A) \[\int_{a-1}^{b-1}{f(x)dx}\]
B) \[\int_{a+1}^{b+1}{f(x)dx}\]
C) \[\int_{a-1}^{b-1}{f(x+1)dx}\]
D) \[\int_{a+1}^{b+1}{f(x+1)dx}\]
Correct Answer: A , B , C , D
Solution :
(BONUS) |
\[\text{l}=\frac{1}{a+b}\int_{a}^{b}{x\left[ f(x)+f(x+1) \right]dx...(1)}\] |
Replace x by \[(a+b-x)\] |
\[\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(a+b-x)+f(a+b+1-x) \right]}dx\] |
As \[f(a+b+1-x)=f(x)\] |
\[\Rightarrow f(a+b+1-a-b+x)=f(a+b-x)\] |
\[\Rightarrow f(1+x)=f(a+b-x)\] |
\[\therefore \,\,\,\,\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(1+x)+f(x) \right]dx...(2)}\] |
Add (1) & (2) |
\[2l=\frac{1}{a+b}\int_{a}^{b}{(a+b)\left[ f(1+x)+f(x) \right]dx}\] |
\[\Rightarrow \frac{1}{2}\int_{a}^{b}{\left[ f(1+x)+f(x) \right]dx}\] |
\[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\] |
\[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+a+b-x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\] |
\[=\int_{a}^{b}{f(x)dx}\] |
Put \[x=t+1\] |
\[=\int_{a-1}^{b-1}{f(t+1)dt}\] |
Further, \[\text{l}=\int_{a-1}^{b-1}{f(a+b-1-t)dt=\int_{a-1}^{b-1}{f(t-2)dt}}\] |
\[=\int_{a+1}^{b+1}{f(x)dx}\] |
Solution :
(BONUS) |
\[\text{l}=\frac{1}{a+b}\int_{a}^{b}{x\left[ f(x)+f(x+1) \right]dx...(1)}\] |
Replace x by \[(a+b-x)\] |
\[\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(a+b-x)+f(a+b+1-x) \right]}dx\] |
As \[f(a+b+1-x)=f(x)\] |
\[\Rightarrow f(a+b+1-a-b+x)=f(a+b-x)\] |
\[\Rightarrow f(1+x)=f(a+b-x)\] |
\[\therefore \,\,\,\,\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(1+x)+f(x) \right]dx...(2)}\] |
Add (1) & (2) |
\[2l=\frac{1}{a+b}\int_{a}^{b}{(a+b)\left[ f(1+x)+f(x) \right]dx}\] |
\[\Rightarrow \frac{1}{2}\int_{a}^{b}{\left[ f(1+x)+f(x) \right]dx}\] |
\[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\] |
\[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+a+b-x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\] |
\[=\int_{a}^{b}{f(x)dx}\] |
Put \[x=t+1\] |
\[=\int_{a-1}^{b-1}{f(t+1)dt}\] |
Further, \[\text{l}=\int_{a-1}^{b-1}{f(a+b-1-t)dt=\int_{a-1}^{b-1}{f(t-2)dt}}\] |
\[=\int_{a+1}^{b+1}{f(x)dx}\] |
Solution :
(BONUS) |
\[\text{l}=\frac{1}{a+b}\int_{a}^{b}{x\left[ f(x)+f(x+1) \right]dx...(1)}\] |
Replace x by \[(a+b-x)\] |
\[\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(a+b-x)+f(a+b+1-x) \right]}dx\] |
As \[f(a+b+1-x)=f(x)\] |
\[\Rightarrow f(a+b+1-a-b+x)=f(a+b-x)\] |
\[\Rightarrow f(1+x)=f(a+b-x)\] |
\[\therefore \,\,\,\,\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(1+x)+f(x) \right]dx...(2)}\] |
Add (1) & (2) |
\[2l=\frac{1}{a+b}\int_{a}^{b}{(a+b)\left[ f(1+x)+f(x) \right]dx}\] |
\[\Rightarrow \frac{1}{2}\int_{a}^{b}{\left[ f(1+x)+f(x) \right]dx}\] |
\[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\] |
\[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+a+b-x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\] |
\[=\int_{a}^{b}{f(x)dx}\] |
Put \[x=t+1\] |
\[=\int_{a-1}^{b-1}{f(t+1)dt}\] |
Further, \[\text{l}=\int_{a-1}^{b-1}{f(a+b-1-t)dt=\int_{a-1}^{b-1}{f(t-2)dt}}\] |
\[=\int_{a+1}^{b+1}{f(x)dx}\] |
Solution :
(BONUS) |
\[\text{l}=\frac{1}{a+b}\int_{a}^{b}{x\left[ f(x)+f(x+1) \right]dx...(1)}\] |
Replace x by \[(a+b-x)\] |
\[\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(a+b-x)+f(a+b+1-x) \right]}dx\] |
As \[f(a+b+1-x)=f(x)\] |
\[\Rightarrow f(a+b+1-a-b+x)=f(a+b-x)\] |
\[\Rightarrow f(1+x)=f(a+b-x)\] |
\[\therefore \,\,\,\,\text{l}=\frac{1}{a+b}\int_{a}^{b}{(a+b-x)\left[ f(1+x)+f(x) \right]dx...(2)}\] |
Add (1) & (2) |
\[2l=\frac{1}{a+b}\int_{a}^{b}{(a+b)\left[ f(1+x)+f(x) \right]dx}\] |
\[\Rightarrow \frac{1}{2}\int_{a}^{b}{\left[ f(1+x)+f(x) \right]dx}\] |
\[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\] |
\[\text{l}=\frac{1}{2}\int_{a}^{b}{f(1+a+b-x)dx+\frac{1}{2}\int_{a}^{b}{f(x)dx}}\] |
\[=\int_{a}^{b}{f(x)dx}\] |
Put \[x=t+1\] |
\[=\int_{a-1}^{b-1}{f(t+1)dt}\] |
Further, \[\text{l}=\int_{a-1}^{b-1}{f(a+b-1-t)dt=\int_{a-1}^{b-1}{f(t-2)dt}}\] |
\[=\int_{a+1}^{b+1}{f(x)dx}\] |
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