Given:
[tex]\int_3^5(6x^2+1)dx\text{ and n=4.}[/tex]Required:
We need to find the trapezoid approximation and exact value.
Explanation:
[tex]Let\text{ f\lparen x\rparen=}6x^2+1.[/tex]The given interval is {3,5} and n=4.
Consider the formula.
[tex]\Delta x=\frac{b-a}{n}[/tex]Substitute a=5, b=5, and n=4 in the formula,
[tex]\Delta x=\frac{5-3}{4}=\frac{2}{4}=0.5[/tex][tex]x_0=3,x_1=3.5,x_2=4,x_3=4.\text{5 and }x_4=5.[/tex]Consider the trapezoid formula.
[tex]\int_a^bf(x)dx\approx\frac{1}{2}\Delta x(f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4))[/tex][tex]Substitute\text{ }\Delta x=0.5,x_0=3,x_1=3.5,x_2=4,x_3=4.\text{5 and }x_4=5\text{ in the formula,}[/tex][tex]\int_3^5f(x)dx\approx\frac{1}{2}(0.5)(f(3)+2f(3.5)+2f(4)+2f(4.5)+f(5))[/tex][tex]Use\text{ f\lparen x\rparen=}6x^2+1.[/tex][tex]=\frac{1}{2}(0.5)((6(3)^2+1)+2(6(3.5)^2+1)+2(6(4)^2+1)+2(6(4.5)^2+1)+(6(5)^2+1)[/tex][tex]=\frac{1}{2}(0.5)(55+149+194+245+151)[/tex][tex]\int_3^5(6x^2+1)dx\approx198.5[/tex]Consider the given integral.
[tex]\int_3^5(6x^2+1)dx=\int_3^56x^2dx+\int_3^51dx[/tex]Integrate the given integral.
[tex]\int_3^5(6x^2+1)dx=[\frac{6x^3}{3}]^5_3+[x]^5_3[/tex][tex]=[2x^3]_3^5+[x]_3^5[/tex][tex]=[2(5)^3-2(3)^3]+[5-3][/tex][tex]=198[/tex][tex]\int_3^5(6x^2+1)dx=198[/tex]Final answer:
[tex]\text{ Trapezoidal Approximation}\approx198.5[/tex][tex]Exact\text{ value=198}[/tex]