the instructions for the given integral have two parts, one for the trapezoidal rule and one for simpsons…

the instructions for the given integral have two parts, one for the trapezoidal rule and one for simpsons rule. complete the following parts\n int_{0}^{2}left(4 t^{3}+t\right) d t \n\n a. estimate the integral with ( n = 4 ) steps and find an upper bound for ( left|e_{t}\right| ).\n( t = 90.5625 )\n(simplify your answer.)\nthe upper bound for ( left|e_{t}\right| ) is 10.125.\n(simplify your answer.)\n b. evaluate the integral directly and find ( left|e_{t}\right| ).\n int_{0}^{2}left(4 t^{3}+t\right) d t = 85.5 \n

the instructions for the given integral have two parts, one for the trapezoidal rule and one for simpsons rule. complete the following parts\n int_{0}^{2}left(4 t^{3}+t\right) d t \n\n a. estimate the integral with ( n = 4 ) steps and find an upper bound for ( left|e_{t}\right| ).\n( t = 90.5625 )\n(simplify your answer.)\nthe upper bound for ( left|e_{t}\right| ) is 10.125.\n(simplify your answer.)\n b. evaluate the integral directly and find ( left|e_{t}\right| ).\n int_{0}^{2}left(4 t^{3}+t\right) d t = 85.5 \n

Answer

Explanation:

Step1: Calculate the integral directly

Use the power rule for integration (\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)). For (\int_{0}^{2}(4t^{3}+t)dt), we have (\int_{0}^{2}(4t^{3}+t)dt=\left[4\times\frac{t^{4}}{4}+\frac{t^{2}}{2}\right]{0}^{2}). Simplify the antiderivative: (\left[t^{4}+\frac{t^{2}}{2}\right]{0}^{2}).

Step2: Evaluate the definite integral

Substitute the upper and lower limits. (\left(2^{4}+\frac{2^{2}}{2}\right)-\left(0^{4}+\frac{0^{2}}{2}\right)). Calculate (2^{4}=16), (\frac{2^{2}}{2} = 2). So (16 + 2-0=18).

Step3: Calculate (|E_T|)

We know that (T = 90.5625) (from part a, assume this is a wrong value in the original problem setup, but following the process). The actual value of the integral (\int_{0}^{2}(4t^{3}+t)dt=\left[t^{4}+\frac{t^{2}}{2}\right]{0}^{2}=16 + 2=18). (|E_T|=|T - \text{actual value}|). If (T = 90.5625) (wrongly calculated in part a, but using the formula (|E_T|=|T-\int{a}^{b}f(x)dx|)), (|E_T|=|90.5625 - 18|=72.5625). But if we calculate the integral correctly: (\int_{0}^{2}(4t^{3}+t)dt=\left[4\times\frac{t^{4}}{4}+\frac{t^{2}}{2}\right]_{0}^{2}=(16 + 2)-0 = 18). If (T) is calculated correctly using trapezoidal rule: (\Delta x=\frac{b - a}{n}=\frac{2-0}{4}=0.5), (x_0 = 0), (x_1=0.5), (x_2 = 1), (x_3=1.5), (x_4 = 2). (f(t)=4t^{3}+t), (f(x_0)=0), (f(x_1)=4\times(0.5)^{3}+0.5=4\times0.125 + 0.5=1), (f(x_2)=4\times1^{3}+1=5), (f(x_3)=4\times(1.5)^{3}+1.5=4\times3.375+1.5 = 15), (f(x_4)=4\times2^{3}+2=34). (T=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)]=\frac{0.5}{2}(0 + 2\times1+2\times5+2\times15+34)=\frac{0.25}(0 + 2+10+30+34)=18). Then (|E_T|=|18 - 18|=0).

Answer:

(\int_{0}^{2}(4t^{3}+t)dt = 18), (|E_T| = 0)