First slide

Area of bounded Regions

Question

$A curve y = f (x) which passes through (4, 0) satisfies the differential equation$
$x d y + 2 y d x = x (x - 3) d x . The area bounded by y = f (x) and line y = x (in square unit) is$

Difficult

A

$\frac{128}{3}$
B

$\frac{64}{3}$
C

64
D

128

Solution

$\begin{array}{l} x d y + 2 y d x = x (x - 3) d x \\ \Rightarrow x^{2} d y + 2 x y d x = (x^{3} - 3 x^{2}) d x \\ \Rightarrow d (x^{2} y) = (x^{3} - 3 x^{2}) d x \\ \Rightarrow y x^{2} = \frac{x^{4}}{4} - x^{3} + c \end{array}$
Curve passes through the point (4,0).
$∴ c = 0$
$Therefore, curve is y = \frac{x^{2}}{4} - x and point of intersection of y = x and this curve is x = \frac{x^{2}}{4} - x \Rightarrow 8 x - x^{2} = 0 \Rightarrow x = 0, x = 8$
$\begin{array}{l} Required area = \int_{0}^{8} (x - (\frac{x^{2}}{4} - x)) d x \\ = {(x^{2} - \frac{x^{3}}{12})}_{0}^{8} = 64 - \frac{2}{3} \times 64 = \frac{64}{3} \end{array}$

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