Let g(x) = cos x2 , f(x) = , and α, β (α < β) be the roots of the quadratic equation 18x2 – 9πx + π2 = 0. Then the area (in sq. units) bounded by the curve y = (gof) (x) and the lines x = α, x = β and y = 0, is

Engineering

Mathematics

Introduction to Area

Question

Let g(x) = cos x² , f(x) = $\sqrt{x}$ , and α, β (α < β) be the roots of the quadratic equation 18x² – 9πx + π² = 0. Then the area (in sq. units) bounded by the curve y = (gof) (x) and the lines x = α, x = β and y = 0, is

$\frac{1}{2} (\sqrt{2} - 1)$

$\frac{1}{2} (\sqrt{3} + 1)$

$\frac{1}{2} (\sqrt{3} - 1)$

$\frac{1}{2} (\sqrt{3} - \sqrt{2})$

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$∵$ 18x² – 9πx + π² = 0

18x² – 6πx – 3πx + π² = 0

⇒ 6x (3x – π) – π (3x – π) = 0 ⇒ (3x – π) (6x – π) = 0

$\Rightarrow x = \frac{π}{3} or \frac{π}{6}$

$∴ α = \frac{π}{6}, β = \frac{π}{3}$

gof (x) = cos x

⇒ Area bounded $= \int_{\frac{π}{6}}^{\frac{π}{3}} cosx dx = {(sinx)}_{π / 6}^{π / 3} = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2}$ .