Standard problems of calculus.

Calculus Level 4

We have a continous and differentiable curve f(x)=y passing through (1,1) satisfying the property that the tangent drawn to the curve at the point (x,f(x)) intersects the x-axis at the point (2x,0).

Find the value of ∫ 1 2 f ( x ) d x \int _{ 1 }^{ 2 }{ f(x)dx }


The answer is 0.693.

Ronak Agarwal by Ronak Agarwal , 23, India

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1 solution

Ayush Garg
Sep 14, 2014

Tangent at a general point x 1 , f ( x ) x_{1},f\left( x \right) is given by

y − f ( x ) = d y d x ( x − x 1 ) y-f\left( x \right) =\frac { dy }{ dx } \left( x-x_{1} \right)

At y=0,x=2 x 1 x_{1}

− f ( x ) -f\left( x \right) = x d y d x x\frac { dy }{ dx }

∫ − f ( x ) = ∫ x d y d x \int { -f\left( x \right) } =\int { x\frac { dy }{ dx } }

∫ d y y = − ∫ d x x \int { \frac { dy }{ y } =-\int { \frac { dx }{ x } } }

ln ⁡ y \ln { \quad y } = - ln ⁡ x \ln { \quad x } +c

Since curve passes through (1,1)

xy = 1 is the required curve

∫ 1 2 f ( x ) d x \int _{ 1 }^{ 2 }{ f\left( x \right) dx } = ∫ 1 2 d x x \int _{ 1 }^{ 2 }{ \frac { dx }{ x } }

l n 2 = 0.693 \boxed{ln 2 = 0.693}

11 Helpful 0 Interesting 0 Brilliant 0 Confused

The problem should mention upto how many decimal places the answer should be! @Ronak Agarwal

Satyajit Mohanty - 5 years, 11 months ago

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