The area bounded by the curve y = x 4 - 2 x 3 + x 2 + 3 with the X -axis and ordinate corresponding to the minimum of y is:

Question

The area bounded by the curve $y = x^{4} - 2 x^{3} + x^{2} + 3$ with the $X$ -axis and ordinate corresponding to the minimum of $y$ is:

Smart Achievers · Accepted Answer

(2)

Ans. $\frac{91}{30}$ sq. units

Explanation:

$y = x^{4} - 2 x^{3} + x^{2} + 3$

$\frac{d y}{d x} = 4 x^{3} - 6 x^{2} + 2 x$

$\frac{d y}{d x} = 0$

$\Rightarrow$ $4 x^{3} - 6 x^{2} + 2 x = 0$

$\Rightarrow$ $x (4 x^{2} - 6 x + 2) = 0$

$\Rightarrow$ $2 x (2 x^{2} - 3 x + 1) = 0$

$\Rightarrow$ $x (2 x^{2} - 2 x - x + 1) = 0$

$\Rightarrow$ $x (2 x - 1) (x - 1) = 0$

$x = 0, \frac{1}{2}, 1$

$\frac{d^{2} y}{d x^{2}} = 12 x^{2} - 12 x + 2$

${(\frac{d^{2} y}{d x^{2}})}_{x = 0} = 2 > 0,$

${(\frac{d^{2} y}{d x^{2}})}_{x = \frac{1}{2}} = - 1 < 0,$

${(\frac{d^{2} y}{d x^{2}})}_{x = 1} = 2 > 0$

Hence, minimum occurs at $x = 0$ and $x = 1$

$∴$ $Area = \int_{0}^{1} (x^{4} - 2 x^{3} + x^{2} + 3) d x$

$= {[\frac{x^{5}}{5} - \frac{2 x^{4}}{4} + \frac{x^{3}}{3} + 3 x]}_{0}^{1}$

$= \frac{91}{30} sq. units$

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