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Application of Integration

Section: Mathematics

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Question : 25 of 30

Marks: +1, -0

Consider the polynomial f (x) =

1 + 2x + 3x^2 + 4x^3

. Let s be the sum of all distinct real roots of f (x) and let t = s .

The area bounded by the curve y = f (x) and the lines x = 0, y = 0 and x = t, lies in the interval

[JEE Adv 2010 P2]

$\left( \frac{3}{4}, 3 \right)$
$\left( \frac{21}{64}, \frac{11]}{16} \right)$
(9 , 10)
$\left( 0, \frac{21}{64} \right)$

Solution:

We have

-\frac{3}{4}

< s <

-\frac{1}{2}

\frac{1}{2}

< t <

\frac{3}{4}

\int\limits_{0}^{\frac{1}{2}} (4x^3+3x^2+2x+1) dx

< area <

\int\limits_{0}^{\frac{3}{4}} (4x^3+3x^2+2x+1) dx

\left[ x^4+x^3+x^2+x \right]_{0}^{\frac{1}{2}}

< area <

\left[ x^4+x^3+x^2+x \right]_{0}^{\frac{3}{4}}

\frac{1}{16} + 18 + \frac{1}{4} + \frac{1}{2}

< area <

\frac{81}{256} + \frac{27}{64} + \frac{9}{16} + \frac{3}{4}

\frac{15}{16}

< area <

\frac{525}{256}

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