The area of the region (in square units) bounded by x=1, x=2 and the curve y^2 = 4x in the first quadrant is Option 1: 4/3[2sqrt(2) - 1] Option 2: 2/3[2sqrt(2) - 1] Option 3: 2/3[2sqrt(2) + 1] Option 4: 4/3[2sqrt(2) + 1]

The answer is '4/3[2sqrt(2) - 1]'. View full solution:

Calculus > Application of Integrals

Medium

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The area of the region (in square units) bounded by $x=1, x=2$ and the curve $y^2 = 4x$ in the first quadrant is

$\frac{4}{3}[2\sqrt{2} - 1]$

$\frac{2}{3}[2\sqrt{2} - 1]$

$\frac{2}{3}[2\sqrt{2} + 1]$

$\frac{4}{3}[2\sqrt{2} + 1]$

✅ Correct Option: 1

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