The area of the region bounded by the lines $y = 2 x + 1, y = 3 x + 1$ and $x = 4$ is

16 s q . u n i t

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\frac{121}{3} s q . u n i t

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\frac{121}{6} s q . u n i t

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8 s q . u n i t

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Solution

The correct option is C $8 s q . u n i t$
$A$ (Shaded region) $= A (△ A B D) - A (△ A B C) = \frac{1}{2} [4 \times 12 - 4 \times 8] = \frac{1}{2} (48 - 32) = 8 s q . u n i t s$ .
OR
$A (△ A C D) = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 1 4 & 9 & 1 4 & 13 & 1 \end{matrix} ∣ ∣ ∣ ∣$
$= \frac{1}{2} \times 16 = 8$

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