Answer:
see below
Step-by-step explanation:
Part-A:
we want to find the quotient and remainder when 4x²+4x is divided by 2x+1 in other words we want to find the quotient and remainder when:
[tex] \displaystyle \frac{4 {x}^{2} + 4x}{2x + 1} [/tex]
To do so, I would prefer using simple algebra rather than using troublesome polynomial long division. anyway dividing it would yield the form [tex]p(x)+\frac{k}{q(x)}[/tex] where:
- p(x) is the quotation
- k is the remainder
- q(x) is the divisor
Therefore,In order to derive the quotation and remainder, rewrite the numerator which yields:
[tex] \displaystyle \frac{(2x + 1 {)}^{2} - 1}{2x + 1} \\ \\ \frac{(2x + 1 {)}^{2} - 1}{2x + 1} \\ \\ \frac{(2x + 1 {)}^{2} }{2x + 1} + \frac{ - 1}{2x + 1} \\ \\ \boxed{2x + 1 + \frac{ - 1}{2x + 1} }[/tex]
Compering it to the mentioned form, we can consider:
- 2x+1, The quotient
- -1, The remainder
Part-B:
we are asked to show that,
[tex] \displaystyle \int _{0} ^{1} \frac{4 {x}^{2} + 4x}{2x + 1} \, dx = 2 - \frac{1}{2} \ln(3) [/tex]
Well,we can start with integrating the indefinite integral and the first step to do so is to decompose the fraction, integrand. As we have already done it in part-a, we can simply skip the steps:
[tex] \displaystyle \implies \int 2x + 1 + \frac{ - 1}{2x + 1} \, dx \stackrel{ ? }{= }2 - \frac{1}{2} \ln(3) [/tex]
utilizing sum integration rule yields:
[tex] \displaystyle \implies \int 2x \, dx+ \int 1 \,dx+ \int \frac{ - 1}{2x + 1} \, dx \stackrel{ ? }{= }2 - \frac{1}{2} \ln(3) [/tex]
apply constant integration rule which yields:
[tex] \displaystyle \implies 2\int x \, dx+ \int 1 \,dx - \int \frac{ 1}{2x + 1} \, dx \stackrel{ ? }{= }2 - \frac{1}{2} \ln(3) [/tex]
recall that,
- integration of xⁿ is xⁿ+¹/n+1
- integration of a constant,k is kx
so we derive from utilizing the rules is that,
[tex] \displaystyle \implies 2 \cdot\frac{ {x}^{2} }{2} + x- \int \frac{ 1}{2x + 1} \, dx \stackrel{ ? }{= }2 - \frac{1}{2} \ln(3) [/tex]
Now integrating [tex]\frac{1}{2x+1}[/tex] would require u-substitution . In order to perform the substitution, let
To perform the substitution multiply the integrand and integral by 2 and ½ respectively:
[tex] \displaystyle \implies { {x}^{2} }+ x- \frac{1}{2} \int \frac{ 2}{2x + 1} \, dx \stackrel{ ? }{= }2 - \frac{1}{2} \ln(3) [/tex]
perform the substitution:
[tex] \displaystyle \implies x^2+ x- \frac{1}{2} \int \frac{ 1}{u} \, du \stackrel{ ? }{= }2 - \frac{1}{2} \ln(3) [/tex]
integrating yields:
[tex] \displaystyle \implies x^2+ x- \frac{1}{2} \ln(u) \stackrel{ ? }{= }2 - \frac{1}{2} \ln(3) [/tex]
back-substitute:
[tex] \displaystyle \implies x^2 + x- \frac{1}{2} \ln(2x + 1) \stackrel{ ? }{= }2 - \frac{1}{2} \ln(3) [/tex]
To evaluate the define integral , return the limits of integration:
[tex] \displaystyle \implies x^2 \Bigg | _{0} ^{1} + x \Bigg | _ {0}^{1} - \frac{1}{2} \ln(2x + 1)\Bigg | _ {0}^{1} \stackrel{ ? }{= }2 - \frac{1}{2} \ln(3) [/tex]
remember fundamental theorem
- [tex]f(x)\Bigg | _ {a}^{b} = f(b) - f(a)[/tex]
utilize it and simplify which yields:
[tex] \displaystyle \implies 1 + 1 - \frac{1}{2} \ln(3) \stackrel{ ? }{= }2 - \frac{1}{2} \ln(3) \\ \\ 2 - \frac{1}{2} \ln(3)\stackrel{ \checkmark }{= }2 - \frac{1}{2} \ln(3) \\ \\ \rm \rightarrow \: hence,showed[/tex]
and we are done!
