Q:

SOMEONE PLEASE HELP!!!

Accepted Solution

A:
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:a = 1b = 3n = 5[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) = 8.72To find the maximum possible error:Identify the last non-zero digit to the right of the decimal point → 2The precision of the number is the place value: 0.01Divide the precision by 2Therefore, 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]