Math  /  Calculus

Questionlist \mid \leftarrow Use the equation, abf(x)dx=limnk=1nf(a+kban)(ban)\int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} f\left(a+k \frac{b-a}{n}\right)\left(\frac{b-a}{n}\right), to evaluate the following definite integral. 14(3x22x+6)dx\int_{1}^{4}\left(3 x^{2}-2 x+6\right) d x
Evaluate the Riemann Sum. Choose the correct choice below. Aabf(x)dx=limP0k=1nf(xk)Δxk=limnk=1n[(3kn)2+6](3n)A^{-} \int_{a}^{b} f(x) d x=\lim _{\|P\| \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left[\left(\frac{3 k}{n}\right)^{2}+6\right]\left(\frac{3}{n}\right) B. abf(x)dx=limP0k=1nf(xk)Δxk=limnk=1n[3(1+3kn)22(1+3kn)+6](3n)\int_{a}^{b} f(x) d x=\lim _{\|P\| \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left[3\left(1+\frac{3 k}{n}\right)^{2}-2\left(1+\frac{3 k}{n}\right)+6\right]\left(\frac{3}{n}\right) c. abf(x)dx=limP0k=1nf(xk)Δxk=limnk=1n[(1+nk)22(1+nk)+6](n3)\int_{a}^{b} f(x) d x=\lim _{\|P\| \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left[\left(1+\frac{n}{k}\right)^{2}-2\left(1+\frac{n}{k}\right)+6\right]\left(\frac{n}{3}\right) D. abf(x)dx=limP0k=1nf(xk)Δxk=limnk=1n(1+2kn)((1+kn)2nk)(2n)\int_{a}^{b} f(x) d x=\lim _{\|P\| \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(1+\frac{2 k}{n}\right)\left(\left(1+\frac{k}{n}\right)-2 \frac{n}{k}\right)\left(\frac{2}{n}\right)
Evaluate the integral. 14(3x22x+6)dx=\int_{1}^{4}\left(3 x^{2}-2 x+6\right) d x= \square (Simplify your answer.)

Studdy Solution
The Riemann sum evaluates to 66\mathbf{66}, so 14(3x22x+6)dx=66\int_{1}^{4} (3x^2 - 2x + 6) dx = \mathbf{66}.

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