Answer to Question #242622 in Real Analysis for saduni

Question #242622

Show that |2πœ‹-2πœ‹x2sin8 (𝑒 π‘₯ ) 𝑑π‘₯| ≀ 16πœ‹ 3/ 3 .

1
Expert's answer
2021-09-28T14:38:58-0400

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}\n 2\\pi \\\\\n -2\\pi\n\\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}"


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