JEE PYQ: Area Under Curves Question 1
Question 1 - 2021 (16 Mar Shift 1)
Let the curve $y = y(x)$ be the solution of the differential equation, $\frac{dy}{dx} = 2(x+1)$. 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 ____.
Type: Numerical
Show Answer
Answer: 2
Solution
$\frac{dy}{dx} = 2(x+1) \Rightarrow y = x^2 + 2x + C$. Area $= \frac{4\sqrt{8}}{3}$. Solving with the boundary conditions gives $C = -1$. So $y = x^2 + 2x - 1$, $y(1) = 1 + 2 - 1 = 2$.
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