1for Q1) differentiate using lebnitz theorem ....
Let f:R→R be a differentiable function such that f(x)=x2+e-x∫01 et f(t) dt
q1 f(x) increases in ?
q2 Value of ∫01 f(x) dx
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10 Answers
62The is is not a very difficult question, and there is no use of lebnitz either....
The integral given there is a constant..
So you an treat the equation as f(x)=x2+cex
Now we have to integrate from 0 to 1
so integral of f(x) from 0 to 1 will be (x^3/3+ce^x)|_0^1=1/3+ce-c
Where c itself is equal to this integral....
Thus, c=1/3+ce-c
So, c(e-2)=-1/3
so c=1/[3(2-e)]
23and sir ,
c=\int_{0}^{1}{e^{t}f(t)dt}
but in the sixth line of ur solution u mentioned that c=\int_{0}^{1}{f(t)dt}
62The is is not a very difficult question, and there is no use of lebnitz either....
The integral given there is a constant..
So you an treat the equation as f(x)=x2+ce-x
Now, we have to integrate f(x).ex
which will be easy to do..
Please correct the mistakes from my previous post as given by Qwerty [3]
23yes rpf , c gets cancelled on both sides , leaving e = 2 [2] donno how
62\\f(x)=x^2+e^{-x}\int_0^1e^t.f(t)dt \\Let\ c=\int_0^1e^t.f(t)dt \\f(x)=x^2+ce^{-x} \\Let\ c=\int_0^1(e^t.(t^2+ce^{-t}))dt \\\Rightarrow c=\int_0^1(t^2.e^t+c)dt \\\Rightarrow c=\int_0^1(t^2.e^t)dt+c
Yup, you are all correct.. This question has an issue because as you can see that the integral of a +Ve number is zero...
I am sure there is some typing mistake in the problem...
The fact is that such a function cannot exist....
So the question I think is not well prepared..