A) \[\int_{-2}^{2}{f(x)dx=\int_{0}^{2}{[f(x)-f(-x)]dx}}\]
B) \[\int_{-3}^{5}{2f(x)dx=\int_{-6}^{10}{f(x-1)dx}}\]
C) \[\int_{-3}^{5}{f(x)dx=\int_{-4}^{4}{f(x-1)dx}}\]
D) \[\int_{-3}^{5}{f(x)dx=\int_{-2}^{6}{f(x-1)dx}}\]
E) \[\int_{-3}^{5}{f(x)dx=\int_{-6}^{10}{f(x/2)]dx}}\]
Correct Answer: D
Solution :
Since, f is continues function. Let \[x=t-1\] \[\therefore \]\[dx=dt\]. When \[x=-3\to 5\], then \[t=-2\to 6\] Therefore, \[\int_{-3}^{5}{f(x)dx}\]\[=\int_{-2}^{6}{f(t-1)dt=}\int_{-2}^{6}{f(x-1)dx}\].
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