QuestionJuly 15, 2025

Find int _(0)^3vert x^2-4vert dx -(23)/(3) (23)/(3) 3 -3

Find int _(0)^3vert x^2-4vert dx -(23)/(3) (23)/(3) 3 -3
Find int _(0)^3vert x^2-4vert dx -(23)/(3) (23)/(3) 3 -3

Solution
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Answer

\frac{19}{3} Explanation 1. Identify the critical points The expression x^2 - 4 changes sign at x = \pm 2. Within the interval [0, 3], the critical point is x = 2. 2. Split the integral Split the integral at x = 2: \int_{0}^{3} |x^2 - 4| \, dx = \int_{0}^{2} (4 - x^2) \, dx + \int_{2}^{3} (x^2 - 4) \, dx. 3. Evaluate the first integral Calculate \int_{0}^{2} (4 - x^2) \, dx = \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} = \left(8 - \frac{8}{3}\right) = \frac{16}{3}. 4. Evaluate the second integral Calculate \int_{2}^{3} (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_{2}^{3} = \left(\frac{27}{3} - 12\right) - \left(\frac{8}{3} - 8\right) = \frac{19}{3} - \left(-\frac{16}{3}\right) = \frac{3}{3} = 1. 5. Sum the integrals Add the results: \frac{16}{3} + 1 = \frac{19}{3}.

Explanation

1. Identify the critical points<br /> The expression $x^2 - 4$ changes sign at $x = \pm 2$. Within the interval $[0, 3]$, the critical point is $x = 2$.<br /><br />2. Split the integral<br /> Split the integral at $x = 2$: $\int_{0}^{3} |x^2 - 4| \, dx = \int_{0}^{2} (4 - x^2) \, dx + \int_{2}^{3} (x^2 - 4) \, dx$.<br /><br />3. Evaluate the first integral<br /> Calculate $\int_{0}^{2} (4 - x^2) \, dx = \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} = \left(8 - \frac{8}{3}\right) = \frac{16}{3}$.<br /><br />4. Evaluate the second integral<br /> Calculate $\int_{2}^{3} (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_{2}^{3} = \left(\frac{27}{3} - 12\right) - \left(\frac{8}{3} - 8\right) = \frac{19}{3} - \left(-\frac{16}{3}\right) = \frac{3}{3} = 1$.<br /><br />5. Sum the integrals<br /> Add the results: $\frac{16}{3} + 1 = \frac{19}{3}$.
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