A) \[\int\limits_{x+5}^{5}{g}(t)dt\]
B) \[5\int\limits_{x+5}^{5}{g}(t)dt\]
C) \[\int\limits_{5}^{x+5}{g}(t)dt\]
D) \[2\int\limits_{5}^{x+5}{g}(t)dt\]
Correct Answer: A
Solution :
\[f(x)=\int\limits_{0}^{x}{g}(t)dt\] \[f(-x)=\int\limits_{0}^{-x}{g}(t)dt\] put \[t=-u\] \[=-\int\limits_{0}^{x}{g}(-u)du\] \[=-\int\limits_{0}^{x}{g}(u)d(u)=-f(x)\] \[\Rightarrow \]\[f(-x)=-f(x)\] \[\Rightarrow \]?(x) is an odd function Also \[\left( 5+x \right)=g\left( x \right)\] \[\left( 5x \right)=g\left( x \right)=g\left( x \right)=\left( 5+x \right)\] \[\Rightarrow \left( 5x \right)=\left( 5+x \right)\] Now \[I=\int\limits_{0}^{x}{f}(t)dt\] \[t=u+5\] \[I=\int\limits_{-5}^{x-5}{f}(u+5)du\] \[=\int\limits_{-5}^{x-5}{g}(u)du\] \[=\int\limits_{-5}^{x-5}{f'}(u)du\] \[=\left( x5 \right)\left( 5 \right)\] \[=\left( 5x \right)+\left( 5 \right)\] \[=\left( 5 \right)\left( 5+x \right)\] \[=\int\limits_{5+x}^{5}{f'}(t)dt=\int\limits_{5+x}^{5}{g}(t)dt\]
You need to login to perform this action.
You will be redirected in
3 sec
