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5. Evaluate \( \int_{-1}^{3}\left(x^{2}-3\right)\left(x^{3}-9 x-6\right)^{3} d x \)

Ask by Perkins Lynch. in Malaysia
Jan 09,2025

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The integral evaluates to \( \frac{320}{3} \).

Solución

To evaluate the integral \[ \int_{-1}^{3}\left(x^{2}-3\right)\left(x^{3}-9 x-6\right)^{3} \, dx, \] we can use substitution. Let's set: \[ u = x^3 - 9x - 6. \] Next, compute the derivative \( du \): \[ \frac{du}{dx} = 3x^2 - 9 = 3(x^2 - 3). \] This implies: \[ (x^2 - 3)\, dx = \frac{du}{3}. \] Substituting into the integral, we get: \[ \int_{u(-1)}^{u(3)} u^3 \cdot \frac{du}{3} = \frac{1}{3} \int_{u(-1)}^{u(3)} u^3 \, du. \] Now, determine the new limits of integration: - When \( x = -1 \): \[ u = (-1)^3 - 9(-1) - 6 = -1 + 9 - 6 = 2. \] - When \( x = 3 \): \[ u = 3^3 - 9(3) - 6 = 27 - 27 - 6 = -6. \] So, the integral becomes: \[ \frac{1}{3} \int_{2}^{-6} u^3 \, du = \frac{1}{3} \left[ \frac{u^4}{4} \right]_2^{-6}. \] Evaluate the definite integral: \[ \frac{1}{3} \left( \frac{(-6)^4}{4} - \frac{2^4}{4} \right) = \frac{1}{3} \left( \frac{1296}{4} - \frac{16}{4} \right) = \frac{1}{3} \left( 324 - 4 \right) = \frac{1}{3} \times 320 = \frac{320}{3}. \] **Final Answer:** \(\boxed{\dfrac{320}{3}}\)

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To evaluate the integral \[ I = \int_{-1}^{3}(x^{2}-3)(x^{3}-9x-6)^{3} \, dx, \] we can start by considering the properties of the integrand, particularly focus on symmetry. Notice that the limits of integration are from \(-1\) to \(3\). Let's first inspect the polynomial \(x^{3} - 9x - 6\) at the endpoints of the integration. Calculate the value of the polynomial at the limits: - For \(x = -1\): \[ x^{3} - 9x - 6 = (-1)^{3} - 9(-1) - 6 = -1 + 9 - 6 = 2. \] - For \(x = 3\): \[ x^{3} - 9x - 6 = (3)^{3} - 9(3) - 6 = 27 - 27 - 6 = -6. \] Next, observe the behavior of \(x^{2} - 3\) over our range of integration. The function \(x^{2} - 3\) is negative when \(-1 < x < \sqrt{3}\) and positive when \(\sqrt{3} < x < 3\). Next, we compute the integral over the specified bounds. Note that the function \(x^3 - 9x - 6\) is a continuous function and we can check if it has roots within the range \(-1\) to \(3\). The roots will help determine the sign of \((x^3 - 9x - 6)^3\). A critical insight is to find out if there is a transformation we can apply that reflects the polynomial about some midpoint, potentially helping to simplify the computation. Let’s define a substitution \(u = x - 1\), which will shift our limits of integration from \([-1, 3]\) to \([-2, 2]\). Substitute into our integral: \[ I = \int_{-2}^{2}((u+1)^{2}-3)((u+1)^{3}-9(u+1)-6)^{3} \, du. \] This form can become quite complex. However, instead, we can look at the original polynomial used, noticing that it is an odd polynomial when correctly formatted. There’s a symmetry between the integral bounds (the center of integration is at \(1\), which lies in the domain, lending to potential cancellation of positive and negative areas. The integrand \((x^{2}-3)(x^{3}-9x-6)^{3}\) has odd symmetry around the midpoint \(x=1\). In particular, we can show: \[ I = f(-x) + f(x) \quad \text{(odd functions) implies } I \text{ evaluates to } 0. \] Thus, we conclude that \[ \int_{-1}^{3}(x^{2}-3)(x^{3}-9x-6)^{3} \,dx = 0. \] The final answer is: \[ \boxed{0}. \]

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