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Area between x=g(y) and y Axis

If fx=a+bx+...

Question

If $f (x) = a + b x + c x^{2},$ where $c > 0$ and $b^{2} - 4 a c < 0$ then the area enclosed by the co-ordinate axes, the line $x = 2$ and the curve $y = f (x)$ is given by $\frac{1}{3} {f (0) + λ f (1) + f (2)}$ .sq.unit,then $λ$ is

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Solution

Area of $O A B L O = \int_{0}^{2} y . d x$
$= \int_{0}^{2} (a + b x + c x^{2}) d x$
$= {[a x + \frac{b x^{2}}{2} + \frac{c x^{3}}{3}]}_{0}^{2}$
$= (2 a + 2 b + \frac{8}{3} c)$
$= \frac{1}{3} [6 a + 6 b + 8 c]$ .............. $(1)$
But $f (x) = a + b x + c x^{2}$
$f (0) = a, \int_{0}^{2} y . d x$
$\Rightarrow f (0) = a, f (1) = a + b + c, f (2) = a + 2 b + 4 c$
$\Rightarrow f (0) = a, f (1) = f (0) + b + c, f (2) = f (0) + 2 b + 4 c$
$\Rightarrow f (0) = a, f (1) - f (0) = b + c, f (2) - f (0) = 2 b + 4 c$
$\Rightarrow f (0) = a, f (1) - f (0) = b + c, \frac{f (2) - f (0)}{2} = b + 2 c$
$\Rightarrow f (0) = a, b + 2 c - b - c = \frac{f (2) - f (0)}{2} - f (1) + f (0) = \frac{f (2) - 2 f (1) + f (0)}{2}$
$\Rightarrow f (0) = a, c = \frac{f (2) - 2 f (1) + f (0)}{2}, b = f (1) - f (0) - c$
$\Rightarrow f (0) = a, c = \frac{f (2) - 2 f (1) + f (0)}{2}, b = f (1) - f (0) - \frac{f (2) - 2 f (1) + f (0)}{2} = \frac{f (2) + f (0) - 2 f (1)}{2}$ (on simplification)
$∴ a = f (0), b = \frac{4 f (1) - f (2) - 3 f (0)}{2}$ and $c = \frac{f (2) + f (0) - 2 f (1)}{2}$
Then, from eqn $(1)$ we get
$\frac{1}{3} [6 a + 6 b + 8 c] = \frac{1}{3} [6 (f (0)) + 6 (\frac{4 f (1) - f (2) - 3 f (0)}{2}) + 8 (\frac{f (2) + f (0) - 2 f (1)}{2})]$
On simplification,we get
$= \frac{1}{3} [f (0) + 4 f (1) + f (2)]$
On comparing the above with $\frac{1}{3} {f (0) + λ f (1) + f (2)}$ .sq.unit we get $λ = 4$
Hence, $λ = 4$

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If f(x)=a+bx+cx2, where c>0 and b2−4ac<0 then the area enclosed by the co-ordinate axes, the line x=2 and the curve y=f(x) is given by 13{f(0)+λf(1)+f(2)}.sq.unit,then λ is

If $f (x) = a + b x + c x^{2},$ where $c > 0$ and $b^{2} - 4 a c < 0$ then the area enclosed by the co-ordinate axes, the line $x = 2$ and the curve $y = f (x)$ is given by $\frac{1}{3} {f (0) + λ f (1) + f (2)}$ .sq.unit,then $λ$ is