Definite integral (e^x) *(x-1)^n=16-6e find n.

[vkash]

Homework Statement

this is definite integral question.
lower limit 0 ; upper limit 1 ; integral (e^x)(x-1)^n = 16-6e
find n (n<6)
here e is euler constant value around 2.7 (irrational)
hope you understood.

Homework Equations

as much as first year student know.

The Attempt at a Solution

tried many substitutions like x-1 = t and some others.
replaced x by 1-x and tried integration by parts. and many more.
but failed

 

[icystrike]

Use integration by parts and you will be able to write the integral in summative notation

 

[vkash]

Use integration by parts and you will be able to write the integral in summative notation[/color]

does not understand.
can you please explain it more briefly.
thanks for reply.

 

[hunt_mat]

Integrate by parts:
$$\int_{0}^{1}e^{x}(x-1)^{n}dx=\Bigg[ e^{x}(x-1)^{n}\Bigg]_{0}^{1}-n\int_{0}^{1}e^{x}(x-1)^{n-1}dx=(-1)^{n}-n\int_{0}^{1}e^{x}(x-1)^{n-1}dx$$
If your original integral is $I_{n}$, then you have a recurrence relation:
$$I_{n}=(-1)^{n}-nI_{n-1}$$

 

[vkash]

Integrate by parts:
$$\int_{0}^{1}e^{x}(x-1)^{n}dx=\Bigg[ e^{x}(x-1)^{n}\Bigg]_{0}^{1}-n\int_{0}^{1}e^{x}(x-1)^{n-1}dx=(-1)^{n}-n\int_{0}^{1}e^{x}(x-1)^{n-1}dx$$
If your original integral is $I_{n}$, then you have a recurrence relation:
$$I_{n}=(-1)^{n}-nI_{n-1}$$

that's called answer thanks!
can you tell me how to use itex tex tags.

 

[hunt_mat]

To open the tex just do tex in square brackets [ tex] (but without the spaces). To close of the tex do [ /tex ] but take off the spaces. The same with itex, which is in-line tex.

 

[Dickfore]

As was pointed out:
$$I_{n} \equiv \int_{0}^{1}{(x - 1)^{n} \, e^{x} \, dx}$$

$$I_{n} = \left. (x - 1)^{n} \, e^{x} \right|^{1}_{0} - n \int_{0}^{1}{(x - 1)^{n - 1} \, e^{x} \, dx}$$

$$I_{n} = (-1)^{n - 1} - n I_{n - 1}$$

$$I_{0} = \int_{0}^{1}{e^{x} \, dx} = e - 1$$

In general: $I_{n} = A_{n} + B_{n} e$. The recursion gives:
$$A_{n} = (-1)^{n - 1} - n A_{n - 1}$$
$$B_{n} = -n B_{n - 1}$$
with the initial conditions $A_{0} = -1$ and $B_{0} = 1$. You need to find such an n that $A_{n} = 16$ and $B_{n} = -6$.

 

[Dickfore]

The general solution to the B recursive relations is:
$$B_{n} = (-1)^{n} n!$$