Find the area of the region bounded by $y = 2 x + 1 y = 3 x + 1$ and $x = 4$ .

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Solution

$y = 3 x + 1 \Rightarrow \frac{y}{1} + \frac{x}{\frac{- 1}{3}} = 1 y = 2 x + 1 \Rightarrow \frac{y}{1} + \frac{x}{\frac{- 1}{2}} = 1$

$A r e a (A B C) = A r e a (A B E D) - A r e a (A C E D) = \int_{x = 0}^{4} (3 x + 1) - 0 d x - \int_{x = 0}^{4} (2 x + 1) - 0 d x = \int_{x = 0}^{4} x d x = {[\frac{x^{2}}{2}]}_{0}^{4} = 8$

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Similar questions

y = 2 x + 1, y = 3 x + 1

x = 4

y^{2} = 2 x

x - y = 4

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x - 3 y + 5 = 0

y^{2} = 2 x

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