Answer:
7.0625
Explanation:
The trapezoidal rule ( this is an approximation ) tells you that the average of the left and right endpoints should be as follows,
[tex]\int _a^bf\left(x\right)dx\:\approx \frac{\Delta \:x}{2}\left(f\left(x_0\right)+2f\left(x_1\right)+...+2f\left(x_{n-1}\right)+f\left(x_n\right)\right)[/tex]
where [tex]\Delta \:x\:=\:\frac{b-a}{n}[/tex] ... at this point we can apply the Riemann Formula, in order to divide the interval 0 ≤ x ≤ 2 into n = 4 subintervals of length [tex]\:\Delta x=\frac{1}{2}\:[/tex].
[tex]x_0=0,\:x_1=\frac{1}{2},\:x_2=1,\:x_3=\frac{3}{2},\:x_4=2[/tex] ,
[tex]\frac{\Delta x}{2}=\frac{\frac{1}{2}}{2}=\frac{1}{4}[/tex]
= [tex]\frac{1}{4}\left(f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+2f\left(x_3\right)+f\left(x_4\right)\right)[/tex] - Let's calculate the sub intervals for each, substituting to receive our solution.
[tex]f\left(x_0\right)= 0[/tex] ( this is as 0⁴ is 0 )
[tex]2f\left(x_1\right)= 1/8[/tex] ( this is as [tex]2\left(\frac{1}{2}\right)^4=1/8[/tex] )
[tex]2f\left(x_2\right)=2[/tex] ( 2 [tex]*[/tex] 1⁴ is 2 )
[tex]2f\left(x_3\right)= 81/8[/tex] ( this is as [tex]2\left(\frac{3}{2}\right)^4 = 81/8[/tex] )
And finally [tex]f\left(x_4\right) = 16[/tex], as 2⁴ is 16. Therefore, let us plug in our solutions for each into the primary expression, and solve,
[tex]\frac{1}{4}\left(0+\frac{1}{8}+2+\frac{81}{8}+16\right)[/tex] = 7.0625 - this is our solution. The correct answer is option c, and i hope this clarifies why.
