Final answer:
The integral of x³√(4-x²) from -2 to 2 is equal to zero due to the symmetry of the integrand function being odd. Only the positive interval from 0 to 2 needs to be considered for the area calculation, and the integral from -2 to 0 would be its negative counterpart. By understanding the odd-function property, we can simplify the evaluation of the integral to recognize that it equals zero.
Step-by-step explanation:
Evaluating the Integral
You’ve been asked to evaluate the integral ∫ x³sqrt(4-x²) dx from -2 to 2 by splitting it into two integrals and interpreting one as an area. The function √(4 - x²) represents a semi-circle when plotted. Since the integrand x³√(4-x²) has an odd exponent of x in it (x³), half of the integral from -2 to 2 will be the negative of the other half; hence, the two halves cancel each other out, making the integral equal to zero. It’s like finding the net area between a curve and the x-axis.
However, if we only look at the interval from 0 to 2, we are essentially calculating the area under the curve of the function x³√(4-x²) above the x-axis from 0 to 2. This area equals the integral from 0 to 2 of the function above, and is the only part we need to actually compute, as the integral from -2 to 0 will be the negative of this value.
By symmetry, the integral over [-2, 0] cancels out with the integral over [0, 2], so you can write the integral as:
∫_{-2}^{2} x³√(4-x²) dx = ∫_{-2}^{0} x³√(4-x²) dx + ∫_{0}^{2} x³√(4-x²) dx
But since the integrand is odd:
∫_{-2}^{2} x³√(4-x²) dx = 0
The significance comes in recognizing the odd-function property of the integrand, leading to an integral with a result of zero. Recognizing this can greatly simplify the evaluation of certain types of definite integrals.