8.4 Integration as the limit of a sum — Practise

Work through these on your own. Write the strip sum first, simplify it to a fraction in $n$ , and only then take the limit. Check the final integral against the Fundamental Theorem of Calculus.

The questions and the answers below come from a set that is checked by computer every time the book is built.

Write down the value of $\displaystyle\sum_{i=1}^{n} i$ in terms of $n$ .
Write down the value of $\displaystyle\sum_{i=1}^{n} i^2$ in terms of $n$ .
Find $\displaystyle\lim_{n \to \infty} \frac{n+1}{2n}$ .
Find $\displaystyle\lim_{n \to \infty} \frac{4(n+1)(2n+1)}{3n^2}$ .
Use a limit of a sum to evaluate $\displaystyle\int_0^1 x\,dx$ .
Use a limit of a sum to evaluate $\displaystyle\int_0^2 x^2\,dx$ .
Use a limit of a sum to evaluate $\displaystyle\int_0^3 (x + 1)\,dx$ .
Use a limit of a sum to evaluate $\displaystyle\int_0^2 (x^2 + 1)\,dx$ .
Use a limit of a sum to show that $\displaystyle\int_0^a x^2\,dx = \frac{a^3}{3}$ for $a > 0$ .
Evaluate $\displaystyle\int_1^2 x\,dx$ by any method.
Write down the value of $\displaystyle\sum_{i=1}^{n} 1$ .
Evaluate $\displaystyle\int_0^1 x^3\,dx$ using a limit of a sum, given that $\displaystyle\sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4}$ .

Answers and concise solutions

1. The sum

Answer: $\displaystyle\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}$ . This is the standard sum used for the height $x_i$ when the function is $y = x$ .

2. The sum

Answer: $\displaystyle\sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}$ . This is the standard sum used for the height $x_i$ when the function is $y = x^2$ .

3. The limit

Answer: Divide top and bottom by $n$ : $\dfrac{1 + 1/n}{2} \to \dfrac{1}{2}$ as $n \to \infty$ . This is the limit that gives $\displaystyle\int_0^1 x\,dx = \dfrac{1}{2}$ .

4. The limit

Answer: The top expands to $8n^2 + 12n + 4$ . Dividing by $n^2$ and letting $n \to \infty$ leaves $\dfrac{8}{3}$ . This is the limit that gives $\displaystyle\int_0^2 x^2\,dx = \dfrac{8}{3}$ .

5. The value of the integral

Answer: With $n$ strips, $S_n = \dfrac{n(n+1)}{2n^2} = \dfrac{n+1}{2n}$ , which tends to $\dfrac{1}{2}$ . So the integral is $\dfrac{1}{2}$ .

6. The value of the integral

Answer: With $n$ strips, $S_n = \dfrac{4(n+1)(2n+1)}{3n^2}$ , which tends to $\dfrac{8}{3}$ . So the integral is $\dfrac{8}{3}$ .

7. The value of the integral

Answer: With $n$ strips, $S_n = \dfrac{9(n+1)}{2n} + 3$ , which tends to $\dfrac{9}{2} + 3 = \dfrac{15}{2}$ . So the integral is $\dfrac{15}{2}$ .

8. The value of the integral

Answer: The $x^2$ part tends to $\dfrac{8}{3}$ and the constant part contributes $2$ , so the integral is $\dfrac{8}{3} + 2 = \dfrac{14}{3}$ .

9. The value of the integral

Answer: With $n$ strips, $S_n = \dfrac{a^3(n+1)(2n+1)}{6n^2}$ , which tends to $\dfrac{a^3 \times 2}{6} = \dfrac{a^3}{3}$ . So the integral is $\dfrac{a^3}{3}$ .

10. The value of the integral

Answer: Using the Fundamental Theorem of Calculus, $\displaystyle\int_1^2 x\,dx = \left[\dfrac{x^2}{2}\right]_1^2 = 2 - \dfrac{1}{2} = \dfrac{3}{2}$ . A limit of a sum gives the same value with $x_i = 1 + \dfrac{i}{n}$ .

11. The sum

Answer: $\displaystyle\sum_{i=1}^{n} 1 = n$ , because the number $1$ is added $n$ times. This is the constant part of a strip sum.

12. The value of the integral

Answer: With $n$ strips, $S_n = \dfrac{1}{n^4}\sum i^3 = \dfrac{n^2(n+1)^2}{4n^4} = \dfrac{(n+1)^2}{4n^2}$ , which tends to $\dfrac{1}{4}$ . So $\displaystyle\int_0^1 x^3\,dx = \dfrac{1}{4}$ .