For each of the following limits, find a function [math]f(x)[/math] such that the limit is equal to [math]\int_0^1 f(x) \; dx[/math]. Evaluate the limit.

• [math]\lim_{n\goesto\infty} \frac{1+2^2+3^2+\cdots+n^2}{n^3}[/math].
• [math]\lim_{n\goesto\infty} \frac{(1+n^2)+(2^2+n^2)+(3^2+n^2)+\cdots+(n^2+n^2)} {n^3}[/math].
• [math]\lim_{n\goesto\infty} \frac{\sqrt{1+n}+\sqrt{2+n}+\sqrt{3+n}+\cdots+\sqrt{n+n}} {n^{\frac32}}[/math].
• [math]\lim_{n\goesto\infty} \frac1{\sqrt n} \left(\frac1{\sqrt{1+n}}+\frac1{\sqrt{2+n}}+ \frac1{\sqrt{3+n}}+\cdots+\frac1{\sqrt{n+n}} \right)[/math].

Prove that

• [math]\ln 2 = \lim_{n\goesto\infty} \left( \frac1{n+1}+\frac1{n+2}+\cdots+\frac1{n+n} \right)[/math].
• [math]\pi = \lim_{n\goesto\infty} \frac4{n^2} \left(\sqrt{n^2-1}+\sqrt{n^2-2}+\cdots+\sqrt{n^2-n^2} \right)[/math].
• [math]\int_1^3(x^2+1)\;dx= \lim_{n\goesto\infty} \frac4{n^3} \sum_{i=1}^n(n^2+2in+2i^2)[/math].
• [math]\frac{\pi}6 = \lim_{n\goesto\infty} \left(\frac{1}{\sqrt{4n^2-1}}+\frac{1}{\sqrt{4n^2-2^2}}+ \cdots+\frac{1}{\sqrt{4n^2-n^2}}\right)[/math].

Use the Trapezoid Rule with [math]n=4[/math] to compute approximations to the following integrals. In \ref{ex8.2.3a}, \ref{ex8.2.3b}, \ref{ex8.2.3c}, and \ref{ex8.2.3d}, compare the approximation obtained with the true value.

• lab{8.2.3a} [math]\int_0^1 (x^2+1) \; dx[/math]
• lab{8.2.3b} [math]\int_0^2 (x^2+1) \; dx[/math]
• lab{8.2.3c} [math]\int_{-1}^3 (4x-1) \; dx[/math]
• lab{8.2.3d} [math]\int_1^3 \frac1{x^2} dx[/math]
• [math]\int_0^1 \frac1{1+x} dx[/math]
• [math]\int_0^1 \frac{dx}{1+x^2}[/math]
• [math]\int_0^1 e^{-x^2} dx[/math]
• [math]\int_0^1 \frac1{x^2+x+1} dx[/math]
• [math]\int_0^1 \frac{x^2-1}{x^2+1} dx[/math]
• [math]\int_0^{\pi} \frac{\sin x}{x} dx[/math].

Show geometrically, without appealing to Theorem \ref{thm 8.2.4}, that the approximation [math]T_n[/math] obtained with the Trapezoid Rule has the following properties.

• If [math]f^{\prime\prime}(x) \geq 0[/math] for every [math]x[/math] is [math][a,b][/math], then [math]T_n \geq \int_a^b f[/math].
• If [math]f^{\prime\prime}(x) \leq 0[/math] for every [math]x[/math] in [math][a,b][/math], then [math]T_n \leq \int_a^b f[/math].

For each of the following integrals, use Theorem \ref{thm 8.2.4} as the basis for finding the smallest integer [math]n[/math] such that the error [math]|\int_a^b f - T_n|[/math] in applying the Trapezoid Rule is less than (i) [math]\frac1{100}[/math], (ii) [math]\frac1{10000}[/math], and (iii) [math]10^{-8}[/math].

• [math]\int_1^4 \frac1{6x^2} dx[/math]
• [math]\int_0^1 (8x^3-5x+3) \; dx[/math]
• [math]\int_{-1}^2 (3x+1)\;dx[/math]
• [math]\int_1^2 \frac1x dx[/math]
• [math]\int_0^{12} \frac1{16x+2}dx[/math]
• [math]\int_0^1 e^{-x^2} dx[/math].

(For those who have access to a high-speed digital computer and know how to use it.) Compute the Trapezoid approximation [math]T_n[/math] to each of the following integrals.

• [math]\int_0^1 \frac1{1+x^3} dx[/math], for [math]n=10[/math], [math]100[/math], and [math]1000[/math].
• [math]\int_0^1 \frac1{1+x^2} dx[/math], for successive values of [math]n = 10,100,1000,\ldots[/math], until the error is less that [math]10^{-6}[/math].
• [math]\int_0^1 \sqrt{1+x^3} \; dx[/math], for [math]n = 5[/math], [math]50[/math], and [math]500[/math].
• [math]\int_0^{\pi} \frac{\sin x}x dx[/math], for [math]n = 2[/math], [math]4[/math], [math]8[/math], [math]16[/math], and [math]100[/math].
• [math]\int_0^1 e^{-x^2}dx[/math], for [math]n=10[/math], [math]100[/math], and [math]1000[/math].