int_0^3 left(sin left(frac{pi}{3} xright) - c | vedclass

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Question

(D) Let $I = \int_0^3 \left(\sin \left(\frac{\pi}{3} x\right) - \cos \left(\frac{\pi}{3} x\right)\right) dx$. Integrating term by term: $I = \int_0^3

Vedclass · Accepted Answer

$\frac{6}{\pi}$

Vedclass · Answer

(D) Let $I = \int_0^3 \left(\sin \left(\frac{\pi}{3} x\right) - \cos \left(\frac{\pi}{3} x\right)\right) dx$. Integrating term by term: $I = \int_0^3 \sin \left(\frac{\pi}{3} x\right) dx - \int_0^3 \cos \left(\frac{\pi}{3} x\right) dx$. Using the formula $\int \sin(ax) dx = -\frac{1}{a} \cos(ax)$ and $\int \cos(ax) dx = \frac{1}{a} \sin(ax)$: $I = \left[ -\frac{3}{\pi} \cos \left(\frac{\pi}{3} x\right) \right]_0^3 - \left[ \frac{3}{\pi} \sin \left(\frac{\pi}{3} x\right) \right]_0^3$. $I = -\frac{3}{\pi} [\cos(\pi) - \cos(0)] - \frac{3}{\pi} [\sin(\pi) - \sin(0)]$. $I = -\frac{3}{\pi} [-1 - 1] - \frac{3}{\pi} [0 - 0]$. $I = -\frac{3}{\pi} [-2] = \frac{6}{\pi}$.

$\int_0^3 \left(\sin \left(\frac{\pi}{3} x\right) - \cos \left(\frac{\pi}{3} x\right)\right) dx =$

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