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F(x) = = between x=4 and x = 7 using the "midpoint rule" with two rectangles of equal width.

F(x) = = between x=4 and x = 7 using the "midpoint rule" with two rectangles-example-1
User Belrog
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Step-by-step explanation

Since we have the function f(x)= 1/x


\int _a^bf\left(x\right)dx\:\approx \Delta \:x\left(f\left((x_0+x_1)/(2)\right)+f\left((x_1+x_2)/(2)\right)+f\left((x_2+x_3)/(2)\right)+...+f\left((x_(n-1)+x_n)/(2)\right)\right)
\mathrm{where}\:\Delta \:x\:=\:(b-a)/(n)

Given a=4, b=7, n=2


Δx=(7-4)/(2)=(3)/(2)

Divide the interval, 4<=x<=7 into 2 subintervals:


x_0=4,\:x_1=(11)/(2),\:x_2=7
=(3)/(2)\left(f\left((x_0+x_1)/(2)\right)+f\left((x_1+x_2)/(2)\right)\right)

Calculate the subintervals:


f\left((x_0+x_1)/(2)\right)=f\left((4+(11)/(2))/(2)\right)
=f\left((19)/(4)\right)
=(1)/((19)/(4))
=(4)/(19)
f\left((x_1+x_2)/(2)\right)=f\left(((11)/(2)+7)/(2)\right)
=(4)/(25)
=(3)/(2)\left((4)/(19)+(4)/(25)\right)

Applying the distributive property and adding fractions:


=(264)/(475)

In conclusion, the solution is 264/475

User Jonrsharpe
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