Question

the collection of exercises marked in red could be used as a chapter test. exercises 1 - 6 refer to the region r in the first quadrant enclosed by the x - axis and the graph of the function y = 4x - x^3. 1. sketch r and partition it into four subregions, each with a base of length δx = 1/2. 2. sketch the rectangles and compute (by hand) the area for the lram4 approximation. 3. sketch the rectangles and compute (by hand) the area for the mram4 approximation. 4. sketch the rectangles and compute (by hand) the area for the rram4 approximation. 5. sketch the trapezoids and compute (by hand) the area for the t4 approximation.

Explanation:

Step1: Find the intersection with x - axis

Set $y = 4x - x^{3}=x(4 - x^{2})=x(2 - x)(2 + x)=0$. In the first - quadrant, the roots are $x = 0$ and $x = 2$.

Step2: Partition the interval

The interval is $[0,2]$ and $\Delta x=\frac{1}{2}$. The sub - intervals are $[0,\frac{1}{2}],[\frac{1}{2},1],[1,\frac{3}{2}],[\frac{3}{2},2]$.

Step3: LRAM (Left - hand Rule)

The left - hand endpoints are $x_0 = 0,x_1=\frac{1}{2},x_2 = 1,x_3=\frac{3}{2}$.
$y_0=4(0)-(0)^{3}=0$
$y_1=4(\frac{1}{2})-(\frac{1}{2})^{3}=2-\frac{1}{8}=\frac{15}{8}$
$y_2=4(1)-(1)^{3}=3$
$y_3=4(\frac{3}{2})-(\frac{3}{2})^{3}=6-\frac{27}{8}=\frac{21}{8}$
$LRAM_4=\sum_{i = 0}^{3}y_i\Delta x=\Delta x(y_0 + y_1+y_2 + y_3)=\frac{1}{2}(0+\frac{15}{8}+3+\frac{21}{8})=\frac{1}{2}(\frac{15 + 24+21}{8})=\frac{1}{2}\times\frac{60}{8}=\frac{15}{4}$

Step4: MRAM (Mid - point Rule)

The mid - points are $x_{0.5}=\frac{1}{4},x_{1.5}=\frac{3}{4},x_{2.5}=\frac{5}{4},x_{3.5}=\frac{7}{4}$
$y_{0.5}=4(\frac{1}{4})-(\frac{1}{4})^{3}=1-\frac{1}{64}=\frac{63}{64}$
$y_{1.5}=4(\frac{3}{4})-(\frac{3}{4})^{3}=3-\frac{27}{64}=\frac{165}{64}$
$y_{2.5}=4(\frac{5}{4})-(\frac{5}{4})^{3}=5-\frac{125}{64}=\frac{195}{64}$
$y_{3.5}=4(\frac{7}{4})-(\frac{7}{4})^{3}=7-\frac{343}{64}=\frac{105}{64}$
$MRAM_4=\sum_{i = 0}^{3}y_{i + 0.5}\Delta x=\frac{1}{2}(\frac{63+165 + 195+105}{64})=\frac{1}{2}\times\frac{528}{64}=\frac{33}{8}$

Step5: RRAM (Right - hand Rule)

The right - hand endpoints are $x_1=\frac{1}{2},x_2 = 1,x_3=\frac{3}{2},x_4 = 2$
$y_1=4(\frac{1}{2})-(\frac{1}{2})^{3}=2-\frac{1}{8}=\frac{15}{8}$
$y_2=4(1)-(1)^{3}=3$
$y_3=4(\frac{3}{2})-(\frac{3}{2})^{3}=6-\frac{27}{8}=\frac{21}{8}$
$y_4=4(2)-(2)^{3}=8 - 8=0$
$RRAM_4=\sum_{i = 1}^{4}y_i\Delta x=\frac{1}{2}(\frac{15}{8}+3+\frac{21}{8}+0)=\frac{1}{2}(\frac{15 + 24+21}{8})=\frac{15}{4}$

Step6: Trapezoidal Rule ($T_4$)

$T_4=\frac{\Delta x}{2}(y_0 + 2y_1+2y_2+2y_3 + y_4)$
$y_0 = 0,y_1=\frac{15}{8},y_2=3,y_3=\frac{21}{8},y_4 = 0$
$T_4=\frac{1}{4}(0 + 2\times\frac{15}{8}+2\times3+2\times\frac{21}{8}+0)=\frac{1}{4}(\frac{15}{4}+6+\frac{21}{4})=\frac{1}{4}(\frac{15 + 24+21}{4})=\frac{15}{4}$

1. For sketching $R$:

• Plot the function $y = 4x - x^{3}$. Find the derivative $y'=4 - 3x^{2}$. Set $y'=0$, then $4 - 3x^{2}=0$, so $x=\pm\frac{2}{\sqrt{3}}$. In the first - quadrant, the critical point is $x=\frac{2}{\sqrt{3}}\approx1.15$. The function is increasing on $(0,\frac{2}{\sqrt{3}})$ and decreasing on $(\frac{2}{\sqrt{3}},2)$. Then partition the interval $[0,2]$ into four sub - intervals of length $\Delta x=\frac{1}{2}$.

1. For $LRAM_4$:

• Sketch rectangles with height equal to the function value at the left - hand endpoint of each sub - interval and base $\Delta x=\frac{1}{2}$. The area is $\frac{15}{4}$.

1. For $MRAM_4$:

• Sketch rectangles with height equal to the function value at the mid - point of each sub - interval and base $\Delta x=\frac{1}{2}$. The area is $\frac{33}{8}$.

1. For $RRAM_4$:

• Sketch rectangles with height equal to the function value at the right - hand endpoint of each sub - interval and base $\Delta x=\frac{1}{2}$. The area is $\frac{15}{4}$.

1. For $T_4$:

• Sketch trapezoids connecting the points on the function at the endpoints of each sub - interval. The area is $\frac{15}{4}$.

Answer:

1. Sketch the region $R$ by plotting $y = 4x - x^{3}$ in the first - quadrant and partitioning $[0,2]$ into four sub - intervals of length $\frac{1}{2}$.
2. $LRAM_4=\frac{15}{4}$
3. $MRAM_4=\frac{33}{8}$
4. $RRAM_4=\frac{15}{4}$
5. $T_4=\frac{15}{4}$