[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\rm \: f(a + b - x) = f(x) \\ [/tex]
and
[tex]\rm \: \displaystyle\int_{a}^{b}\rm \: \:x \: f(x) \: dx \\ [/tex]
Let assume that
[tex]\rm \:I \: = \: \displaystyle\int_{a}^{b}\rm \: \:x \: f(x) \: dx - - - (1) \\ [/tex]
We know that,
[tex]\boxed{ \rm{ \:\displaystyle\int_{a}^{b} \rm \: f(x) \: dx \: = \: \displaystyle\int_{a}^{b} \:\rm \: f(a + b - x) \: dx \: }} \\ \\ [/tex]
So, using this property of definite integrals, we have
[tex]\rm \:I \: = \: \displaystyle\int_{a}^{b}\rm \: \:(a + b - x) \: f(a + b - x) \: dx \\ [/tex]
As it is given that,
[tex] \red{\rm \: f(a + b - x) = f(x)} \\ \\ [/tex]
So, using this, above integral can be rewritten as
[tex]\rm \:I \: = \: \displaystyle\int_{a}^{b}\rm \: \:(a + b - x) \: f(x) \: dx - - - (2) \\ \\ [/tex]
On adding equation (1) and (2), we get
[tex]\rm \:2I \: = \: \displaystyle\int_{a}^{b}\rm \: \:x \: f(x) \: dx + \displaystyle\int_{a}^{b}\rm \: \:(a + b - x) \: f(x) \: dx \\ \\ [/tex]
[tex]\rm \:2I \: = \: \displaystyle\int_{a}^{b}\rm \: \:(x + a + b - x) \: f(x) \: dx \\ \\ [/tex]
[tex]\rm \:2I \: = \: \displaystyle\int_{a}^{b}\rm \: \:( a + b) \: f(x) \: dx \\ \\ [/tex]
[tex]\rm \:2I \: = \: (a + b) \: \displaystyle\int_{a}^{b}\rm \: \:f(x) \: dx \\ \\ [/tex]
[tex]\rm \:I \: = \: \dfrac{a + b}{2} \: \displaystyle\int_{a}^{b}\rm \: \:f(x) \: dx \\ \\ [/tex]
Hence,
[tex]\bf\implies \:\displaystyle\int_{a}^{b}\bf \: \: x \: f(x) \: dx = \: \dfrac{a + b}{2} \: \displaystyle\int_{a}^{b}\bf \:f(x) \: dx \\ \\ [/tex]
[tex]\rule{190pt}{2pt} \\ [/tex]
[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = - \: \: \displaystyle \int_{b}^{a}\sf \: f(x) \: dx}} \\ [/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{a}^{b}\sf \: f(y) \: dy}} \\ [/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{a}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{0}^{a}\sf \: f(a - x) \: dx}} \\ [/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{a}^{b}\sf \: f(a + b - x) \: dx}} \\ [/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{2a}\sf \: f(x) \: dx \: = \:2 \displaystyle \int_{0}^{2}\sf f(x)dx \: \: if \: f(2a - x) = f(x)}} \\ [/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{2a}\sf \: f(x) \: dx \: = \:0 \: \: if \: f(2a - x) = - \: f(x)}} \\ [/tex]
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Answers & Comments
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\rm \: f(a + b - x) = f(x) \\ [/tex]
and
[tex]\rm \: \displaystyle\int_{a}^{b}\rm \: \:x \: f(x) \: dx \\ [/tex]
Let assume that
[tex]\rm \:I \: = \: \displaystyle\int_{a}^{b}\rm \: \:x \: f(x) \: dx - - - (1) \\ [/tex]
We know that,
[tex]\boxed{ \rm{ \:\displaystyle\int_{a}^{b} \rm \: f(x) \: dx \: = \: \displaystyle\int_{a}^{b} \:\rm \: f(a + b - x) \: dx \: }} \\ \\ [/tex]
So, using this property of definite integrals, we have
[tex]\rm \:I \: = \: \displaystyle\int_{a}^{b}\rm \: \:(a + b - x) \: f(a + b - x) \: dx \\ [/tex]
As it is given that,
[tex] \red{\rm \: f(a + b - x) = f(x)} \\ \\ [/tex]
So, using this, above integral can be rewritten as
[tex]\rm \:I \: = \: \displaystyle\int_{a}^{b}\rm \: \:(a + b - x) \: f(x) \: dx - - - (2) \\ \\ [/tex]
On adding equation (1) and (2), we get
[tex]\rm \:2I \: = \: \displaystyle\int_{a}^{b}\rm \: \:x \: f(x) \: dx + \displaystyle\int_{a}^{b}\rm \: \:(a + b - x) \: f(x) \: dx \\ \\ [/tex]
[tex]\rm \:2I \: = \: \displaystyle\int_{a}^{b}\rm \: \:(x + a + b - x) \: f(x) \: dx \\ \\ [/tex]
[tex]\rm \:2I \: = \: \displaystyle\int_{a}^{b}\rm \: \:( a + b) \: f(x) \: dx \\ \\ [/tex]
[tex]\rm \:2I \: = \: (a + b) \: \displaystyle\int_{a}^{b}\rm \: \:f(x) \: dx \\ \\ [/tex]
[tex]\rm \:I \: = \: \dfrac{a + b}{2} \: \displaystyle\int_{a}^{b}\rm \: \:f(x) \: dx \\ \\ [/tex]
Hence,
[tex]\bf\implies \:\displaystyle\int_{a}^{b}\bf \: \: x \: f(x) \: dx = \: \dfrac{a + b}{2} \: \displaystyle\int_{a}^{b}\bf \:f(x) \: dx \\ \\ [/tex]
[tex]\rule{190pt}{2pt} \\ [/tex]
[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]
[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = - \: \: \displaystyle \int_{b}^{a}\sf \: f(x) \: dx}} \\ [/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{a}^{b}\sf \: f(y) \: dy}} \\ [/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{a}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{0}^{a}\sf \: f(a - x) \: dx}} \\ [/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{a}^{b}\sf \: f(a + b - x) \: dx}} \\ [/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{2a}\sf \: f(x) \: dx \: = \:2 \displaystyle \int_{0}^{2}\sf f(x)dx \: \: if \: f(2a - x) = f(x)}} \\ [/tex]
[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{2a}\sf \: f(x) \: dx \: = \:0 \: \: if \: f(2a - x) = - \: f(x)}} \\ [/tex]