Calculate $M_6$ for the integral $\displaystyle \int_{-1}^2 e^{x^2}\ dx$. This should be a numerical answer (calculator required).
We need to break this into $6$ rectangles. What are the endpoints and midpoints of these rectangles?
The midpoints are $-.75,\ -.25,\ .25,\ .75,\ 1.25,\ 1.75$.
Plugging these into our function will give the height, and then we need to multiply by the $\Delta x$, which is $0.5$
Calculate $T_5$ for the integral $\displaystyle \int_1^2 \sqrt{x^4 + 1}\ dx$ (calculator required).
I want to break this into 5 trapezoids. What are the x endpoints of these trapezoids?
Since I want 5 trapezoids, I will end up with 6 endpoints, and they should all be $\frac{2-1}{5} = 0.2$ apart. Thus, my endpoints are 1, 1.2, 1.4, 1.6, 1.8, 2.0.
We then need to plug these values into the function, and combine them appropriately for the trapezoid rule.
The sum you should get is $0.1(\sqrt{2} + 2\sqrt{1.2^4 + 1} + 2\sqrt{1.4^4 + 1} + 2\sqrt{1.6^4 + 1} + 2\sqrt{1.8^4 + 1} + \sqrt{2^4 + 1})$
Calculate $S_6$ for the integral $\int_1^4 e^{-x}\ dx$ and compare to the actual value of the integral.
Using Simpsons rule with $N=6$ means we need $6$ rectangles that are all of width $\frac{4-1}{6} = 0.5$. What are the endpoints here?
The endpoints should be 1, 1.5, 2, 2.5, 3, 3.5, 4, which then need to be plugged into the function to apply Simpson's Rule.
The sum you need for Simpsons Rule is $\displaystyle \frac{0.5}{3}\left(e^{-1} + 4e^{-1.5} + 2e^{-2} + 4e^{-2.5} + 2e^{-3} + 4e^{-3.5} + e^{-4} \right)$.
The actual integral is $e^{-1} - e^{-4}$
Simpsons rule gives 0.34968, while the actual integral is 0.34956.
Find the smallest value of $N$ for with the error in the Trapezoid rule in approximating the integral $\displaystyle \int_3^7 \frac{1}{x}\ dx $ is less than $10^{-6}$.
We need to use the Error Bound for the Trapezoid Rule here. What do we need for this?
The only tricky part here is the bound on the second derivative of the function. Since $f(x) = \frac{1}{x}$, the second derivative is $\frac{2}{x^3}$. What is the biggest this can be on the interval from $3$ to $7$?
We can use $K_2 = \frac{2}{27}$, and $b-a = 4$ to solve for $N$ in the error bound.
Set up $\displaystyle \frac{(2/27) 4^3}{12} = 10^{-6}$ and solve for $N^2$, then round up.
$629$
Find the smallest value of $N$ for with the error in the Simpsons rule in approximating the integral $\displaystyle \int_3^7 \frac{1}{x}\ dx $ is less than $10^{-6}$.
We need to use the Error Bound for the Simpsons Rule here. What do we need for this?
The only tricky part here is the bound on the fourth derivative of the function. Since $f(x) = \frac{1}{x}$, the fourth derivative is $\frac{24}{x^5}$. What is the biggest this can be on the interval from $3$ to $7$?
We can use $K_4 = \frac{24}{243} = \frac{8}{81}$, and $b-a = 4$ to solve for $N$ in the error bound.
Set up $\displaystyle \frac{(8/81) 4^5}{180N^2} = 10^{-6}$ and solve for $N$, then round up.
$59$
Compute $S_6$ for the integral $\int_5^8 \sin\left(\frac{1}{x}\right)\ dx$.
Find a number $N$ so that the error in approximating the integral $\displaystyle \int_2^5 5x^4 - x^5\ dx$ by $M_N$ is less than $10^{-4}$.