Exponential Integration

Calculus Level 4

$\large \int e^{x^2} \ e^x \ (2x^2+x+1) \, dx=e^{x^2} f(x)+C$

Above shows an indefinite integral for some non-constant function $f(x)$ and arbitrary constant $C$ .

If the minimum value of $f (x)$ is equal to $m$ , then find the value of $\left \lfloor -\dfrac{1}{m} \right \rfloor$ .

4 5 -3 0 -2 2 1

2 solutions

Tanishq Varshney
May 13, 2015

$\displaystyle \int e^{x^2+x}(x(2x+1)+1)$

using $\large{\displaystyle \int e^{f(x)}(g(x)\times f^{\prime}(x)+g^{\prime}(x))=g(x).e^{f(x)}+c}$

the integral is $x.e^{x^2+x}+c$

$\phi(x)=xe^{x}$

on differentiating and equating to zero we get

$x=-1$ and the double derivative gives a positive value at $x=-1$

hence the minimum value is $\large{m=\frac{-1}{e}}$

and $[\frac{-1}{m}]=[e]$

$\large{\boxed{[e]=[2.7]=2}}$

I also did the exactly same,
I am astonished to see that this question has got level 4, I will rate it to level 1

Akhil Bansal - 5 years, 8 months ago

Santhosh Talluri
Feb 19, 2020

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