How do you prove that # int_0^1 \ x(1-x)^n \ dx = 1/((n+1)(n+2))# ?

Answer 1

#I=int_0^1 x*(1-x)^n*dx=int_0^1 (1-y)*y^n*dy=1/[(n+1)*(n+2)]#

#I=int_0^1 x*(1-x)^n*dx#
After using #x=1-y# and #dx=-dy# transforms, #I# became,
#I=int_1^0 (1-y)*y^n(-dy)#
=#int_0^1 (1-y)*y^n*dy#
=#int_0^1 y^n*dy#-#int_0^1 y^(n+1)*dy#
=#[y^(n+1)/(n+1)]_0^1#-#[y^(n+2)/(n+2)]_0^1#
=#1/(n+1)-1/(n+2)#
=#1/[(n+1)*(n+2)]#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

We aim to prove that:

# int_0^1 \ x(1-x)^n = 1/((n+1)(n+2)) \ dx #

We can then apply Integration By Parts:

Let # { (u,=x, => (du)/dx,=1), ((dv)/dx,=(1-x)^n, => v,=-(1-x)^(n+1)/(n+1) ) :}#

Then plugging into the IBP formula:

# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #

We have:

# int \ (x)(1-x)^n \ dx = (x)(-(1-x)^(n+1)/(n+1)) - int \ (-(1-x)^(n+1)/(n+1))(1) \ dx #

And applying the limits of integration we get:

# int_0^1 \ x(1-x)^n \ dx = [ -(x(1-x)^(n+1))/(n+1) ]_0^1 + int_0^1 \ (1-x)^(n+1)/(n+1) \ dx #
# " " = (0-0) + 1/(n+1) \ int_0^1 \ (1-x)^(n+1) \ dx #
# " " = 1/(n+1) \ [ -(1-x)^(n+2)/(n+2) ]_0^1 #
# " " = -1/((n+1)(n+2)) \ [ (1-x)^(n+2) ]_0^1 #
# " " = -1/((n+1)(n+2)) (0-1) #
# " " = 1/((n+1)(n+2)) \ \ \ \ \ # QED
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

See explanation...

Note that:

#x(1-x)^n = (1-(1-x))(1-x)^n = (1-x)^n - (1-x)^(n+1)#
#d/(dx) (1-x)^n = -n(1-x)^(n-1)#

So:

#int (1-x)^n dx = -1/(n+1) (1-x)^(n+1) + C#

and:

#int_0^1 x(1-x)^n dx = int_0^1 (1-x)^n - (1-x)^(n+1) dx#
#color(white)(int_0^1 x(1-x)^n dx) = [-1/(n+1) (1-x)^(n+1)]_0^1 - [-1/(n+2) (1-x)^(n+2)]_0^1#
#color(white)(int_0^1 x(1-x)^n dx) = 1/(n+1)-1/(n+2)#
#color(white)(int_0^1 x(1-x)^n dx) = ((n+2)-(n+1))/((n+1)(n+2))#
#color(white)(int_0^1 x(1-x)^n dx) = 1/((n+1)(n+2))#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 4

To prove that (\int_{0}^{1} x(1-x)^n , dx = \frac{1}{(n+1)(n+2)}), we can use integration by parts.

  1. Let (u = x) and (dv = (1-x)^n , dx).
  2. Calculate (du) and (v).
  3. Apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Here's the calculation:

Given: [ u = x ] [ dv = (1-x)^n , dx ]

Calculate: [ du = dx ] [ v = \frac{(1-x)^{n+1}}{n+1} ]

Apply integration by parts: [ \int_{0}^{1} x(1-x)^n , dx = \left. \frac{x(1-x)^{n+1}}{n+1} \right|{0}^{1} - \int{0}^{1} \frac{(1-x)^{n+1}}{n+1} , dx ]

Evaluate the definite integral: [ = \frac{(1-1)^{n+1}}{n+1} - \frac{(0)^{n+1}}{n+1} - \int_{0}^{1} \frac{(1-x)^{n+1}}{n+1} , dx ] [ = 0 - 0 - \frac{1}{n+1} \int_{0}^{1} (1-x)^{n+1} , dx ] [ = - \frac{1}{n+1} \cdot \frac{(1-x)^{n+2}}{n+2} \Bigg|_{0}^{1} ] [ = - \frac{1}{n+1} \left( \frac{(1-1)^{n+2}}{n+2} - \frac{(1-0)^{n+2}}{n+2} \right) ] [ = - \frac{1}{n+1} \left( 0 - \frac{1}{n+2} \right) ] [ = \frac{1}{(n+1)(n+2)} ]

Therefore, (\int_{0}^{1} x(1-x)^n , dx = \frac{1}{(n+1)(n+2)}).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7