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int_(-2)^(2)(x^(3)cos ((x)/(2))+(1)/(2))sqrt(4-x^(2)dx)

∫−22(x3cos⁡x2+12)4−x2dx \int_{-2}^{2}\left(x^{3} \cos \frac{x}{2}+\frac{1}{2}\right) \sqrt{4-x^{2} d x}

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Q. ∫−22(x3cos⁡x2+12)4−x2dx \int_{-2}^{2}\left(x^{3} \cos \frac{x}{2}+\frac{1}{2}\right) \sqrt{4-x^{2} d x}
  1. Given Integral: We are given the integral to evaluate:\newline∫−22(x3cos⁡(x2)+12)4−x2 dx\int_{-2}^{2} (x^3 \cos(\frac{x}{2}) + \frac{1}{2}) \sqrt{4 - x^2} \, dx\newlineFirst, we notice that the function has a square root of 4−x24 - x^2, which suggests a trigonometric substitution might be useful. However, before we proceed with any substitution, let's examine the integrand for any symmetry properties that might simplify the calculation.
  2. Symmetry Properties: We observe that the function x3cos⁡(x2)x^3 \cos(\frac{x}{2}) is an odd function because replacing xx with −x-x changes the sign of the function, and the function 4−x2\sqrt{4 - x^2} is even because replacing xx with −x-x does not change the value of the function. The product of an odd function and an even function is an odd function. The term 12\frac{1}{2} is a constant and does not affect the symmetry of the first term.\newlineSince the integral is over a symmetric interval from −2-2 to 22, the integral of the odd function part will be zero. Therefore, we only need to evaluate the integral of the constant term (12)(\frac{1}{2}) multiplied by 4−x2\sqrt{4 - x^2}.
  3. Evaluate Remaining Term: Now we evaluate the integral of the remaining term: \newline∫−22124−x2 dx\int_{-2}^{2} \frac{1}{2} \sqrt{4 - x^2} \, dx\newlineThis integral represents the area of a semicircle with radius 22, because the function 4−x2\sqrt{4 - x^2} describes a semicircle. The area of a full circle with radius rr is πr2\pi r^2, so the area of a semicircle is 12πr2\frac{1}{2}\pi r^2.
  4. Calculate Semicircle Area: We calculate the area of the semicircle with radius 22: \newlineArea = (1/2)π(2)2=(1/2)π(4)=2π(1/2)\pi(2)^2 = (1/2)\pi(4) = 2\pi\newlineThis is the value of the integral of the constant term (1/2)(1/2) multiplied by 4−x2\sqrt{4 - x^2} from −2-2 to 22.
  5. Final Integral Value: Therefore, the value of the original integral is also 2π2\pi, since the integral of the odd function part is zero.

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