which of the following is a left riemann sum approximation of ∫₁⁷ (4 ln x + 2) dx with n subintervals of…

which of the following is a left riemann sum approximation of ∫₁⁷ (4 ln x + 2) dx with n subintervals of equal length?\na ∑ₖ₌₁ⁿ (4 ln (1 + (k - 1)/n) + 2) 1/n\nb ∑ₖ₌₁ⁿ (4 ln (6k/n) + 2) 6/n\nc ∑ₖ₌₁ⁿ (4 ln (1 + 6(k - 1)/n) + 2) 6/n\nd ∑ₖ₌₁ⁿ (4 ln (1 + 6k/n) + 2) 6/n

which of the following is a left riemann sum approximation of ∫₁⁷ (4 ln x + 2) dx with n subintervals of equal length?\na ∑ₖ₌₁ⁿ (4 ln (1 + (k - 1)/n) + 2) 1/n\nb ∑ₖ₌₁ⁿ (4 ln (6k/n) + 2) 6/n\nc ∑ₖ₌₁ⁿ (4 ln (1 + 6(k - 1)/n) + 2) 6/n\nd ∑ₖ₌₁ⁿ (4 ln (1 + 6k/n) + 2) 6/n

Answer

Explanation:

Step1: Recall left - Riemann sum formula

The left - Riemann sum for $\int_{a}^{b}f(x)dx$ with $n$ sub - intervals of equal length $\Delta x=\frac{b - a}{n}$ is given by $\sum_{k = 1}^{n}f(x_{k-1})\Delta x$, where $x_{k}=a + k\Delta x$ and $x_{k-1}=a+(k - 1)\Delta x$. Here, $a = 1$, $b = 7$, so $\Delta x=\frac{7 - 1}{n}=\frac{6}{n}$.

Step2: Find $x_{k-1}$

$x_{k-1}=1+(k - 1)\frac{6}{n}$.

Step3: Substitute into $f(x)$

The function $f(x)=4\ln x+2$. Substituting $x = x_{k-1}=1+\frac{6(k - 1)}{n}$ into $f(x)$, we get $f(x_{k-1})=4\ln(1+\frac{6(k - 1)}{n})+2$.

Step4: Form the left - Riemann sum

The left - Riemann sum is $\sum_{k = 1}^{n}(4\ln(1+\frac{6(k - 1)}{n})+2)\frac{6}{n}$.

Answer:

C. $\sum_{k = 1}^{n}(4\ln(1+\frac{6(k - 1)}{n})+2)\frac{6}{n}$