int_(0)^(4)|x^(2)-9|dx

Identify sign change points: We need to evaluate the integral of the absolute value function |x^2 - 9| from 0 to 4. To do this, we first need to find the points where the function inside the absolute value, x^2 - 9, changes sign, as this will affect the integral.Split integral into intervals: The function x^2 - 9 equals zero when x^2 = 9, which happens at x = -3 and x = 3. Since we are only interested in the interval from 0 to 4, we only need to consider x = 3. This is the point where the function changes from negative to positive.Evaluate first integral: We split the integral into two parts, from 0 to 3 and from 3 to 4, and remove the absolute value by considering the sign of the function in each interval. For 0 to 3, x^2 - 9 is negative, so we take the negative of the function. For 3 to 4, x^2 - 9 is positive, so we take the function as it is.Evaluate second integral: Now we write the integral as the sum of two integrals: int_(0)^(3)(-(x^2 - 9))dx + int_(3)^(4)(x^2 - 9)dx.Add results: We evaluate the first integral: int_(0)^(3)(-(x^2 - 9))dx = -int_(0)^(3)(x^2 - 9)dx = -[(1/3)x^3 - 9x] from 0 to 3.Plugging in the limits for the first integral, we get -[(1/3)(3)^3 - 9(3) - (0 - 0)] = -[(1/3)(27) - 27] = -[9 - 27] = 18.We evaluate the second integral: int_(3)^(4)(x^2 - 9)dx = [(1/3)x^3 - 9x] from 3 to 4.Plugging in the limits for the second integral, we get [(1/3)(4)^3 - 9(4)] - [(1/3)(3)^3 - 9(3)] = [(64/3) - 36] - [(27/3) - 27] = [(64/3) - (27/3)] - [36 - 27] = (37/3) - 9.Now we add the results of the two integrals: 18 + (37/3) - 9 = 18 + 12.333... - 9 = 21.333...

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$27$ . $\int_{0}^{4}\left|x^{2}-9\right| d x$

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Evaluate definite integrals using the chain rule

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27

\int_{0}^{4}\left|x^{2}-9\right| d x

Identify sign change points: We need to evaluate the integral of the absolute value function $|x^2 - 9|$ from $0$ to $4$ . To do this, we first need to find the points where the function inside the absolute value, $x^2 - 9$ , changes sign, as this will affect the integral.
Split integral into intervals: The function $x^2 - 9$ equals zero when $x^2 = 9$ , which happens at $x = -3$ and $x = 3$ . Since we are only interested in the interval from $0$ to $4$ , we only need to consider $x = 3$ . This is the point where the function changes from negative to positive.
Evaluate first integral: We split the integral into two parts, from $0$ to $3$ and from $3$ to $4$ , and remove the absolute value by considering the sign of the function in each interval. For $0$ to $3$ , $x^2 - 9$ is negative, so we take the negative of the function. For $3$ to $4$ , $x^2 - 9$ is positive, so we take the function as it is.
Evaluate second integral: Now we write the integral as the sum of two integrals: $\int_{0}^{3}(-(x^2 - 9))\,dx + \int_{3}^{4}(x^2 - 9)\,dx$ .
Add results: We evaluate the first integral: $\int_{0}^{3}(-(x^2 - 9))\,dx = -\int_{0}^{3}(x^2 - 9)\,dx = -[\frac{1}{3}x^3 - 9x]$ from $0$ to $3$ .
Add results: We evaluate the first integral: $\int_{0}^{3}(-(x^2 - 9))dx = -\int_{0}^{3}(x^2 - 9)dx = -[\frac{1}{3}x^3 - 9x]$ from $0$ to $3$ .Plugging in the limits for the first integral, we get $-[\frac{1}{3}(3)^3 - 9(3) - (0 - 0)] = -[\frac{1}{3}(27) - 27] = -[9 - 27] = 18$ .
Add results: We evaluate the first integral: $\int_{0}^{3}(-(x^2 - 9))dx = -\int_{0}^{3}(x^2 - 9)dx = -[\frac{1}{3}x^3 - 9x]$ from $0$ to $3$ . Plugging in the limits for the first integral, we get $-[\frac{1}{3}(3)^3 - 9(3) - (0 - 0)] = -[\frac{1}{3}(27) - 27] = -[9 - 27] = 18$ . We evaluate the second integral: $\int_{3}^{4}(x^2 - 9)dx = [\frac{1}{3}x^3 - 9x]$ from $3$ to $4$ .
Add results: We evaluate the first integral: $\int_{0}^{3}(-(x^2 - 9))dx = -\int_{0}^{3}(x^2 - 9)dx = -[\frac{1}{3}x^3 - 9x]$ from $0$ to $3$ . Plugging in the limits for the first integral, we get $-[\frac{1}{3}(3)^3 - 9(3) - (0 - 0)] = -[\frac{1}{3}(27) - 27] = -[9 - 27] = 18$ . We evaluate the second integral: $\int_{3}^{4}(x^2 - 9)dx = [\frac{1}{3}x^3 - 9x]$ from $3$ to $4$ . Plugging in the limits for the second integral, we get $[(\frac{1}{3})(4)^3 - 9(4)] - [(\frac{1}{3})(3)^3 - 9(3)] = [(\frac{64}{3}) - 36] - [(\frac{27}{3}) - 27] = [(\frac{64}{3}) - (\frac{27}{3})] - [36 - 27] = (\frac{37}{3}) - 9$ .
Add results: We evaluate the first integral: $\int_{0}^{3}(-(x^2 - 9))dx = -\int_{0}^{3}(x^2 - 9)dx = -[\frac{1}{3}x^3 - 9x]$ from $0$ to $3$ .Plugging in the limits for the first integral, we get $-[\frac{1}{3}(3)^3 - 9(3) - (0 - 0)] = -[\frac{1}{3}(27) - 27] = -[9 - 27] = 18$ .We evaluate the second integral: $\int_{3}^{4}(x^2 - 9)dx = [\frac{1}{3}x^3 - 9x]$ from $3$ to $4$ .Plugging in the limits for the second integral, we get $[(\frac{1}{3}(4)^3 - 9(4)] - [\frac{1}{3}(3)^3 - 9(3)] = [(\frac{64}{3}) - 36] - [(\frac{27}{3}) - 27] = [(\frac{64}{3}) - (\frac{27}{3})] - [36 - 27] = (\frac{37}{3}) - 9$ .Now we add the results of the two integrals: $18 + (\frac{37}{3}) - 9 = 18 + 12.333... - 9 = 21.333...$