Let the curve y = y(x) be the solution of the differential equation, dy/dx = 2(x + 1). If the numerical value of area bounded by the curve y = y(x) and x-axis is 4√8/3, then the value of y(1) is equal to........

dy/dx = 2 (x + 1)
dy = 2 (x + 1)dx.
y = (x + 1)² - c (1)
For y = 0, (x+1)² = c ⇒ x = -1 ± √c.
Area A = ∫? √c? ¹? √c (0 - [ (x+1)² - c])dx = ∫? √c? ¹? √c (c - (x+1)²)dx
A = [cx - (x+1)³/3] from -1-√c to -1+√c
= (c (-1+√c) - (√c)³/3) - (c (-1-√c) - (-√c)³/3)
= -c+c√c - c√c/3 + c+c√c - c√c/3 = 2c√c - 2c√c/3 = (4/3)c√c.
Given A = 4√8 / 3 = 8√2 / 3.
(4/3)c^ (3/2) = 8√2 / 3 ⇒ c^ (3/2) = 2√2 = 2^ (3/2) ⇒ c = 2.
∴ y = (x + 1)² - 2.
&there

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dy/dx = 2 (x + 1)
dy = 2 (x + 1)dx.
y = (x + 1)² - c (1)
For y = 0, (x+1)² = c ⇒ x = -1 ± √c.
Area A = ∫? √c? ¹? √c (0 - [ (x+1)² - c])dx = ∫? √c? ¹? √c (c - (x+1)²)dx
A = [cx - (x+1)³/3] from -1-√c to -1+√c
= (c (-1+√c) - (√c)³/3) - (c (-1-√c) - (-√c)³/3)
= -c+c√c - c√c/3 + c+c√c - c√c/3 = 2c√c - 2c√c/3 = (4/3)c√c.
Given A = 4√8 / 3 = 8√2 / 3.
(4/3)c^ (3/2) = 8√2 / 3 ⇒ c^ (3/2) = 2√2 = 2^ (3/2) ⇒ c = 2.
∴ y = (x + 1)² - 2.
∴ y (1) = (1 + 1)² - 2 = 2.

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<div class="photo-widget-full"><div class="figure"><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1756892466php8O6uQB.jpeg" alt="" width="183" height="64" loading="lazy"></picture></div></div><p>dy/dx = 2 (x + 1)<br>dy = 2 (x + 1)dx.<br>y = (x + 1)² - c (1)<br>For y = 0, (x+1)² = c ⇒ x = -1 ± √c.<br>Area A = ∫? √c? ¹? √c (0 - [ (x+1)² - c])dx = ∫? √c? ¹? √c (c - (x+1)²)dx<br>A = [cx - (x+1)³/3] from -1-√c to -1+√c<br>= (c (-1+√c) - (√c)³/3) - (c (-1-√c) - (-√c)³/3)<br>= -c+c√c - c√c/3 + c+c√c - c√c/3 = 2c√c - 2c√c/3 = (4/3)c√c.<br>Given A = 4√8 / 3 = 8√2 / 3.<br> (4/3)c^ (3/2) = 8√2 / 3 ⇒ c^ (3/2) = 2√2 = 2^ (3/2) ⇒ c = 2.<br>∴ y = (x + 1)² - 2.<br>∴ y (1) = (1 + 1)² - 2 = 2.</p></div></div>

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