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Determine a positive integer $n\leq5$,such that $\int_{0}^{1}e^x(x-1)^ndx=16-6e$.

I tried to solve it.But since $n$ is given to be $\leq$ 5,my calculations went lengthy.

Applying integration by parts repeatedly,we get
\begin{align} \int e^x(x-1)^n \, dx &= \left[(x-1)^ne^x-n(x-1)^{n-1}e^x+n(n-1)(x-1)^{n-2}e^x \right. \\ & \hspace{5mm} \left.-n(n-1)(n-2)(x-1)^{n-3}e^x+n(n-1)(n-2)(n-3)(x-1)^{n-4}e^x \right. \\ & \hspace{5mm} \left.-n(n-1)(n-2)(n-3)(n-4)e^x\right] \end{align} \begin{align} \int_{0}^{1}e^x(x-1)^n \, dx &= -n(n-1)(n-2)(n-3)(n-4)e-(-1)^n+n(-1)^{n-1}-n(n-1)(-1)^{n-2} \\ & \hspace{5mm} +n(n-1)(n-2)(-1)^{n-3}-n(n-1)(n-2)(n-3)(-1)^{n-4} \\ & \hspace{5mm} +n(n-1)(n-2)(n-3)(n-4) \\ &=16-6e \end{align}

Now solving this is very difficult,is there another simple and elegant method to find $n=3.$

Leucippus
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Brahmagupta
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5 Answers5

4

HINT:

Let $x-1=y$

$$\int_0^1e^x(x-1)^n\ dx=\int_{-1}^0e^{y+1}y^n\ dy=e\int_{-1}^0e^yy^n\ dy$$

Now integrating by parts, $$I_n=\int_{-1}^0e^yy^n\ dy=[y^n\int e^y\ dy]_{-1}^0-\int_{-1}^0\left(\dfrac{d(y^n)}{dy}\int e^y\ dy\right)dy$$

$\implies I_n=-(-1)^n\dfrac1e-nI_{n-1}$

Now $I_0=\int_{-1}^0e^y\ dy=1-\dfrac1e$

$\implies I_1=\dfrac1e-I_0$ and so on

lab bhattacharjee
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3

HINT:

Notice, the following property of definite integral $$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$

Now, we have

$$\int_{0}^{1}e^x(x-1)^ndx=16-6e$$ $$\int_{0}^{1}e^{1-x}(1-x-1)^ndx=16-6e$$ $$\int_{0}^{1}e^{1-x}(-x)^ndx=16-6e$$ $$\int_{0}^{1}e \cdot e^{-x}(-1)^nx^n dx=16-6e$$ $$(-1)^n e\int_{0}^{1} e^{-x}x^n dx=16-6e$$

I hope you can proceed further.

Harish Chandra Rajpoot
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2

If we set $$ I_n = \int_{0}^{1}e^x(1-x)^n\,dx = \int_{0}^{1}x^n e^{1-x}\,dx\tag{1}$$ integration by parts gives: $$ I_{n+1} = -1 + (n+1)\, I_n\tag{2}$$ Since $I_0 = -1+e$, by using the previous formula and induction we have: $$ \forall n\in\mathbb{N},\qquad I_n = A_n + B_n e,\quad A_n,B_n\in\mathbb{Z}\tag{3}$$ as well as $$ B_n = n! \tag{4} $$ so the solution to the original problem is clearly $n=\color{red}{3}$.

Jack D'Aurizio
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1

I would define

$$a_n:=\int_0^1 e^x (x-1)^n dx.$$

Then we simply get $a_0=e-1$ and for $n>0$ using integrations by parts we get $$a_n = [e^x(x-1)^n]_0^1 - n\int_0^1 e^x (x-1)^{n-1}dx = (-1)^{n+1}-n a_{n-1}.$$

With that recursive formula you compute $a_1$, $a_2$ and $a_3$ very quickly and get your result $a_3=16-6e$.

principal-ideal-domain
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0

If you have reason to believe the equation is satisfied by a positive integer $n\le5$, we can rule out the even integers $n=2$ and $4$ since $16-6e\approx-0.30969$ is negative. For odd integers, a change of variable leaves the equation

$$I_n=\int_0^1 x^ne^{-x}dx=6-{16\over e}\approx0.1139$$

Now

$$1-x\le e^{-x}\le1-x+{1\over2}x^2\quad\text{for }0\le x\le1$$

which means

$$I_1\ge\int_0^1(x-x^2)dx={1\over2}-{1\over3}={1\over6}=0.1666\ldots$$

is too big, and

$$I_5\le\int_0^1(x^5-x^6+{1\over2}x^7)dx={1\over6}-{1\over7}+{1\over16}\approx0.0863$$

is too small. So by the Goldilocks Principle, $n=3$ must be just right.

Barry Cipra
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