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You may refer to the link below:
https://www.youtube.com/watch?v=6hi7d4ZZ6UA
In this question, the indefinite integral of product of a function f(x) and its derivative f'(x) is given. Also value of function at x = 1 is 4 (f(1) = 4). Find the function f(x).
Solution:
Let us assume f(x) = u
Then f'(x)dx = du
As we know,
\[\int u.du=\frac{u^2}{2}+c\]
\[\Rightarrow\int f(x)f'(x).dx=\frac{[f(x)]^2}{2}+c\]
Now the equation becomes,
\[\frac{[f(x)]^2}{2}+c = x\]
Also given that f(1) = 4,
\[\Rightarrow\frac{[f(1)]^2}{2}+c = 1\]
\[\Rightarrow\frac{16}{2}+c = 1\]
\[\Rightarrow c=-7\]
Put this value in our equation,
\[\frac{[f(x)]^2}{2}-7 = x\]
\[\Rightarrow[f(x)]^2 =2x+14\]
\[\Rightarrow f(x)=\sqrt{2x+14}\]
Question related to topics: algebra, function, derivatives, integration.
