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Questionlist ∣←\mid \leftarrow Use the equation, ∫abf(x)dx=lim⁡n→∞∑k=1nf(a+kb−an)(b−an)\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(3x2−2x+6)dx\int_{1}^{4}\left(3 x^{2}-2 x+6\right) d x
Evaluate the Riemann Sum. Choose the correct choice below. A−∫abf(x)dx=lim⁡∥P∥→0∑k=1nf(xk∗)Δxk=lim⁡n→∞∑k=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=lim⁡∥P∥→0∑k=1nf(xk∗)Δxk=lim⁡n→∞∑k=1n[3(1+3kn)2−2(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=lim⁡∥P∥→0∑k=1nf(xk∗)Δxk=lim⁡n→∞∑k=1n[(1+nk)2−2(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=lim⁡∥P∥→0∑k=1nf(xk∗)Δxk=lim⁡n→∞∑k=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(3x2−2x+6)dx=\int_{1}^{4}\left(3 x^{2}-2 x+6\right) d x= □\square (Simplify your answer.)

Studdy Solution

STEP 1

What is this asking? We need to use the Riemann sum definition of a definite integral to evaluate the integral of 3x2−2x+63x^2 - 2x + 6 from x=1x = 1 to x=4x = 4. Watch out! Don't forget to carefully substitute the correct values for aa, bb, and f(x)f(x) into the Riemann sum formula.
Also, remember to simplify your final answer!

STEP 2

1. Set up the Riemann Sum
2. Expand and Simplify the Summand
3. Evaluate the Sum
4. Take the Limit

STEP 3

We're given the integral ∫14(3x2−2x+6)dx\int_{1}^{4} (3x^2 - 2x + 6) dx.
Here, a=1a = \mathbf{1}, b=4b = \mathbf{4}, and f(x)=3x2−2x+6f(x) = \mathbf{3x^2 - 2x + 6}.

STEP 4

The Riemann sum formula is ∫abf(x)dx=lim⁡n→∞∑k=1nf(a+kb−an)⋅b−an\int_{a}^{b} f(x) dx = \lim_{n \rightarrow \infty} \sum_{k=1}^{n} f\left(a + k \frac{b-a}{n}\right) \cdot \frac{b-a}{n}.
Substituting our values, we get: ∫14(3x2−2x+6)dx=lim⁡n→∞∑k=1nf(1+k4−1n)⋅4−1n \int_{1}^{4} (3x^2 - 2x + 6) dx = \lim_{n \rightarrow \infty} \sum_{k=1}^{n} f\left(1 + k \frac{4-1}{n}\right) \cdot \frac{4-1}{n} =lim⁡n→∞∑k=1nf(1+3kn)⋅3n = \lim_{n \rightarrow \infty} \sum_{k=1}^{n} f\left(1 + \frac{3k}{n}\right) \cdot \frac{3}{n} =lim⁡n→∞∑k=1n[3(1+3kn)2−2(1+3kn)+6]⋅3n = \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left[ 3\left(1 + \frac{3k}{n}\right)^2 - 2\left(1 + \frac{3k}{n}\right) + 6 \right] \cdot \frac{3}{n} This matches option B.

STEP 5

(1+3kn)2=1+6kn+9k2n2 \left(1 + \frac{3k}{n}\right)^2 = 1 + \frac{6k}{n} + \frac{9k^2}{n^2}

STEP 6

[3(1+6kn+9k2n2)−2(1+3kn)+6]⋅3n \left[ 3\left(1 + \frac{6k}{n} + \frac{9k^2}{n^2}\right) - 2\left(1 + \frac{3k}{n}\right) + 6 \right] \cdot \frac{3}{n} =[3+18kn+27k2n2−2−6kn+6]⋅3n = \left[ 3 + \frac{18k}{n} + \frac{27k^2}{n^2} - 2 - \frac{6k}{n} + 6 \right] \cdot \frac{3}{n} =[7+12kn+27k2n2]⋅3n=21n+36kn2+81k2n3 = \left[ 7 + \frac{12k}{n} + \frac{27k^2}{n^2} \right] \cdot \frac{3}{n} = \frac{21}{n} + \frac{36k}{n^2} + \frac{81k^2}{n^3}

STEP 7

lim⁡n→∞∑k=1n(21n+36kn2+81k2n3)=lim⁡n→∞(∑k=1n21n+∑k=1n36kn2+∑k=1n81k2n3) \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left( \frac{21}{n} + \frac{36k}{n^2} + \frac{81k^2}{n^3} \right) = \lim_{n \rightarrow \infty} \left( \sum_{k=1}^{n} \frac{21}{n} + \sum_{k=1}^{n} \frac{36k}{n^2} + \sum_{k=1}^{n} \frac{81k^2}{n^3} \right)

STEP 8

Recall ∑k=1nc=nc\sum_{k=1}^{n} c = nc, ∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}, and ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}. lim⁡n→∞(21n⋅n+36n2⋅n(n+1)2+81n3⋅n(n+1)(2n+1)6) \lim_{n \rightarrow \infty} \left( \frac{21}{n} \cdot n + \frac{36}{n^2} \cdot \frac{n(n+1)}{2} + \frac{81}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \right)

STEP 9

lim⁡n→∞(21+18(n+1)n+27(n+1)(2n+1)2n2) \lim_{n \rightarrow \infty} \left( 21 + \frac{18(n+1)}{n} + \frac{27(n+1)(2n+1)}{2n^2} \right)

STEP 10

=21+18lim⁡n→∞n+1n+272lim⁡n→∞2n2+3n+1n2=21+18(1)+272(2)=21+18+27=66 = 21 + 18\lim_{n \rightarrow \infty}\frac{n+1}{n} + \frac{27}{2}\lim_{n \rightarrow \infty}\frac{2n^2 + 3n + 1}{n^2} = 21 + 18(1) + \frac{27}{2}(2) = 21 + 18 + 27 = 66

STEP 11

The Riemann sum evaluates to 66\mathbf{66}, so ∫14(3x2−2x+6)dx=66\int_{1}^{4} (3x^2 - 2x + 6) dx = \mathbf{66}.

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