If the area of region bounded by the parabola $y = x^{2} - 4 x + 3$ and the straight lines touching it at the points with abscissae $x_{1} = 1$ and $x_{2} = 3 (in sq. units)$ is $A,$ then the value of $3 A$ is

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Solution

Given curve : $y = x^{2} - 4 x + 3$
Point corresponding to abscissae $x = 1, 3$ on the parabola are $A (1, 0)$ and $B (3, 0)$
Tangent at points $A$ and $B$ are $2 x + y = 2$ and $y - 2 x = - 6$ respectively.
The bounded region is shown below :
So, Required area $=$ Area of $△ A B C - | 3 \int 1 (x^{2} - 4 x + 3) d x |$
$= \frac{1}{2} \cdot 2 \cdot 2 - ∣ ∣ ∣ ∣ {[\frac{x^{3}}{3} - 2 x^{2} + 3 x]}_{1}^{2} ∣ ∣ ∣ ∣ A = \frac{2}{3}$
$∴ 3 A = 2$

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