Find the area of the region bounded by the given curves. $y=\sin \left(\frac{\pi x}{6}\right), y=x^{2}-6 x$
A) $16+\frac{12}{\pi}$
B) None of these
C) $\frac{-54}{3}-\frac{12}{\pi}$
D) $-54+\frac{\pi}{12}$
E) $\frac{-54}{3}+\frac{12}{\pi}$

Question

Quizplus · Accepted Answer

The area of the region bounded by the curves is given by the integral of the upper curve minus the integral of the lower curve. So, the area is:$$\int_0^6 \left(\sin \left(\frac{\pi x}{6}\right) - (x^2-6x)\right) dx$$$$=\left[-\frac{6}{\pi}\cos \left(\frac{\pi x}{6}\right) - \frac{x^3}{3} + 3x^2\right]_0^6$$$$=\left[-\frac{6}{\pi}\cos \pi - \frac{6^3}{3} + 3(6)^2\right] - \left[-\frac{6}{\pi}\cos 0 - \frac{0^3}{3} + 3(0)^2\right]$$$$=\left[-\frac{6}{\pi} - 72 + 108\right] - \left[-\frac{6}{\pi}\right]$$$$=\left[\frac{-54}{3}+\frac{12}{\pi}\right]$$

Find the Area of the Region Bounded by the Given y=sin⁡(πx6),y=x2−6xy=\sin \left(\frac{\pi x}{6}\right), y=x^{2}-6 xy=sin(6πx​),y=x2−6x

Find the Area of the Region Bounded by the Given $y=\sin \left(\frac{\pi x}{6}\right), y=x^{2}-6 x$