How much time it would take you?

Calculus Level pending

Find the value of the following integral, if x=0

∫ − ∞ 0 ( x 6 e x ) d x . \int_{-\infty} ^ 0 ( x^6 e^x)dx.


The answer is 720.

Mauricio Ceballos by Mauricio Ceballos , 26, USA

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1 solution

∫ ( x 6 e x ) d x = ∫ ( x 6 ) d ( e x ) = x 6 ∗ e x − 6 ∫ ( x 5 e x ) d x . Going on with integration by parts and noting that x=0, we finaly come to + 6 ! ∫ ( e x ) d x and putting x=0, we get 6! = 720 \int( x^6 e^x)dx=\int( x^6)d( e^x) = x^6* e^x- 6 \int( x^5 e^x)dx.\\\text{Going on with integration by parts and noting that x=0,}\\ \text{ we finaly come to } +6!\int( e^x)dx \text{ and putting x=0, we get 6! }=~~~\huge \color{#D61F06}{720}
Just to show what the answer would look if x ≠ 0 x 6 ∗ e x − 6 x 5 ∗ e x + 6 ∗ 5 ∗ x 4 ∗ e x − 6 ∗ 5 ∗ 4 x 3 ∗ e x + 6 ∗ 5 ∗ 4 ∗ 3 ∗ x 2 ∗ e x − 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 x 1 ∗ e x + 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 ∗ e x = ∑ n = 0 6 6 ! ∗ x 6 − n ∗ e x ( 6 − n ) ! W i t h x = 0 , I n t . = 6 ! ∗ x 0 ∗ e 0 0 ! = 6 ! . \text {Just to show what the answer would look if } x \neq 0\\ x^6* e^x- 6 x^5* e^x+6*5*x^4* e^x- 6*5*4 x^3* e^x+\\6*5*4*3*x^2* e^x- 6*5*4*3*2 x^1* e^x+6*5*4*3*2*1*e^x\\\ { \Large\displaystyle = \sum_{n=0}^6 \frac{6!*x^{6-n}*e^x}{(6-n)!}~}~~With~ x=0, Int.=\dfrac {6!*x^0*e^0}{0!}=6!.

2 Helpful 0 Interesting 1 Brilliant 0 Confused

Note that it is not sufficient to say "assume that the constant of integration is 0". Because the integral could be both F ( x ) F(x) and F ( x ) + sin ⁡ 2 x + cos ⁡ 2 x F(x) + \sin ^2 x + \cos ^2 x , and you would be hard-pressed to argue for one instead of the other.

I've updated the problem statement to deal with this issue.

Calvin Lin Staff - 6 years ago

But this is indefinite integration... what about constant of integration?

Vishwak Srinivasan - 6 years ago

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It would be better if the problem adds theses words, "constant of integration is 0".

Niranjan Khanderia - 6 years ago

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I have submitted a report, sir. @Brilliant Mathematics please check.

Vishwak Srinivasan - 6 years ago

I think the faster way to do it is to substitute t = -x ,so that we arrive at an integral with the same form as Γ(7) , where Γ is the Gamma function , from there the initial integral is easily (7-1)!=6! = 720 .Judging by the way the question is frased i assume this is the intented way to solve it . Excuse my lack of Latex , it is my first comment on brilliant

Gregory Kafanelis - 5 months, 3 weeks ago

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Great observation!

Calvin Lin Staff - 5 months, 3 weeks ago

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