Question

72. Between the sine and inverse sine Find the area of the region bound by the curves y = sin x and y = sin⁻¹x on the interval [0, 1/2].

Accepted Answer

Identify the two functions given: $$y = \sin x$$ and $$y = \sin$$^{-1}$$ x ($$also written as $$y = \arcsin x). We $$are interested in the area between these curves on the interval $$[0, \frac{1}{2}]. $$Determine which function is on top and which is on the bottom over the interval $$[0, \frac{1}{2}]. $$Since $$\sin x is $$increasing and $$\arcsin x is $$also increasing but defined differently, evaluate or reason about their values at key points (like $$x=0$$ and $$x=\frac{1}{2}) to $$find which curve lies above the other. Set up the integral expression for the area between the curves. The area $$A is $$given by the integral of the difference of the top function minus the bottom function over the interval $$[0, \frac{1}{2}]$$:

$$A = \$$int_0$$^{\frac{1}{2}} \left( 	ext{top function} - 	ext{bottom function} ight) \, dx. $$Write the integral explicitly using the functions identified in step 2. For example, if $$\arcsin x is on $$top, then

$$A = \$$int_0$$^{\frac{1}{2}} \left( \arcsin x - \sin x ight) \, dx. $$Evaluate the integral by integrating each term separately. Recall that the integral of $$\arcsin x$$ can be found using integration by parts, and the integral of $$\sin x is $$straightforward. Set up the integration by parts for $$\int \arcsin x \, dx$$ and then subtract the integral of $$\sin x$$ over the interval.

72. Between the sine and inverse sine Find the area of the region bound by the curves y = sin x and y = sin⁻¹x on the interval [0, 1/2].

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Key Concepts

Understanding the Functions y = sin x and y = sin⁻¹ x

Finding the Area Between Two Curves

Integration Techniques and Limits