use equal width rectangles to estimate the area under the graph of f(x)=2 + 2^(x/5) from x = 0 to x = 8…

use equal width rectangles to estimate the area under the graph of f(x)=2 + 2^(x/5) from x = 0 to x = 8. round your answers to three decimal places.\na) estimate the area using four rectangles that take the sample points from the left endpoints.\nb) estimate the area using four rectangles that take the sample points from the right endpoints.\nc) estimate the area using eight rectangles that take the sample points from the left endpoints.\nd) estimate the area using eight rectangles that take the sample points from the right endpoints.

use equal width rectangles to estimate the area under the graph of f(x)=2 + 2^(x/5) from x = 0 to x = 8. round your answers to three decimal places.\na) estimate the area using four rectangles that take the sample points from the left endpoints.\nb) estimate the area using four rectangles that take the sample points from the right endpoints.\nc) estimate the area using eight rectangles that take the sample points from the left endpoints.\nd) estimate the area using eight rectangles that take the sample points from the right endpoints.

Answer

Explanation:

Step1: Calculate the width of rectangles

The interval is from $x = 0$ to $x=8$. For $n$ rectangles, the width $\Delta x=\frac{b - a}{n}$, where $a = 0$, $b = 8$.

Part (a)

For $n = 4$, $\Delta x=\frac{8-0}{4}=2$. The left - endpoints are $x_0 = 0,x_1=2,x_2 = 4,x_3=6$. $f(x)=2 + 2^{\frac{x}{5}}$ $f(x_0)=2+2^{0}=3$ $f(x_1)=2+2^{\frac{2}{5}}\approx2 + 1.3207 = 3.3207$ $f(x_2)=2+2^{\frac{4}{5}}\approx2+1.7411=3.7411$ $f(x_3)=2+2^{\frac{6}{5}}\approx2 + 2.3014=4.3014$ The area $A\approx\sum_{i = 0}^{3}f(x_i)\Delta x=\Delta x(f(x_0)+f(x_1)+f(x_2)+f(x_3))$ $A\approx2(3 + 3.3207+3.7411+4.3014)=2\times14.3632 = 28.726$

Part (b)

For $n = 4$, $\Delta x = 2$. The right - endpoints are $x_1=2,x_2 = 4,x_3=6,x_4=8$. $f(x_1)=2+2^{\frac{2}{5}}\approx3.3207$ $f(x_2)=2+2^{\frac{4}{5}}\approx3.7411$ $f(x_3)=2+2^{\frac{6}{5}}\approx4.3014$ $f(x_4)=2+2^{\frac{8}{5}}\approx2+3.0314=5.0314$ The area $A\approx\sum_{i = 1}^{4}f(x_i)\Delta x=\Delta x(f(x_1)+f(x_2)+f(x_3)+f(x_4))$ $A\approx2(3.3207+3.7411+4.3014+5.0314)=2\times16.3946=32.789$

Part (c)

For $n = 8$, $\Delta x=\frac{8 - 0}{8}=1$. The left - endpoints are $x_0 = 0,x_1=1,x_2 = 2,\cdots,x_7=7$. $f(x_0)=2+2^{0}=3$ $f(x_1)=2+2^{\frac{1}{5}}\approx2 + 1.1487=3.1487$ $f(x_2)=2+2^{\frac{2}{5}}\approx3.3207$ $f(x_3)=2+2^{\frac{3}{5}}\approx3.5158$ $f(x_4)=2+2^{\frac{4}{5}}\approx3.7411$ $f(x_5)=2+2^{\frac{5}{5}}\approx4$ $f(x_6)=2+2^{\frac{6}{5}}\approx4.3014$ $f(x_7)=2+2^{\frac{7}{5}}\approx4.6651$ The area $A\approx\sum_{i = 0}^{7}f(x_i)\Delta x=\Delta x\sum_{i = 0}^{7}f(x_i)$ $A\approx1\times(3 + 3.1487+3.3207+3.5158+3.7411+4+4.3014+4.6651)=29.693$

Part (d)

For $n = 8$, $\Delta x = 1$. The right - endpoints are $x_1=1,x_2 = 2,\cdots,x_8=8$. $f(x_1)=2+2^{\frac{1}{5}}\approx3.1487$ $f(x_2)=2+2^{\frac{2}{5}}\approx3.3207$ $f(x_3)=2+2^{\frac{3}{5}}\approx3.5158$ $f(x_4)=2+2^{\frac{4}{5}}\approx3.7411$ $f(x_5)=2+2^{\frac{5}{5}}\approx4$ $f(x_6)=2+2^{\frac{6}{5}}\approx4.3014$ $f(x_7)=2+2^{\frac{7}{5}}\approx4.6651$ $f(x_8)=2+2^{\frac{8}{5}}\approx5.0314$ The area $A\approx\sum_{i = 1}^{8}f(x_i)\Delta x=\Delta x\sum_{i = 1}^{8}f(x_i)$ $A\approx1\times(3.1487+3.3207+3.5158+3.7411+4+4.3014+4.6651+5.0314)=32.724$

Answer:

a) $28.726$ b) $32.789$ c) $29.693$ d) $32.724$