Calculus ALL IN ONE!

Calculus Level 3

Given a function f : ( − ∞ , 3 ] → R f:\left(-\infty,3\right]\rightarrow\mathbb{R} such that f ( x ) = 3 16 ( ∫ 0 1 f ( x ) d x ) x 2 − 9 10 ( ∫ 0 2 f ( x ) d x ) x + 2 ( ∫ 0 3 f ( x ) d x ) + 4 \displaystyle f\left( x\right)=\frac{3}{16}\left( \int_{0}^{1}\, f\left( x\right)\, dx \right)x^{2} -\frac{9}{10}\left( \int_{0}^{2}\, f\left( x\right)\, dx \right)x +2\left( \int_{0}^{3}\, f\left( x\right)\, dx \right) +4 Then find the value of lim ⁡ t → 0 2 t − ∫ f ( 2 ) + 2 f − 1 ( t ) [ f ′ ( x ) ] 2 d x 1 − cos ⁡ t cosh ⁡ 2 t cos ⁡ 3 t \displaystyle \lim_{t\rightarrow 0}\, \frac{\displaystyle 2t-\int_{f\left(2\right)+2}^{f^{-1}\left(t\right)}\, \left[ {f}'\left(x\right)\right]^2 \, dx }{1-\cos t \cosh 2t \cos 3t}


The answer is 0.16666.

Uzumaki Nagato Tenshou Uzumaki by uzumaki nagato tenshou u… , 26

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1 solution

C Anshul
Jul 13, 2018

H i n t s : Hints:

put these unknown integrals ∫ 0 1 f ( x ) d x = a \displaystyle\int_{0}^{1}\, f\left( x\right)\, dx =a and others two equal to b b and c c respectively.

Now integrate the given equation from 0 0 to 1 1 then 0 0 to 2 2 and then 0 0 to 3 3 .

U will get three equations in terms of a , b , c a,b,c On solving we get a = − 16 3 , b = − 20 3 , c = − 6 a=-\frac{16}{3},b=-\frac{20}{3},c=-6 .

Now our f ( x ) = − x 2 + 6 x − 8 f(x)=-x^{2}+6x-8 .

Also f − 1 ( x ) = 3 − ( 1 − x ) 1 2 f^{-1}(x)=3-(1-x)^{\frac{1}{2}}

Also on solving in numerator we will get

2 t + 4 3 ( 1 − t ) 3 2 − 4 3 2t+\frac{4}{3}(1-t)^{\frac{3}{2}}-\frac{4}{3} . Then i just use L'hopital rule 2 times to get limit equal to 1 6 \frac{1}{6} .

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