using rectangles whose height is given by the value of the function at the mid - point of the rectangles…

using rectangles whose height is given by the value of the function at the mid - point of the rectangles base, estimate the area under the graph using first two and then four rectangles. f(x)=x³ between x = 1 and x = 3. using two rectangles to estimate, the area under f(x) is approximately

using rectangles whose height is given by the value of the function at the mid - point of the rectangles base, estimate the area under the graph using first two and then four rectangles. f(x)=x³ between x = 1 and x = 3. using two rectangles to estimate, the area under f(x) is approximately

Answer

Explanation:

Step1: Calculate width for 2 - rectangles case

The interval is from $x = 1$ to $x=3$. For $n = 2$ rectangles, the width $\Delta x=\frac{3 - 1}{2}=1$. The sub - intervals are $[1,2]$ and $[2,3]$. The mid - points are $x_1 = 1.5$ and $x_2=2.5$.

Step2: Calculate the sum for 2 - rectangles case

We use the mid - point rule $A\approx\sum_{i = 1}^{n}f(x_i)\Delta x$. Here, $n = 2$, $\Delta x = 1$, $f(x)=x^{3}$. $f(1.5)=(1.5)^{3}=3.375$ and $f(2.5)=(2.5)^{3}=15.625$. $A_2\approx f(1.5)\times1 + f(2.5)\times1=3.375+15.625 = 19$.

Step3: Calculate width for 4 - rectangles case

For $n = 4$ rectangles, $\Delta x=\frac{3 - 1}{4}=0.5$. The sub - intervals are $[1,1.5]$, $[1.5,2]$, $[2,2.5]$, $[2.5,3]$. The mid - points are $x_1 = 1.25$, $x_2 = 1.75$, $x_3=2.25$, $x_4 = 2.75$.

Step4: Calculate the sum for 4 - rectangles case

$f(1.25)=(1.25)^{3}=1.953125$, $f(1.75)=(1.75)^{3}=5.359375$, $f(2.25)=(2.25)^{3}=11.390625$, $f(2.75)=(2.75)^{3}=20.796875$. $A_4\approx\sum_{i = 1}^{4}f(x_i)\Delta x=0.5\times(1.953125 + 5.359375+11.390625 + 20.796875)$ $=0.5\times39.5000 = 19.75$.

Answer:

Using two rectangles to estimate, the area under $f(x)$ is approximately $19$. Using four rectangles to estimate, the area under $f(x)$ is approximately $19.75$.