Answer
$133.5$
Work Step by Step
Step 1. Given the x-interval of $[1,4]$, divide the region into $n$ parts, the width of each part is $\Delta x=\frac{3}{n}$
Step 2. Use the right endpoints, the x-coordinates can be found as $x_k=1+k\Delta x=1+\frac{3k}{n}$
Step 3. Based on the given function $f(x)=x+6x^2$, the height of the $k$th rectangle is given by $f(x_k)=(1+\frac{3k}{n})+6(1+\frac{3k}{n})^2=1+\frac{3k}{n}+6(1+\frac{6k}{n}+\frac{9k^2}{n^2})=7+\frac{39k}{n}+\frac{54k^2}{n^2}$
Step 4. Approximate the area by adding up the areas of all the triangles and evaluate the limit when $n\to\infty$
$$A=\lim_{n\to\infty}\sum^n_{k=1}f(x_k)\Delta x=\lim_{n\to\infty}\sum^n_{k=1}(7+\frac{39k}{n}+\frac{54k^2}{n^2})(\frac{3}{n})=\lim_{n\to\infty} (\frac{3}{n}\sum^n_{k=1} 7+\frac{117}{n^2}\sum^n_{k=1}k+\frac{162}{n^3}\sum^n_{k=1}k^2)=\lim_{n\to\infty} (21+\frac{117}{n^2}\frac{n(n+1)}{2}+\frac{162}{n^3}\frac{n(n+1)(2n+1)}{6})=21+\frac{117}{2}+54=133.5$$
Step 5. The area of the region that lies under the graph over the given interval is $A=133.5$
