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  • One of the values of ‘a’ for which the area bounded by the curve  ,

One of the values of ‘a’ for which the area bounded by the curve y = 8x^{2}-x^{5} , straight lines x = 1, x = a  and x-axis is equal to \frac{16}{3}, is

Option: 1

1


Option: 2

2


Option: 3

-1

 


Option: 4

\frac{1}{2}


Answers (1)

best_answer

 

Geometrical interpretation of a definite integral -

An algebraic sum of the area of the figure bounded by the curve y = f(x), the x axis and the striaght lines x=a and x=b. The areas above x axis are taken as positive and the areas below x axis are taken as negative.  
 

- wherein

Where a< b

Hence

\int_{a}^{b}f\left ( x \right )dx=

ar\left ( PAQP \right )-ar\left ( QTRQ \right )+ar\left ( RBSR \right )

 

 

y=8x^{2}-x^{5}=x^{2}(8-x^{3})

Case I a < 1

A= \int_{a}^{l}(8x^{2}-x^{5})dx=\frac{16}{3}

or \frac{8}{3}-\frac{1}{6}-\frac{8a^{3}}{3}+\frac{a^{6}}{6}=\frac{16}{3}

or (a^{3}-17)(a^{3}+1)=0

\Rightarrow a=-1,a=(17)^{1/3} is not possible

Case II a\in [1,2]

A= \int_{a}^{l}(8x^{2}-x^{5})dx=\frac{16}{3}

or 16a^{3}-a^{6}-15=32

or a^{6}-16a^{3}+47=0

This equation is not satisfied by a=1,a=2

Case III a>2

There is no option

Hence one solution is -1

 

Posted by

Kuldeep Maurya

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