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Find the area of the propeller-shaped region enclosed between the graphs of $$y=\sin x$$ and $$y=x^{3}$$ .

Short Answer

Expert verified
The final answer is approximately 1.670.

Step by step solution

01

Find the points of intersection

Set the two equations equal to each other and solve for x. Hence, \(\sin x = x^{3}\). This equation doesn't have a simple algebraic solution, so numerical methods should be used to solve this equation, obtaining approximately x=0, and x=±0.92.
02

Set up the integral

The area A between the graphs of the functions is given by the integral from a to b of the absolute value of the difference of the functions, \(A = \int_a^b |f(x)-g(x)| dx\). Since within the limits -0.92 to 0.92, the graph of \(f(x) = \sin x\) is above \(g(x) = x^{3}\), the equation becomes \(A = \int_{-0.92}^{0.92} [\sin x - x^{3}] dx\). As the function is symmetric around x=0, this integral can be converted to \(A = 2\int_{0}^{0.92} [\sin x - x^{3}] dx\) to simplify the integrand.
03

Evaluate the integral

By individually integrating \(\sin x\) and \(x^{3}\), and evaluating it at the limits 0 and 0.92, we obtain the final solution for A. Remember that the integral of \(\sin x\) is \(-\cos x\) and the integral of \(x^{3}\) is \(1/4 x^{4}\).
04

Final Calculation

Perform the calculations and approximate the answer to a suitable number of decimal places.

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