The area bounded by the curve y = x^{4} – 2x^{3} + x^{2} + 3 with x-axis and ordinates corresponding to the minima of y is

y = x^{4} – 2x^{3} + x^{2} + 3

⇒ y’ = 4x^{3} – 6x^{2} + 2x

At extrema of y, y’ = 0

i.e., 4x^{3} – 6x^{2} + 2x = 0

or, 2x^{3} – 3x^{2} + x = 0

⇒ x(2x – 1)(x – 1) = 0

i.e., x = 0, x = 1/2 and x = 1 correspond to extrema of y

Now, y’’ = 12x^{2} – 12x + 2

y’’_{0} = 2 > 0 ⇒ x = 0 corresponds to a minima

y’’_{1/2} = -1 < 0 ⇒ x = 1/2 corresponds to a maxima

y’’_{1} = 2 > 0 ⇒ x = 1 corresponds to a minima

So, x = 0 and x = 1 are the ordinates we need.

So, the area A is –

A =

=

(Ans)

1