First slide

Evaluation of definite integrals

Question

The area bounded by the curve $y = f (x) = x^{4} - 2 x^{3} + x^{2} + 3$ , x-axis and ordinates corresponding to
minimum of the function f(x) is

Moderate

A

1
B

91/30
C

30/9
D

4

Solution

$f^{'} (x) = 4 x^{3} - 6 x^{2} + 2 x = 2 x (2 x^{2} - 3 x + 1) = 2 x (2 x - 1) (x - 1)$ .
Since $f$ is a differentiable function, so extremum points of $f (x)$ we must have $f^{'} (x) = 0$ so $x = 0, 1 / 2$ ,1. Now $f^{''} (x) = 12 x^{2}$
$- 12 x + 2, f^{''} (0) = 2, f^{''} (1) = 2$ and $f^{''} (1 / 2) = 3 - 6 + 2 = - 1$
Thus the function has minimum at $x = 0$ and $x = 1$ . Therefore, the required area $= \int_{0}^{1} (x^{4} - 2 x^{3} + x^{2} + 3) d x$ (See Chapter 14 for area as definite integral)
$= {(\frac{x^{5}}{5} - \frac{x^{4}}{2} + \frac{x^{3}}{3} + 3 x)|}_{0}^{1} = \frac{1}{5} - \frac{1}{2} + \frac{1}{3} + 3 = \frac{91}{30}$

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