It can be shown that the area enclosed between the x-axis and one arch of the sine curve is 2. Use this fact to compute the
definite integral.
-$\int _ { 0 } ^ { 6 \pi } \sin \left( \frac { x } { 6 } \right) d x$ (Hint: each rectangle is 6 times as wide.)
A) 72
B) 2
C) 12
D) 6

Question

Quizplus · Accepted Answer

The integral $\int _ { 0 } ^ { 6 \pi } \sin \left( \frac { x } { 6 } \right) d x$ represents the area under the curve of $\sin \left( \frac { x } { 6 } \right)$ from $0$ to $6\pi$. Given that the area under one arch of the sine curve is 2, and considering the transformation of the sine function, we can deduce that the period of $\sin \left( \frac { x } { 6 } \right)$ is $6\pi$, which is 6 times the period of the standard sine function. Since the integral covers $6\pi$ (or one full period of this transformed sine function), it encompasses 6 arches of the standard sine curve (each stretched to be 6 times wider but with the same area under each arch). Therefore, the total area is $6 \times 2 = 12$.

It Can Be Shown That the Area Enclosed Between the X-Axis