Problem
Write the definite integral $\int_{-2}^{6}\left(x^{2}+3\right) d x$ as a limit of a Riemann sum and evaluate. $\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(7+\frac{32 k}{n}+\frac{64 k^{2}}{n^{2}}\right)\left(\frac{8}{n}\right)=\frac{1064}{3}$ $\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(4-\frac{32 k}{n}+\frac{64 k^{2}}{n^{2}}\right)\left(\frac{8}{n}\right)=\frac{224}{3}$ $\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(7-\frac{32 k}{n}+\frac{64 k^{2}}{n^{2}}\right)\left(\frac{8}{n}\right)=\frac{296}{3}$ $\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(4+\frac{64 k^{2}}{n^{2}}\right)\left(\frac{8}{n}\right)=\frac{608}{3}$
Solution
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 \(F(x) = \frac{1}{3}x^3 + 3x\).
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 \(\int_{-2}^{6}\left(x^{2}+3\right) d x\) is \(\boxed{\frac{296}{3}}\).
