Write the definite integral $\int_{- 2}^{6} (x^{2} + 3) d x$ as a limit of a Riemann sum and evaluate.
$lim_{n \to \infty} \sum_{k = 1}^{n} (7 + \frac{32 k}{n} + \frac{64 k^{2}}{n^{2}}) (\frac{8}{n}) = \frac{1064}{3}$
$lim_{n \to \infty} \sum_{k = 1}^{n} (4 - \frac{32 k}{n} + \frac{64 k^{2}}{n^{2}}) (\frac{8}{n}) = \frac{224}{3}$
$lim_{n \to \infty} \sum_{k = 1}^{n} (7 - \frac{32 k}{n} + \frac{64 k^{2}}{n^{2}}) (\frac{8}{n}) = \frac{296}{3}$
$lim_{n \to \infty} \sum_{k = 1}^{n} (4 + \frac{64 k^{2}}{n^{2}}) (\frac{8}{n}) = \frac{608}{3}$

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Final Answer: The definite integral $\int_{- 2}^{6} (x^{2} + 3) d x$ is $\frac{296}{3}$ .

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Step 1 ：The problem is asking for the definite integral of the function $f (x) = x^{2} + 3$ from $- 2$ to $6$ .

Step 2 ：This can be calculated using the Fundamental Theorem of Calculus, which states that the definite integral of a function from $a$ to $b$ is equal to the antiderivative of the function evaluated at $b$ minus the antiderivative of the function evaluated at $a$ .

Step 3 ：The antiderivative of $f (x) = x^{2} + 3$ is $F (x) = \frac{1}{3} x^{3} + 3 x$ .

Step 4 ：So, the definite integral of $f (x)$ from $- 2$ to $6$ is $F (6) - F (- 2)$ .

Step 5 ：Final Answer: The definite integral $\int_{- 2}^{6} (x^{2} + 3) d x$ is $\frac{296}{3}$ .

Write the definite integral ∫−26(x2+3)dx as a limit of a Riemann sum and evaluate.limn→∞∑k=1n(7+32kn+64k2n2)(8n)=10643limn→∞∑k=1n(4−32kn+64k2n2)(8n)=2243limn→∞∑k=1n(7−32kn+64k2n2)(8n)=2963limn→∞∑k=1n(4+64k2n2)(8n)=6083

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