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∫3x(2x−1)2 dx \int 3\sqrt{x}(2x-1)^{2}\,dx

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Full solution

Q. ∫3x(2x−1)2 dx \int 3\sqrt{x}(2x-1)^{2}\,dx
  1. Recognize Integral Independence: First, we need to recognize that the integral provided does not depend on an angle θ\theta, so the limits of integration from 00 to 180180 degrees are not applicable. Instead, we should consider the integral in terms of xx. The integral to solve is ∫3x(2x−1)2 dx\int 3\sqrt{x}(2x-1)^2 \, dx.
  2. Expand Integrands: Next, we will expand the integrand. The term (2x−1)2(2x-1)^2 can be expanded to 4x2−4x+14x^2 - 4x + 1. So the integral becomes ∫3x(4x2−4x+1) dx\int 3\sqrt{x}(4x^2 - 4x + 1) \, dx.
  3. Distribute and Simplify: We can distribute 3x3\sqrt{x} across the terms inside the parentheses to get ∫(3x×4x2−3x×4x+3x×1)dx\int(3\sqrt{x} \times 4x^2 - 3\sqrt{x} \times 4x + 3\sqrt{x} \times 1) dx, which simplifies to ∫(4x5/2−4x3/2+3x1/2)dx\int(4x^{5/2} - 4x^{3/2} + 3x^{1/2}) dx.
  4. Integrate Each Term: Now we can integrate each term separately. The antiderivative of xnx^{n} is (xn+1)/(n+1)(x^{n+1})/(n+1) for n≠−1n \neq -1. Applying this rule to each term, we get:\newline∫4x5/2dx=(4/((5/2)+1))x(5/2)+1=(4/((7/2)))x7/2=(8/7)x7/2\int 4x^{5/2} dx = (4/((5/2)+1))x^{(5/2)+1} = (4/((7/2)))x^{7/2} = (8/7)x^{7/2}\newline∫−4x3/2dx=(−4/((3/2)+1))x(3/2)+1=(−4/((5/2)))x5/2=(−8/5)x5/2\int -4x^{3/2} dx = (-4/((3/2)+1))x^{(3/2)+1} = (-4/((5/2)))x^{5/2} = (-8/5)x^{5/2}\newline∫3x1/2dx=(3/((1/2)+1))x(1/2)+1=(3/(3/2))x3/2=2x3/2\int 3x^{1/2} dx = (3/((1/2)+1))x^{(1/2)+1} = (3/(3/2))x^{3/2} = 2x^{3/2}
  5. Combine Results: Combining these results, the antiderivative of the original integral is (87)x(72)−(85)x(52)+2x(32)+C(\frac{8}{7})x^{(\frac{7}{2})} - (\frac{8}{5})x^{(\frac{5}{2})} + 2x^{(\frac{3}{2})} + C, where CC is the constant of integration.
  6. Evaluate Final Answer: Since there are no limits of integration provided in the original problem, we cannot evaluate the definite integral. The final answer is the indefinite integral in its simplest form.

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