A) \[\int_{a+c}^{b+c}{f(x)dx=\int_{a}^{b}{f(x+c)dx}}\]
B) \[\int_{ac}^{bc}{f(x)dx=c\int_{a}^{b}{f(cx)dx}}\]
C) \[\int_{-a}^{a}{f(x)dx=\frac{1}{2}\int_{-a}^{a}{f(x)+f(-x)dx}}\]
D) None of these
Correct Answer: D
Solution :
[d] \[I=\int\limits_{a+c}^{b+c}{f(x)dx.}\] Putting \[x=t+c\] or \[dx=dt,\]we get \[I=\int_{a}^{b}{f(t+c)dt=\int_{a}^{b}{f(x+c)dx=\int_{ac}^{bc}{f(x)dx}}}\] Putting \[x=tc\] or\[dx=c\,dt\], we get \[I=c\int_{a}^{b}{f(ct)dt=c\int_{a}^{b}{f(cx)dx}}\] \[f(x)=\frac{1}{2}(f(x)+f(-x)+f(x)-f(-x))\] \[\therefore \int_{-a}^{a}{f(x)dx=\frac{1}{2}\int_{-a}^{a}{(f(x)+f(-x)+f(x)-f(-x))dx}}\] \[=\frac{1}{2}\int_{-a}^{a}{(f(x)+f(-x))dx+\frac{1}{2}\int_{-a}^{a}{(f(x)-f(-x))}dx}\]\[=\frac{1}{2}\int_{-a}^{a}{(f(x)+f(-x)dx}\] as \[f(x)+f(-x)\] is even and \[f(x)-f(-x)\] is odd.
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