Let g(x)= x^2, f(x)= x, and , (

Question

Let $g(x)=\cos x^2$, $f(x)=\sqrt x$, and $\alpha,\beta$ $(\alpha<\beta)$ be the roots of the quadratic equation $18x^2-9\pi x+\pi^2=0$. Then the area (in sq. units) bounded by the curve $y=(g\circ f)(x)$ and the lines $x=\alpha$, $x=\beta$, $y=0$, is:

Accepted Answer

$18x^2-9\pi x+\pi^2=0 \Rightarrow (3x-\pi)(6x-\pi)=0 \Rightarrow \alpha=\dfrac{\pi}{6},\ \beta=\dfrac{\pi}{3}$.

$(g\circ f)(x) = \cos\left((\sqrt x)^2\right) = \cos x$.

$$\text{Area} = \int_{\pi/6}^{\pi/3}\cos x\,dx = [\sin x]_{\pi/6}^{\pi/3} = \frac{\sqrt3}{2}-\frac12 = \frac{\sqrt3-1}{2}$$

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