Question

step 5
this gives us
m _ { 4 } = \frac { 1 } { 4 } left 2 cos left( \frac { 1 } { 64 }
ight) + 2 cos left( \frac { 9 } { 64 }
ight) + 2 cos left( \frac { 25 } { 64 }
ight) + 2 cos left( \frac { 49 } { 64 }
ight)
ight
=
(rounded to six decimal places)
therefore, using the midpoint rule with ( n = 4 ) and rounding to six decimal places we have
int _ { 0 } ^ { 1 } 2 cos left( x ^ { 2 }
ight) d x =
step 6
a graph of the integrand shows that ( f ) is concave down on ( 0,1 ). therefore,
t _ { 4 } \text { is an underestimate }
( t _ { 4 } ) is an overestimate
and
( m _ { 4 } \text { is an underestimate } )
( m _ { 4 } \text { is an overestimate } )
step 7
since we know that ( t _ { 4 } = 1.791518 ) is an underestimate and ( m _ { 4 } ) is an overestimate, it is clear that the true value of the integral will fall somewhere between these two values. therefore,
1.791518 < int _ { 0 } ^ { 1 } 2 cos left( x ^ { 2 }
ight) d x <
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Explanation:

Step1: Recall the properties of Mid - point Rule and Trapezoidal Rule for concave - down functions

For a function \(y = f(x)\) that is concave down on the interval \([a,b]\), the Mid - point Rule \(M_n\) overestimates the value of the integral \(\int_{a}^{b}f(x)dx\) and the Trapezoidal Rule \(T_n\) underestimates the value of the integral \(\int_{a}^{b}f(x)dx\).
Here, \(a = 0\), \(b = 1\), \(n = 4\), \(f(x)=2\cos(x^{2})\) and the function \(y = f(x)\) is concave down on \([0,1]\) (given).
We know that \(T_4=1.791518\) (under - estimate) and \(M_4 = 1.817814\) (over - estimate)

Step2: Determine the bounds of the integral

Since \(T_4\) is an underestimate and \(M_4\) is an overestimate of \(\int_{0}^{1}2\cos(x^{2})dx\)

Answer:

\(1.791518<\int_{0}^{1}2\cos(x^{2})dx<1.817814\)