Find the region

Calculus Level 4

Let 2 ( f ( x ) ) 2 − d 2 f ( x ) d x 2 f ( x ) + ( d f ( x ) d x ) 2 = 0 2{ \left( f\left( x \right) \right) }^{ 2 }-\frac { { d }^{ 2 }f\left( x \right) }{ d{ x }^{ 2 } } f\left( x \right) +{ \left( \frac { df\left( x \right) }{ dx } \right) }^{ 2 }=0 and f ( 0 ) = f ( 1 ) = − 1 f\left( 0 \right) =f\left( 1 \right) =-1

Then the area of the region bounded by y=0, x=0, x=1 and y = ( 2 x − 1 ) f ( x ) y=\left( 2x-1 \right) f\left( x \right) is 2 ( 1 − 1 e 1 / a ) 2\left( 1-\frac { 1 }{ { e }^{ { 1 }/{ a } } } \right)

Then the value of a is???


The answer is 4.

Sudipto Banerjee by Sudipto Banerjee , 25, India

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1 solution

Nishant Sharma
Jul 5, 2014

Rearranging the given differential equation we get

2 = d f ′ ( x ) f ( x ) \displaystyle2=\mathrm d \frac{f'(x)}{f(x)}

Integrating twice w.r.t x x we get x 2 + a x + b = ln ⁡ f ( x ) \displaystyle\,x^2+ax+b=\ln f(x) for some a , b ∈ C a,b\in\mathbb{C} . Now using initial conditions of f ( 0 ) = f ( 1 ) = − 1 f(0)=f(1)=-1 we get a = − 1 , b = i π a=-1,\,b=i\pi . So y = ( 1 − 2 x ) e x 2 − x = Y \displaystyle\,y=(1-2x)e^{x^2-x}=Y (say).

Here is the plot of the required area(in red)

Graph Graph

So area = ∫ 0 1 2 Y d x − ∫ 1 2 1 Y d x = 2 ( 1 − 1 e 1 4 ) \displaystyle=\int\limits_{0}^{\frac{1}{2}} Y\,\mathrm d x-\int\limits_\frac{1}{2}^1 Y\,\mathrm d x=2\left(1-\frac{1}{e^{\frac{1}{4}}}\right) . So a = 4 a=\boxed{4} .

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