The Trapezoidal Rule states that a definite integral 

\(\displaystyle \int_{a}^{b} f(x) \textup{d}x\) 

can be approximated by computing the expression

\(\displaystyle T_{N}=\frac{b-a}{2N} \left [f (x_{0} ) +2 f (x_{1}) +2 f ( x_{2} ) + ... 2f ( x_{N-1} )+ f (x_{N} ) \right ]\)

where \(\displaystyle x_{k} = a + \frac{b-a}{N} \cdot k\) for \(\displaystyle k = 0, 1, 2, 3, ...,N\)

Since we are dividing \(\displaystyle \left [ 2,3 \right ]\) into four subintervals, set 

\(\displaystyle a = 2, b =3, N = 4\); the expression becomes

\(\displaystyle T_{4}=\frac{1}{8} \left [ f(2 ) +2 f \left (2.25 \right ) +2 f \left (2.5 \right ) + 2f \left ( 2.75 \right ) + f (3 ) \right ]\)

where

\(\displaystyle f(x)= \log_{3}( x^{2} )\)

\(\displaystyle f(2)= \log_{3}( 2^{2} ) = \log_{3}4= \frac{\ln 4}{\ln 3} \approx 1.2619\)

\(\displaystyle f(2.25)= \log_{3}( 2.25^{2} ) = \log_{3}5.0625= \frac{\ln 5.0625}{\ln 3} \approx 1.4763\)

\(\displaystyle f(2.5)= \log_{3}( 2.5^{2} ) = \log_{3}6.25= \frac{\ln 6.25}{\ln 3} \approx 1.6681\)

\(\displaystyle f(2.75)= \log_{3}( 2.75^{2} ) = \log_{3}7.5625= \frac{\ln 7.5625}{\ln 3} \approx 1.8416\)

\(\displaystyle f(3)= \log_{3}( 3^{2} ) = \log_{3}9= 2\)

\(\displaystyle T_{4}=\frac{1}{8} \left [ 1.2619 +2 (1.4763 )+2 ( 1.6681)+ 2 ( 1.8416) +2 \right]\)

\(\displaystyle \approx 1.6542\)

To use the trapezoidal rule, we apply the following formula:

\(\displaystyle \int_{a}^{b}f(x)dx\approx(b-a)\left [ \frac{f(a)-f(b)}{2}\right ]\)

Using the integral from our problem statement, we get

\(\displaystyle (5-3)\frac{(\cos(e^{15})-\cos(e^{9}))}{2}=\cos(e^{15})-\cos(e^{9})\)

To solve the integral, we will use the formula for the trapezoidal sums method:

\(\displaystyle \int_{a}^{b}f(x)dx=(b-a)\left [ \frac{f(a)+f(b)}{2}\right ]\)

Using the integral from the problem statement, we get

\(\displaystyle (2\pi-\frac{\pi}{2})\left [ \frac{e^{\sin(2\pi)}+e^{\sin(\frac{\pi}{2})}}{2}\right ]\)

Simplifying, we end up with

\(\displaystyle \frac{3\pi}{2}(\frac{1+e}{2})\)

The trapezoidal approximation of a definite integral is given by

\(\displaystyle \int_{a}^{b}f(x)dx\approx (b-a)(\frac{f(b)+f(a)}{2})\)

For our integral, we get

\(\displaystyle \int_{0}^{3}x^2 dx \approx (3-0)(\frac{9+0}{2})=\frac{27}{2}\)

To approximate the integral using trapezoidal sums, we use the following formula:

\(\displaystyle \int_{a}^{b} f(x)dx=(b-a)\left [ \frac{f(b)+f(a)}{2}\right ]\)

Using the integral from the problem statement, we get

\(\displaystyle (1-0)\left [ \frac{(\frac{4-e}{5})+(\frac{-1}{5})}{2} \right ]=\frac{3-e}{10}\)

To find the value of the integral, we use the following formula

\(\displaystyle \int_{a}^{b}f(x)dx=\left [ b-a\right ](\frac{f(a)+f(b)}{2})\)

\(\displaystyle \left [ \frac{\pi}{2}-0 \right ](\frac{\sin(\frac{\pi}{2})e^\frac{\pi}{2}+\sin(0)e^0}{2})=\frac{\pi}{4}e^\frac{\pi}{2}\)

To use the method of trapezoidal sums, we follow the definition

\(\displaystyle \int_{a}^{b}f(x)dx=(b-a)\left [ \frac{f(a)+f(b)}{2}\right ]\)

Using the information from the problem statement, we get

\(\displaystyle (5-3)(\frac{10e^{30}+6e^{14}}{2})=10e^{30}+6e^{14}\)

The trapezoidal approximation to definite integrals is given by

\(\displaystyle \int_{a}^{b} f(x)dx\approx (b-a)(\frac{f(b)+f(a)}{2})\)

Using this formula for our integral, we get

\(\displaystyle \int_{0}^{5} x^4 dx \approx (5-0)(\frac{625-0}{2})=\frac{3125}{2}\)

Actually integrating, we get

\(\displaystyle \int_{0}^{5} x^4 dx = \frac{x^5}{5}\left.\begin{matrix} 5\\0 \end{matrix}\right| = \frac{5^5}{5}=5^4=625\)

The rule used for integration is

\(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\)

The difference between the approximation and the actual answer is

\(\displaystyle \frac{3125}{2}-625=\frac{1875}{2}\)

Trapezoidal sums are found by creating trapezoids whose left and right end points are on the specified function, and whose widths are the step size. We then sum up their areas by remembering that the area of a trapezoid is the base times the average of the heights.

Thus, the calculation of the trapezoidal sum for this example would be \(\displaystyle 2 \cdot [\frac{f(-2)+f(0)}{2}] + 2 \cdot [\frac{f(0)+f(2)}{2}] + 2 \cdot [\frac{f(2)+f(4)}{2}]\)

A more simplified version would be given by \(\displaystyle f(-2) + 2(f(0) + f(2)) + f(4)\)

Which evaluates to 

\(\displaystyle 4 + 2(0 + 4) + 16 = 28\)

The actual answer is found by evaluating the definite integral given, which would just be given by \(\displaystyle \int_{-2}^{4} x^2 dx = \left . \frac{x^3}{3} \right ]^4_{-2} = \frac{4^3 - (-2)^3}{3} = \frac{64 + 8}{3} = \frac{72}{3} = 24\)

The difference between the approximation and the the true answer is thus \(\displaystyle 28-24 = 4\)