So you think you can solve!!

Calculus Level 5

Let $f(x)$ be a differentiable function such that $\displaystyle \dfrac{d}{dx} f(x) = f(x) + \int_0^2 f(x) \, dx$ and $f(0) = \dfrac{4-e^2}3$ , find $f(2)$ .

Round your answer correct to 2 decimal places.

The answer is 5.26.

2 solutions

Rishabh Jain
Feb 11, 2016

Let C= $\int_0^2 f(x) \, dx$
$f'(x)-f(x)=C$ $\Rightarrow f'(x)e^{-x}-f(x)e^{-x}=Ce^{-x}$ $\Rightarrow \dfrac{d(f(x)e^{-x})}{dx} =Ce^{-x}$ Integrating both sides and introducing a arbitrary real constant 'D'. $\Rightarrow f(x)=-C+De^{x}$ Now we can evaluate C: $C=\int_0^2 -C+De^{-x} \, dx =-2C +D(e^2-1)$ $\Rightarrow C=\frac{D}{3}(e^2-1)$ Substituting C in f(x), we get $f(x)=\dfrac{D(3e^x-e^2+1)}{3}$ Using the initial condition on f(0), we get D=1 Hence $\large f(x)=\dfrac{(3e^x-e^2+1)}{3}$ $\Large\therefore~f(2)=\dfrac{2e^2+1}{3}\approx\boxed{\color{#007fff}{5.26}}$

Where did you find this??? In previous IIT paper??

Venkata Nikhil - 5 years, 3 months ago

If you take f(x)= (e^x-1)/3 then it satisfies both the equation given in the question but from this the answer is 2.13 .Kindly tell me how it is so ?

Shivang Gupta - 5 years, 4 months ago

According to your f(x): f(0)=0 but instead in the question it is specified that $f(0) = \dfrac{4-e^2}3$ .
Also check properly, it isn't satisfying any of the given conditions.

Rishabh Jain - 5 years, 4 months ago

Purusharth Verma
Feb 17, 2016

(2e^2 + 1)/3 is it......????

So you think you can solve!!

The answer is 5.26.

2 solutions

0 pending reports