int_{0}^{4/pi} left( 3x^2 sin frac{1}{x} - x | vedclass

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Question

(C) Let $I = \int_{0}^{4/\pi} \left( 3x^2 \sin \frac{1}{x} - x \cos \frac{1}{x} \right) dx$. Consider the integral $\int 3x^2 \sin \frac{1}{x} dx$. Us

Vedclass · Accepted Answer

$\frac{32\sqrt{2}}{\pi^3}$

Vedclass · Answer

(C) Let $I = \int_{0}^{4/\pi} \left( 3x^2 \sin \frac{1}{x} - x \cos \frac{1}{x} \right) dx$. Consider the integral $\int 3x^2 \sin \frac{1}{x} dx$. Using integration by parts,let $u = \sin \frac{1}{x}$ and $dv = 3x^2 dx$. Then $du = -\cos \frac{1}{x} \cdot (-\frac{1}{x^2}) dx = \frac{1}{x^2} \cos \frac{1}{x} dx$ and $v = x^3$. Thus,$\int 3x^2 \sin \frac{1}{x} dx = x^3 \sin \frac{1}{x} - \int x^3 \cdot \frac{1}{x^2} \cos \frac{1}{x} dx = x^3 \sin \frac{1}{x} - \int x \cos \frac{1}{x} dx$. Substituting this into the original integral: $I = \left[ x^3 \sin \frac{1}{x} - \int x \cos \frac{1}{x} dx \right]_{0}^{4/\pi} - \int_{0}^{4/\pi} (-x \cos \frac{1}{x}) dx$. $I = \left[ x^3 \sin \frac{1}{x} \right]_{0}^{4/\pi} - \int_{0}^{4/\pi} x \cos \frac{1}{x} dx + \int_{0}^{4/\pi} x \cos \frac{1}{x} dx$. $I = \left[ x^3 \sin \frac{1}{x} \right]_{0}^{4/\pi} = (4/\pi)^3 \sin(\pi/4) - 0 = \frac{64}{\pi^3} \cdot \frac{1}{\sqrt{2}} = \frac{64}{\sqrt{2} \pi^3} = \frac{32 \cdot 2}{\sqrt{2} \pi^3} = \frac{32\sqrt{2}}{\pi^3}$.

$\int_{0}^{4/\pi} \left( 3x^2 \sin \frac{1}{x} - x \cos \frac{1}{x} \right) dx$ has the value:

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