∫ekx [fkx + fkx]dx will be equal to 1kekx fkx + C only
![∫ekx [fkx + fkx]dx will be equal to 1kekx fkx + C only](/img/relate-questions.png)
|
∫ekx [f(kx) + f'(kx)]dx will be equal to 1kekx f(kx) + C only when k = 1.
A.
True
B.
False
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Let's proceed the same way we found the formula for the integral of ∫ex [f(x) + f'(x)]dx
∫ekx [f(kx) + f'(kx)]dx = ∫ekx f(kx) + ekx f'(kx)]dx
Now apply integration by parts on the first integral. f(kx) will be the first function and ekx will be second function.
ekx.f(kx)k - ∫ekx.f′(kx).kk dx + ∫ekx f'(kx)dx
=ekx.f(kx)k−∫ekx f'(kx)dx + ∫ekx f'(kx)dx + C
=ekx.f(kx)k + C
Here, we see that k can take any real value except zero. So the given statement that it's true for only for k = 1 is not true.