If a function f:[$$0$$, $$27$$]→R is differentiable then for $$some0$$<α<β<$$3$$,∫$$027$$ $$f(x)dx$$ is equal to

Correct option is 33α2fα3+12β2fβ3

Questions

If a function $f : [0, 27] \to R$ is differentiable then for some
$0 < α < β < 3, \int_{0}^{27} f (x) d x$ is equal to

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3α2fα3+β2fβ3

3α2f(α)+β2f(β)

3α2fα3+12β2fβ3

3α2f(α)+12β2f(β)

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detailed solution

Correct option is C

Let g(x)=∫0x3 f(t)dt so g(0)=0g′(x)=3x2fx3∫027 f(t)dt=g(3)=g(3)−g(1)3−1+12g(1)−g(0)1−0=g′(α)+12g′(β)for some α∈(1,3),β∈(0,1)⊆(0,3), but g′(α)=3α2fα3and g′(β)=3β2f(β)3Therefore, ∫027 f(t)dt=3α2fα3+12β2fβ3

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If a function f:[0, 27]→R is differentiable then for some0<α<β<3,∫027 f(x)dx is equal to

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

If a function $f : [0, 27] \to R$ is differentiable then for some
$0 < α < β < 3, \int_{0}^{27} f (x) d x$ is equal to