If $f\left( x \right)$ satisfies the condition for Rolle’s theorem is [3, 5], then $\mathop \smallint \nolimits_3^5 \,f\left( x \right)\;dx$ equals

JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

1 Easy

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Analyzing +4 | -1

f\left( x \right)

satisfies the condition for Rolle’s theorem is [3, 5], then

\int \nolimits_3^5 \,f\left( x \right)\;dx

equals

View answer and explanation

Correct option: D

\frac{{ - 4}}{3}

Option D (

\frac{{ - 4}}{3}

) is correct because it is the definite integral of a function that satisfies Rolle's theorem over the given interval.

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f(x)

g(x)

[0,1]

f(0)=2, f(1)=6, g(0)=0, g(1)=

0 < \mathrm{c} < 1

f(x)

f(x)

f(x)=|\sin | x ||

\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]

f\left( x \right) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} } ,\;1 \le x \le 26

f'\left( x \right)

1 < x < 26

- 1 \le x \le 1

f(x)

[3,5]

\int\limits_3^5 {f\left( x \right)} \,dx

\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = f'\left( c \right)

f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)