Answer
$MP_{4}\approx0.90891$
$T_{4}\approx0.89576$
$T_{4}\leq\int_{0}^{4}cos(x^{2})dx \leq MP_{4}$
Work Step by Step
We are given the endpoints, $(0, 1)$, and $n=4$.
$Δx=\frac{b-a}{n} = \frac{1-0}{4} =\frac{1}{4}$
Now we find $X_{0}$ to $X_{4}$, then $M_{1}$ to $M_{4}$ from there
$X_{0}=0$
$X_{1}=\frac{1}{4}$
$X_{2}=\frac{1}{2}$
$X_{3}=\frac{3}{4}$
$X_{4}=1$
$M_{1}=\frac{X_{0}+X_{1}}{2}=\frac{1}{8}$
$M_{2}=\frac{X_{1}+X_{2}}{2}=\frac{3}{8}$
$M_{3}=\frac{X_{2}+X_{3}}{2}=\frac{5}{8}$
$M_{4}=\frac{X_{3}+X_{4}}{2}=\frac{7}{8}$
$MP_{4} = \frac{1}{4}(cos(\frac{1}{8})^{2}+cos(\frac{3}{8})^{2}+cos(\frac{5}{8})^{2}+cos(\frac{7}{8})^{2}) \approx 0.90891$
$T_{4} = \frac{1}{2\times4}(cos(0)^{2}+2cos(\frac{1}{4})^{2}+2cos(\frac{1}{2})^{2}+2cos(\frac{3}{4})^{2}+cos(1)^{2}) \approx0.89576$
$\int^{1}_{0}cos(x^{2})dx = 0.90452$
We can see that $T_{4}$ is an underestimate, and that $MP_{4}$ is an overestimate, and thus we can conclude that the actual value of the integral is between $T_{4}$ and $MP_{4}$.
$T_{4}\leq\int_{0}^{4}cos(x^{2})dx \leq MP_{4}$
