}} \right)\int_a^b {f\left( x \right)dx} $ and then find the value of $c$.

$  f\left( c \right) = \dfrac{1}{{b - a}}\int_a^b {f\left( x \right)dx}  \\  \left[ {\because {c^3} = {x^3}} \right] \\  \therefore f\left( x \right) = f\left( c \right) \\   \Rightarrow {c^3} = \dfrac{1}{{3 - 0}}\int_0^3 {{x^3}dx}  \\   \Rightarrow {c^3} = \dfrac{1}{3}\left[ {\dfrac{{{x^4}}}{4}} \right]_0^3 \\   \Rightarrow {c^3} = \dfrac{1}{3}\left( {\dfrac{{81}}{4}} \right) \\   \Rightarrow {c^3} = \dfrac{{27}}{4} \\   \Rightarrow c = \dfrac{3}{{\sqrt[3]{4}}} \\  $

 Find the value of cover the interval $\left( {1,4} \right)$ of the function $f\left( x \right) = 6\sqrt x .$

 Apply the Mean Value Theorem i.e. $f\left( c \right) = \left( {\dfrac{1}{{b - a}}} \right)\int_a^b {f\left( x \right)dx} $ to find the value of $c$. 

$  f\left( c \right) = \dfrac{1}{{\left( {b - a} \right)}}\int_a^b {f\left( x \right)dx}  \\   \Rightarrow 6\sqrt c  = \dfrac{1}{{\left( {4 - 1} \right)}}\int_1^4 {\left( {6\sqrt x } \right)dx}  \\   \Rightarrow 6\sqrt c  = \dfrac{1}{3} \times 6\int_1^4 {\left( {{x^{\dfrac{1}{2}}}} \right)dx}  \\   \Rightarrow 6\sqrt c  = 2\left[ {\dfrac{{{x^{\dfrac{3}{2}}}}}{{\left( {\dfrac{3}{2}} \right)}}} \right]_1^4 \\   \Rightarrow 6\sqrt c  = 2 \times \dfrac{2}{3}\left[ {{{\left( 4 \right)}^{\dfrac{3}{2}}} - {{\left( 1 \right)}^{\dfrac{3}{2}}}} \right] \\   \Rightarrow 6\sqrt c  = \dfrac{4}{3}\left( {8 - 1} \right) \\   \Rightarrow 6\sqrt c  = \left( {\dfrac{{28}}{3}} \right) \\   \Rightarrow \sqrt c  = \dfrac{{14}}{9} \\   \Rightarrow \left( c \right) = \left( {\dfrac{{196}}{{81}}} \right) \\  $

 Find the value of c that satisfy the Mean Value Theorem for integrals $f\left( x \right) = \dfrac{{{x^2}}}{2} + x + \dfrac{3}{2},$ over the interval $\left( {3,1} \right).$

 Apply formula $f\left( c \right) = \left( {\dfrac{1}{{b - a}}} \right)\int_a^b {f\left( x \right)dx} $and here find the value of $c$.

$  f\left( c \right) = \dfrac{1}{{b - a}}\int_a^b {f\left( x \right)dx}  \\   \Rightarrow \left( {\dfrac{{{c^2}}}{2} + c + \dfrac{3}{2}} \right) = \dfrac{1}{2}\int_1^3 {\left( {\dfrac{{{x^2}}}{2} + x + \dfrac{3}{2}} \right)dx}  \\   \Rightarrow \left( {\dfrac{{{c^2}}}{2} + c + \dfrac{3}{2}} \right) = \dfrac{1}{2}\left[ {\dfrac{{{x^3}}}{6} + \dfrac{{{x^2}}}{2} + \dfrac{{3x}}{2}} \right]_1^3 \\   \Rightarrow \left( {\dfrac{{{c^2}}}{2} + c + \dfrac{3}{2}} \right) = \dfrac{1}{2}\left[ {\dfrac{{27}}{6} + \dfrac{9}{2} + \dfrac{9}{2} - \dfrac{1}{6} - \dfrac{1}{2} - \dfrac{3}{2}} \right] \\   \Rightarrow \left( {\dfrac{{{c^2}}}{2} + c + \dfrac{3}{2}} \right) = \dfrac{1}{2}\left[ {\dfrac{{26}}{6} + \dfrac{{14}}{2}} \right] \\   \Rightarrow \left( {\dfrac{{{c^2}}}{2} + c + \dfrac{3}{2}} \right) = \dfrac{1}{2}\left[ {\dfrac{{13}}{3} + 7} \right] \\   \Rightarrow \left( {\dfrac{{{c^2}}}{2} + c + \dfrac{3}{2}} \right) = \dfrac{{17}}{3} \\   \Rightarrow \left( {\dfrac{{{c^2}}}{2} + c} \right) = \dfrac{{25}}{6} \\   \Rightarrow \left( {{c^2} + 2c} \right) = \dfrac{{25}}{3} \\   \Rightarrow {c^2} + 2c - \dfrac{{25}}{3} = 0 \\  {\text{Applying }}\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\   \Rightarrow \dfrac{{ - 2 \pm \sqrt {4 - \dfrac{{100}}{3}} }}{2} \\   \Rightarrow \dfrac{{ - 2 \pm 4\sqrt {\dfrac{7}{3}} }}{2} \\   \Rightarrow  - 1 \pm 2\sqrt {\dfrac{7}{3}}  \\  $

 Find average value of the function $f\left( x \right) = \dfrac{1}{x}$ over the interval $\left( {2,3} \right).$

 Apply Mean Value Theorem  $f\left( c \right) = \left( {\dfrac{1}{{b - a}}} \right)\int_a^b {f\left( x \right)dx} $from this find the value of $c$.

$  f\left( c \right) = \dfrac{1}{{b - a}}\int_a^b {\dfrac{1}{x}dx}  \\   \Rightarrow \dfrac{1}{1}\int_2^3 {\dfrac{1}{x}dx}  \\   \Rightarrow \left[ {\log x} \right]_2^3 \\   \Rightarrow \left( {\log 3 - \log 2} \right) \\   \Rightarrow 0.4777 - 0.3010 \\   \Rightarrow 0.1761 \\   \Rightarrow f\left( c \right) = \dfrac{1}{c} = 0.1761 \\   \Rightarrow c = \dfrac{1}{{0.1761}} \\   \Rightarrow c = 5.67859 \\  $ 