Let the curve y - y x be the solution of the differential equation- dydx - 2x - 1- If the numerical value of area bounded by the curve y - y x and x-axis is 4-83- then the value of y1 is equal to

Y = x^2 + 2x + c Area of rectangle (ABCD) = | (C 1)(√(1 c))| Area of parabola and x axis = nbsp;2 (23((1 c)^32 ) ) = 4 √(8)3 1 c = 2 ⇒ c= 1 Equation o ...

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Standard XII

Mathematics

Property 6

Let the curve...

Question

Let the curve $y = y (x)$ be the solution of the differential equation. $\frac{d y}{d x} = 2 (x + 1) .$ . If the numerical value of area bounded by the curve $y = y (x)$ and $x -$ axis is $\frac{4 \sqrt{8}}{3}$ , then the value of $y (1)$ is equal to

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Solution

$y = x^{2} + 2 x + c$
Area of rectangle $(A B C D) = | (C - 1) (\sqrt{1 - c}) |$
Area of parabola and x - axis $= 2 ⎛ ⎜ ⎝ \frac{2}{3} ⎛ ⎜ ⎝ (1 - c)^{\frac{3}{2}} ⎞ ⎟ ⎠ ⎞ ⎟ ⎠ = \frac{4 \sqrt{8}}{3} 1 - c = 2 \Rightarrow c = - 1 Equation of y (x) = x^{2} + 2 x - 1 \Rightarrow y (1) = 1 + 2 - 1 = 2$

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Let the curve y=y(x) be the solution of the differential equation. dydx=2(x+1).. If the numerical value of area bounded by the curve y=y(x) and x−axis is 4√83, then the value of y(1) is equal to

y=x2+2x+c Area of rectangle (ABCD)=|(C−1)(√1−c)| Area of parabola and x - axis =2⎛⎜⎝23⎛⎜⎝(1−c)32⎞⎟⎠⎞⎟⎠=4√831−c=2⇒c=−1Equation of y(x)=x2+2x−1⇒y(1)=1+2−1=2

Let the curve $y = y (x)$ be the solution of the differential equation. $\frac{d y}{d x} = 2 (x + 1) .$ . If the numerical value of area bounded by the curve $y = y (x)$ and $x -$ axis is $\frac{4 \sqrt{8}}{3}$ , then the value of $y (1)$ is equal to