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\(\displaystyle\int_{a+c}^{b+c}f(x)dx = \ ?\)1. \(\displaystyle\int_a^b f(x-c)dx\)2. \(\displaystyle\int_a^b f(x+c)dx\)3. \(\displaystyle\int_a^b f(x)dx\)4. \(\displaystyle\int_{a-c}^{b-c} f(x)dx\) |
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Answer» Correct Answer - Option 2 : \(\displaystyle\int_a^b f(x+c)dx\) Explanation: Given Integral is, \(I~=~\displaystyle\int_{a+c}^{b+c}\ f(x)dx\) Put t = x - c i.e. x = t + c By differentiating we have, dt = dx at x = a + c → t = a + c - c = a ........(1) at x = b + c → t = b + c -c = b .........(2) Now, \(\displaystyle\int_{a+c}^{b+c}f(x)dx =~\int_a^b f(t+c)dt \) .........(3) Also, put t = x then, t = a → x = a ,,,,,,from equation (1) t = b → x = b ......from equation (2) From the equation (3) \(\displaystyle\int_{a+c}^{b+c}f(x)dx =~\int_a^b f(x+c)dt \) Hence, it is proove. |
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