Evaluate the limit as n approaches infinity of the sum from i = 1 to n of (3 + 2i/n)^2 times (2/n) by interpreting it as a definite integral.

Step-by-step solution

1. 1

   Compare the sum with the definition of the integral. The right-endpoint Riemann sum of $f$ on $[a,b]$ is

   $\int_{a}^{b}f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(a+i\,\Delta x)\,\Delta x,\qquad \Delta x=\frac{b-a}{n}.$

   So the job is to match three things: the width $\Delta x$, the sample point $a+i\,\Delta x$, and the rule $f$.

2. 2

   Read off the width, and with it the length of the interval. The factor sitting outside the square is $\dfrac{2}{n}$, so

   $\Delta x=\frac{2}{n}=\frac{b-a}{n}\quad\Longrightarrow\quad b-a=2.$

3. 3

   Read off the left endpoint. The sample point is $3+\dfrac{2i}{n}=3+i\,\Delta x$. Matching this against $a+i\,\Delta x$ gives $a=3$, and then $b=a+2=5$. (The constant that is not multiplied by $i$ is always the left endpoint.)

4. 4

   Read off the function. What is being done to the sample point is squaring it, so $f(x)=x^{2}$. Putting the three pieces together,

   $\lim_{n\to\infty}\sum_{i=1}^{n}\left(3+\frac{2i}{n}\right)^{2}\frac{2}{n}=\int_{3}^{5}x^{2}\,dx.$

5. 5

   Evaluate the integral.

   $\int_{3}^{5}x^{2}\,dx=\left[\frac{x^{3}}{3}\right]_{3}^{5}=\frac{125}{3}-\frac{27}{3}=\frac{98}{3}\approx 32.667.$

   Check: adding the sum directly with $n=10^{6}$ gives $32.66668$, which agrees with $\tfrac{98}{3}$ to five decimals. Note the answer is not unique-looking by accident — writing the same limit with $a=0,\ b=2$ and $f(x)=(3+x)^{2}$ gives the same number.