Show that |∫−2𝜋2𝜋 𝑥 2 sin8 (𝑒 𝑥 ) 𝑑𝑥| ≤ (16𝜋 3 )/3
If f is integrable on [a,b], then |f| is integrable on [a,b] and
∣∫abf(x)dx∣≤∫ab∣f∣dx|\int^b_a f(x)dx|\le \int^b_a |f|dx∣∫abf(x)dx∣≤∫ab∣f∣dx
So:
∣∫−2π2πx2sin8(ex)dx∣≤∫−2π2π∣x2sin8(ex)∣dx=|\int^{2\pi}_{-2\pi} x^2sin^8(e^x)dx|\le \int^{2\pi}_{-2\pi} |x^2sin^8(e^x)|dx=∣∫−2π2πx2sin8(ex)dx∣≤∫−2π2π∣x2sin8(ex)∣dx=
=∫−2π2π∣x2∣∣sin8(ex)∣dx≤∫−2π2π∣x2∣dx=∫−2π2πx2dx=16π33=\int^{2\pi}_{-2\pi} |x^2||sin^8(e^x)|dx\le \int^{2\pi}_{-2\pi} |x^2|dx=\int^{2\pi}_{-2\pi} x^2dx=\frac{16\pi^3}{3}=∫−2π2π∣x2∣∣sin8(ex)∣dx≤∫−2π2π∣x2∣dx=∫−2π2πx2dx=316π3
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