Question

Let \(\alpha\) and \(\beta\), \((\alpha\lt \beta)\), are roots of \(18x^2-9\pi x+\pi^2=0\), \(f(x)=x^2\), \(g(x)=\cos x\). Then \(\displaystyle \int_{\alpha}^{\beta} x(gof(x))dx=\)

• A. \(\dfrac{\sqrt{3}-1}{4}\)
• B. \(\dfrac{\sqrt{3}}{4}\)
• C. \(\dfrac{2+\sqrt{3}}{2}\)
• D. \(\dfrac{1}{2}\left(\sin\dfrac{\pi^2}{9}-\sin\dfrac{\pi^2}{36}\right)\)

Hint:

For integrals of the form \(\int x\cos(x^2)\,dx\), use the substitution \(u=x^2\), so that \(du=2x\,dx\).

Answer 1

The Correct Option is D

Step 1: Simplify the composite function.
Since $f(x)=x^2$ and $g(x)=\cos x$, we have $(g\circ f)(x)=g(f(x))=\cos(x^2)$, so the integrand is $x\cos(x^2)$.
Step 2: Solve the quadratic for the limits.
For $18x^2-9\pi x+\pi^2=0$, the discriminant is $81\pi^2-72\pi^2=9\pi^2$, so $x=\dfrac{9\pi\pm3\pi}{36}$.
Step 3: Identify $\alpha$ and $\beta$.
The roots are $\dfrac{\pi}{3}$ and $\dfrac{\pi}{6}$. Since $\alpha<\beta$, $\alpha=\dfrac{\pi}{6}$, $\beta=\dfrac{\pi}{3}$.
Step 4: Substitute $u=x^2$.
Then $du=2x\,dx$, so $x\,dx=\dfrac{du}{2}$ and $\displaystyle\int x\cos(x^2)\,dx=\dfrac12\sin(x^2)$.
Step 5: Apply the limits.
$\displaystyle\int_{\pi/6}^{\pi/3}x\cos(x^2)\,dx=\dfrac12\Big[\sin(x^2)\Big]_{\pi/6}^{\pi/3}=\dfrac12\Big(\sin\tfrac{\pi^2}{9}-\sin\tfrac{\pi^2}{36}\Big)$.
Step 6: State the answer.
This matches the given option.
\[ \boxed{\dfrac{1}{2}\left(\sin\dfrac{\pi^2}{9}-\sin\dfrac{\pi^2}{36}\right)} \]