which of the following limits is equal to $int_{2}^{5}sqrt{1 + x^{4}}dx$?\n(a) $lim_{n\rightarrowinfty}sum_{k…

which of the following limits is equal to $int_{2}^{5}sqrt{1 + x^{4}}dx$?\n(a) $lim_{n\rightarrowinfty}sum_{k = 1}^{n}left(sqrt{1+(\frac{k}{n})^{4}}\right)\frac{1}{n}$\n(b) $lim_{n\rightarrowinfty}sum_{k = 1}^{n}left(sqrt{1+(\frac{3k}{n})^{4}}\right)\frac{3}{n}$\n(c) $lim_{n\rightarrowinfty}sum_{k = 1}^{n}left(sqrt{1+(2+\frac{k}{n})^{4}}\right)\frac{1}{n}$\n(d) $lim_{n\rightarrowinfty}sum_{k = 1}^{n}left(sqrt{1+(2+\frac{3k}{n})^{4}}\right)\frac{3}{n}$

which of the following limits is equal to $int_{2}^{5}sqrt{1 + x^{4}}dx$?\n(a) $lim_{n\rightarrowinfty}sum_{k = 1}^{n}left(sqrt{1+(\frac{k}{n})^{4}}\right)\frac{1}{n}$\n(b) $lim_{n\rightarrowinfty}sum_{k = 1}^{n}left(sqrt{1+(\frac{3k}{n})^{4}}\right)\frac{3}{n}$\n(c) $lim_{n\rightarrowinfty}sum_{k = 1}^{n}left(sqrt{1+(2+\frac{k}{n})^{4}}\right)\frac{1}{n}$\n(d) $lim_{n\rightarrowinfty}sum_{k = 1}^{n}left(sqrt{1+(2+\frac{3k}{n})^{4}}\right)\frac{3}{n}$

Answer

Explanation:

Step1: Recall the definition of definite integral

The definite - integral $\int_{a}^{b}f(x)dx=\lim_{n\rightarrow\infty}\sum_{k = 1}^{n}f(x_k)\Delta x$, where $\Delta x=\frac{b - a}{n}$ and $x_k=a + k\Delta x$. Here, $a = 2$, $b = 5$, and $f(x)=\sqrt{1 + x^{4}}$.

Step2: Calculate $\Delta x$

$\Delta x=\frac{b - a}{n}=\frac{5 - 2}{n}=\frac{3}{n}$.

Step3: Calculate $x_k$

$x_k=a + k\Delta x=2+\frac{3k}{n}$.

Step4: Substitute into the limit - sum formula

$f(x_k)=\sqrt{1+(2 + \frac{3k}{n})^{4}}$, and $\int_{2}^{5}\sqrt{1 + x^{4}}dx=\lim_{n\rightarrow\infty}\sum_{k = 1}^{n}\sqrt{1+(2+\frac{3k}{n})^{4}}\frac{3}{n}$.

Answer:

D. $\lim_{n\rightarrow\infty}\sum_{k = 1}^{n}\left(\sqrt{1+(2+\frac{3k}{n})^{4}}\right)\frac{3}{n}$