Answer to Question #242622 in Real Analysis for saduni

Question

Show that |2𝜋-2𝜋x2sin8 (𝑒 𝑥 ) 𝑑𝑥| ≤ 16^{𝜋 3}/ 3 .

Answer 1

The function $f(x)=x^2\sin^8(e^x)$ is continuous on $[-2\pi, 2\pi].$ Then

$$|\displaystyle\int_{-2\pi}^{2\pi}x^2\sin^8(e^x)dx|\leq\displaystyle\int_{-2\pi}^{2\pi}|x^2\sin^8(e^x)|dx$$

$$0\leq|sin^8(e^x)|\leq1=>|x^2\sin^8(e^x)|\leq x^2, x\in\R$$

Then

$$\displaystyle\int_{-2\pi}^{2\pi}|x^2\sin^8(e^x)|dx\leq\displaystyle\int_{-2\pi}^{2\pi}x^2dx$$

$$\displaystyle\int_{-2\pi}^{2\pi}x^2dx=[\dfrac{x^3}{3}]\begin{matrix} 2\pi \\ -2\pi \end{matrix}=\dfrac{16\pi^3}{3}$$

Therefore

$$|\displaystyle\int_{-2\pi}^{2\pi}x^2\sin^8(e^x)dx|\leq\displaystyle\int_{-2\pi}^{2\pi}|x^2\sin^8(e^x)|dx$$

$$\leq\displaystyle\int_{-2\pi}^{2\pi}x^2dx=\dfrac{16\pi^3}{3}$$

Therefore

$$|\displaystyle\int_{-2\pi}^{2\pi}x^2\sin^8(e^x)dx|\leq\dfrac{16\pi^3}{3}$$