Q.62 - Calculus

Question

Prove that ${\int }_1^3(3x+4) dx=20$in three different ways:

(a) algebraically, by calculating a limit of Riemann sums;

(b) geometrically, by recognizing the region in question as a trapezoid and calculating its area;

(c) with formulas, by using properties and formulas of definite integrals.

Step-by-Step Solution

Answer

 $$\begin{gathered} \left(a\right) {\int }_1^3(3x+4) dx=\underset{n\to \infty }{\lim }\sum _{k=1}^n\left[3\left(1+\frac{2k}{n}\right)+4\right]\left(\frac{2}{n}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{14}{n}\right)\sum _{k=1}^{n}1+\underset{n\to \infty }{\lim }\left(\frac{12}{n^2}\right)\sum _{k=1}^{n}k \\ =\underset{n\to \infty }{\lim }\left(\frac{14}{n}\right)n+\underset{n\to \infty }{\lim }\left(\frac{12}{n^2}\right)\left[\frac{n\left(n+1\right)}{2}\right] \\ =14+\frac{12}{2}=20 \end{gathered}$$

$$\begin{gathered} \left(b\right) {\int }_1^3(3x+4)dx=3\frac{1}{2}(3^2-1^2)+4(3-1) \\ =\frac{3}{2}\left(8\right)+4\left(2\right)=20 \end{gathered}$$

$$\begin{gathered} \left(c\right) {\int }_1^3(3x+4)dx={\int }_1^{3}3x dx+{\int }_1^{3}4 dx \\ =3\left(\frac{1}{2}\right)\left(3^2-1^2\right)+4(3-1) \\ =12+8=20 \end{gathered}$$

Step 1. Given information.

The given integral is ${\int }_1^3(3x+4) dx=20.$

Step 2. Part (a).

Take the interval $\left[1,3\right].$

$$\begin{gathered} ∆x=\frac{3-1}{n}=\frac{2}{n} \\ {x}_k=a+k∆x \\ {x}_k=1+k\left(\frac{2}{n}\right) \end{gathered}$$

Use Riemann's sum.

$$\begin{gathered} {\int }_1^3(3x+4) dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(\stackrel{ *}{x_k}\right) ∆x \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n\left[3\left(1+\frac{2k}{n}\right)+4\right]\left(\frac{2}{n}\right) \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(7+\frac{6k}{n}\right)\left(\frac{2}{n}\right) \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(\frac{14}{n}\right)+\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(\frac{12k}{n^2}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{14}{n}\right)\sum _{k=1}^{n}1+\underset{n\to \infty }{\lim }\left(\frac{12}{n^2}\right)\sum _{k=1}^{n}k \\ =\underset{n\to \infty }{\lim }\left(\frac{14}{n}\right)n+\underset{n\to \infty }{\lim }\left(\frac{12}{n^2}\right)\left[\frac{n\left(n+1\right)}{2}\right] \\ =\underset{n\to \infty }{\lim }14+\underset{n\to \infty }{\lim }\left[\frac{12n^2+12}{n^2}\right] \\ =14+\frac{12}{2} \\ =20 \end{gathered}$$

Step 2. Part (b).

The limit is the area covered by the trapezoid of widths a and b and height $b-a.$

$\frac{a+b}{2}(b-a)=\frac{1}{2}(b^2-a^2)$

SO the area in the interval $[1,3]$is

$$\begin{gathered} {\int }_1^3(3x+4)dx=3\frac{1}{2}(3^2-1^2)+4(3-1) \\ =\frac{3}{2}\left(8\right)+4\left(2\right) \\ =12+8 \\ =20 \end{gathered}$$

So 

Step 4. Part (c)

use properties and formulas of definite integrals.

$$\begin{gathered} {\int }_1^3(3x+4)dx={\int }_1^{3}3x dx+{\int }_1^{3}4 dx \\ =3\left(\frac{1}{2}\right)\left(3^2-1^2\right)+4(3-1) \\ =3\left(4\right)+4\left(2\right) \\ =12+8 \\ =20 \end{gathered}$$