Answer:
Question 3
Find the range of the function for the domain 2 < x < 3
[tex]f(2)=(2)^2-1=3[/tex]
[tex]f(3)=(3)^2-1=8[/tex]
Therefore, the range of [tex]f(x)=x^2-1[/tex] is [tex]3 < f(x) < 8[/tex] with domain 2 < x < 3
This means that for the function to be continuous, when the domain is x=3, f(x) = 8
[tex]f(3)=2a(3)=6a[/tex]
[tex]\implies 6a=8\implies a=\dfrac43[/tex]
Similarly, when the domain is x > 3, f(x) > 8
[tex]f(3)=b(3)-4=3b-4[/tex]
[tex]\implies 3b-4 > 8 \implies b = 4[/tex]
Therefore, [tex]a=\dfrac43[/tex] and b = 4
Question 5
[tex]\textsf{let}\:u=\ln (x)[/tex]
[tex]\implies \dfrac{du}{dx}=\dfrac{1}{x}\implies dx=x\:du[/tex]
[tex]x=1\implies u=0[/tex]
[tex]x=3 \implies u=\ln 3[/tex]
[tex]\implies \displaystyle \int\limits_{1}^{3} \dfrac{\ln x \sin (\ln x)}{x} \,\:dx=\displaystyle \int\limits_{0}^{\ln 3} \dfrac{u \sin u}{x} \, \cdot x\:du=\displaystyle \int\limits_{0}^{\ln 3} (u \sin u) \, \:du[/tex]
Use integration by parts:
[tex]\displaystyle \int u\dfrac{dv}{dx} \, dx=uv- \displaystyle \int v \dfrac{du}{dx} \, dx[/tex]
[tex]\textsf{where}\:u=u\:\textsf{and}\:\dfrac{dv}{dx}=\sin u[/tex]
[tex]\implies \dfrac{du}{dx}=1\:\textsf{and}\:v=\int \sin u=-\cos u[/tex]
[tex]\implies \displaystyle \int\limits_{0}^{\ln 3} (u \sin u) \, \:du=-u \cos u- \int\limits_{0}^{\ln 3}-\cos u \,\:du[/tex]
[tex]\begin{aligned}\implies \displaystyle \int\limits_{0}^{\ln 3} (u \sin u) \, \:du & =\left[-u \cos u+ \sin u\right]_{0}^{\ln 3}\\ & =[- (\ln 3) \cos (\ln 3)+ \sin (\ln 3)]-[-0 \cos (0)+ \sin (0)]\\\\ & =[- (\ln 3) \cos (\ln 3)+ \sin (\ln 3)]-0\\\\ & = 0.3908925527...\end{aligned}[/tex]
