find the exact value of ∫₃⁴ x²dx. find the error of approximation between the exact value and the value…

Explanation:

Step1: Find the exact value of the integral

Use the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$. $$\int_{3}^{4}x^{2}dx=\left[\frac{x^{3}}{3}\right]_{3}^{4}=\frac{4^{3}}{3}-\frac{3^{3}}{3}=\frac{64}{3}-\frac{27}{3}=\frac{64 - 27}{3}=\frac{37}{3}\approx12.3333$$

Step2: Apply the trapezoidal rule

The trapezoidal rule for $\int_{a}^{b}f(x)dx$ with $n$ sub - divisions is $T_n=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+\cdots+2f(x_{n - 1})+f(x_n)]$, where $\Delta x=\frac{b - a}{n}$, $a = 3$, $b = 4$, and $n = 3$. Then $\Delta x=\frac{4 - 3}{3}=\frac{1}{3}$. $x_0=3$, $x_1=3+\frac{1}{3}=\frac{10}{3}$, $x_2=3 + 2\times\frac{1}{3}=\frac{11}{3}$, $x_3=4$. $f(x)=x^{2}$, so $f(x_0)=3^{2}=9$, $f(x_1)=(\frac{10}{3})^{2}=\frac{100}{9}$, $f(x_2)=(\frac{11}{3})^{2}=\frac{121}{9}$, $f(x_3)=4^{2}=16$. $T_3=\frac{1/3}{2}\left[9 + 2\times\frac{100}{9}+2\times\frac{121}{9}+16\right]=\frac{1}{6}\left[9+\frac{200}{9}+\frac{242}{9}+16\right]=\frac{1}{6}\left[\frac{81 + 200+242 + 144}{9}\right]=\frac{1}{6}\times\frac{667}{9}=\frac{667}{54}\approx12.3519$

Step3: Calculate the error

The error $E$ is the absolute value of the difference between the exact value and the approximate value. $E=\left|\frac{37}{3}-\frac{667}{54}\right|=\left|\frac{666 - 667}{54}\right|=\frac{1}{54}\approx0.0185$

Answer:

The exact value is: $\frac{37}{3}\approx12.3333$ The error in the approximation is: $0.0185$