Answer:
Part (a)
Given:
[tex]\displaystyle \int\limits_{1}^{3} x^2 \:dx[/tex]
Trapezium Rule
[tex]\displaystyle \int\limits_{a}^{b} y \:dx \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}[/tex]
Calculate the width of each strip (h).
Given:
[tex]\implies h=\dfrac{3-1}{5}=0.4[/tex]
Create a table with the x and y values:
[tex]\large\begin{array}{| c | c | c | c | c | c | c |}\cline{1-7} x & 1 & 1.4 & 1.8 & 2.2 & 2.6 & 3\\\cline{1-7} y & 1 & 1.96 & 3.24 & 4.84 & 6.76 & 9\\\cline{1-7}\end{array}[/tex]
Put all the values into the formula:
[tex]\begin{aligned}\displaystyle \int\limits_{1}^{3} x^2 \:dx & \approx \dfrac{1}{2}(0.4)\left[(1+9)+2(1.96+3.24+4.84+6.76)\right]\\& = 0.2\left[10+2(16.8)\right]\\\\& = 0.2\left[10+33.6\right]\\\\& = 0.2\left[43.6\right]\\\\& = 8.72 \end{aligned}[/tex]
Part (b)
[tex]\begin{aligned}\displaystyle \int\limits_{1}^{3} x^2 \:dx &=\left[\dfrac{1}{3}x^3\right]^3_1\\& = \dfrac{1}{3}(3)^3-\dfrac{1}{3}(1)^3\\& = 9-\dfrac{1}{3}\\& = \dfrac{26}{3}\end{aligned}[/tex]
The exact area as a decimal is [tex]\sf 8.\dot{6}[/tex]. Therefore, the estimated answer of 8.72 is an overestimate.
Part (c)
Answer to (a) = 8.72
To find the maximum possible error:
- Identify the last non-zero digit to the right of the decimal point → 2
- The precision of the number is the place value: 0.01
- Divide the precision by 2
Therefore, the maximum possible error for 8.72 is 0.005.
Part (d)
Trapezoid Error formula
[tex]|E| \leq \dfrac{(b-a)^3}{12n^2}\left[\textsf{max}|f''(x)|\right],\quad a \leq x \leq b[/tex]
Calculate the second derivative:
[tex]f(x)=x^2[/tex]
[tex]f'(x)=2x[/tex]
[tex]f''(x)=2[/tex]
Input the values into the formula:
[tex]\implies |E| \leq \dfrac{(3-1)^3}{12n^2}(2) < 0.001[/tex]
[tex]\implies 16 \leq 0.012n^2[/tex]
[tex]\implies 36.5148...\leq n[/tex]
[tex]\implies 37\leq n[/tex]
