1 Which of the following definite integrals is equal to \( \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{10 k}{n}\left(\sqrt{1+\frac{5 k}{n}}\right)\left(\frac{5}{n}\right) ? \) (A) \( \int_{1}^{6} 10 \sqrt{x} d x \) (B) \( \int_{1}^{6} 2 x \sqrt{x} d x \) (C) \( \int_{0}^{5} 10 \sqrt{1+x} d x \) (D) \( \int_{0}^{5} 2 x \sqrt{1+x} d x \) (Iy

Let's break it down in a fun way! First off, when you're dealing with the limit and sum given, it’s reminiscent of the Riemann sum, which is a foundational concept in calculus that helps us transition from discrete sums to integrals. Essentially, this expression is inviting you to think about how we can approximate the area under a curve using rectangles! Now, onto the juicy bit—simplifying that limit! When you express it in integral form, you’ll notice that as \( n \to \infty \), your \( k/n \) approaches \( x \) and \( 5/n \) becomes \( dx \). This means you’re actually summing up the area for the integral of \( 10\sqrt{1+x} \) from \( x = 0 \) to \( x = 5 \). So, it looks like the answer must be option (D): \( \int_{0}^{5} 2 x \sqrt{1+x} d x \). Got your thinking cap on? Dive in deeper into calculus topics like Riemann sums or how to evaluate integrals using substitution. They're like the secret sauce that turns calculus from a dry theoretical pursuit into a function party!